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Physics Part I Chapter 4 Moving Charges And Magnetism

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NCERT Solution For Class 12 Physics Physics Part I

Moving Charges And Magnetism Here is the CBSE Physics Chapter 4 for Class 12 students. Summary and detailed explanation of the lesson, including the definitions of difficult words. All of the exercises and questions and answers from the lesson's back end have been completed. NCERT Solutions for Class 12 Physics Moving Charges And Magnetism Chapter 4 NCERT Solutions for Class 12 Physics Moving Charges And Magnetism Chapter 4 The following is a summary in Hindi and English for the academic year 2021-2022. You can save these solutions to your computer or use the Class 12 Physics.

Question 1

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A circular coil of wire consisting of 100 turns, each of radius 8.0 cm carries a current of 0.40 A. What is the magnitude of the magnetic field B at the centre of the coil?

Solution

Given,
Current carried by wire, I = 0.40 A,
Radius of circular wire, r = 8.0 cm = $8 \times 10^{- 2} m$
Number of turns in the coil, $n = 100$

Using the formula, we get the magnetic field at the centre of coil as

$B = \frac{μ_{0} nI}{2 r} = \frac{4 π \times 10^{- 7} \times 100 \times 0.4}{2 \times 8.0 \times 10^{- 2}} T$
$= 3.1 \times 10^{- 4} T .$

Question 2

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A long straight wire carries a current of 35 A. What is the magnitude of the field B at a point 20 cm from the wire?

Solution

Given,
Current carried by conductor, I = 35 A
Distance, r = 20 cm = 0.2 m

Using Ampere's circuital law we get,

$B = \frac{μ_{0} I}{2 πr} = \frac{4 π \times 10^{- 7} \times 35}{2 π \times 0.20} T = 3.5 \times 10^{- 5} T .$

Question 3

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A long straight wire in the horizontal plane carries a current of 50 A in north to south direction. Give the magnitude and direction of B at a point 2.5 m east of the wire.

Solution

Given,
Current carried by wire, I= 50 A
Distance from the wire, r = 2.5 m
Magnetic field at point r, B = ?

Using ampere's circuital law

B = \frac{μ_{0} I}{2 πr} = \frac{4 π \times 10^{- 7} \times 50}{2 π \times 2.5} = 4 \times 10^{- 6} T

The direction of magnetic field can be found out by applying right hand thumb rule. As the direction of current is from north to south represented by thumb, the direction iof magnetic field is vertically upwards in east direction of wire.

Question 4

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A horizontal overhead power line carries a current of 90 A in east to west direction. What is the magnitude and direction of the magnetic field due to the current 1.5 m below the line?

Solution

Given, current, I= 90 A
distance ,r = 1.5 m
Thus,
the magnitude of magnetic field is given by Ampere's circuital law

$B = \frac{μ_{0} I}{2 πr} = \frac{4 π \times 10^{- 7} \times 90}{2 π \times 1.5} = 1.2 \times 10^{- 5} T$

Applying the right-hand thumb rule, we find that the magnetic field at the observation point is directed towards south.

Question 5

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What is the magnitude of magnetic force per unit length on a wire carrying a current of 8 A and making an angle of 30° with the direction of a uniform magnetic field of 0.15 T?

Solution

Given,
   $Current carried by wire, I = 8 A, Uniform magnetic field, B = 0.15 T, Angle made by the wire with magnetic field, θ = 30 °$

Therefore,

Force acting on wire

   $F = BI l \sin θ$

   $= 0.27 \times 10 \times (3 \times 10^{- 2}) \times \sin 90 ° = 0.27 \times 10 \times 3 \times 10^{- 2} \times 1$

$F = 8.1 \times 10^{- 2} N$

Question 6

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Two long and parallel straight wires A and B carrying currents of 8.0 A and 5.0 A in the same direction are separated by a distance of 4.0 cm. Estimate the force on a 10 cm section of wire A.

Solution

Given,
Current on first conductor, I₁ = 8.0 A
Current on second conductor, I₂ = 5 A
Distance between the conductors, r = 4.0 cm = $4 \times 10^{- 2}$
Length of the wire, $l = 10 cm = 0.1 m$
Force on length l is given by,

$F = \frac{μ_{0}}{2 π} \frac{I_{1} I_{2} l}{r}$
$= \frac{(4 π \times 10^{- 7}) \times 8 \times 5 \times 0.1}{2 π \times 0.04} = 2 \times 10^{- 5} N$

Question 7

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A closely wound solenoid 80 cm long has 5 layers of windings of 400 turns each. The diameter of the solenoid is 1.8 cm. If the current carried is 8.0 A, estimate the magnitude of B inside the solenoid near its centre.

Solution

Here given,
Length of the solenoid, l = 80 cm = 0.80 m
Total number of turns, N = 5 x 400 = 2000
Current passing through, I = 8.0 A
Diameter of the solenoid, D = 1.8 cm

n = no. of turns per unit length

Therfore,

$n = \frac{Total turn}{length} n = \frac{2000}{0.80}$

Magnitude of magnetic field inside a solenoid near its centre
$B = μ_{0} nI = \frac{4 π \times 10^{- 7} \times 2000 \times 8.0}{0.80} = 8 π \times 10^{- 3} T = 2.5 \times 10^{- 2} T .$

Question 8

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A square coil of side 10 cm consists of 20 turns and carries a current of 12 A. The coil is suspended vertically and the normal to the plane of the coil makes an angle of 30° with the direction of a uniform horizontal magnetic field of magnitude 0.80 T. What is the magnitude of torque experienced by the coil?

Solution

Here,
Length of the side of the square, l = 10 cm = 0.10 m
Number of turns in the coil, N = 20
Current carried by the conductor, I = 12 A
Angle made by the coil with the magnetic field, $θ = 30 °$
Magnetic field , B = 0.80 T
Area of the conductor $A = l \times l = 0.1 \times 0.1 = 0.01 m^{2}$
Therefore,

Torque is given by, $τ = NBIA \sin θ = 20 \times 0.80 \times 12 \times (0.1)^{2} \times \sin 90 ° = 0.96 Nm$

Question 9

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Two moving coil meters, M₁ and M₂ have the following particulars:
R₁ = 10 Ω N₁ = 30,
A₁ = 3.6 x 10^–3 m², B₁ = 0.25 T
R₂ = 14 Ω; N₂ = 42,
A₂ = 1.8 x 10^–3 m², B₂ = 0.50 T
(The spring constants k are identical for the two meters).
Determine the ratio of (a) current sensitivity and (b) voltage sensitivity of M₂ and M_1.

Solution

(a) Current sensitivity of a moving coil galvanometer is the amount of deflection per unit current.
Therefore,
Current sensitivity of first meter is given by

  Is = ϕI= BANk
IM1 = ϕI = B1A1N1k      =0.25 × 3.6 × 10-3 × 30k      = 27 × 10-3k                                             ...(i)

Current sensitivity of second meter,
IM2 = ϕI = B2A2N2k      = 0.50 × 1.80 × 10-3 × 42k      = 37.8 × 10-3k                                               ...(ii)

Ratio of current sensitivity is therefore,

IM2IM1 =37.8 × 10-3k27 × 10-3k = 1.4

(b) Voltage sensitivity of a moving coil galvanometer is given as the amount of deflection per unit voltage.
Thus,
Voltage sensitivity of first meter is,
  VM1= ϕV= ϕI. R       = 27 × 10-3k × 10       = 2.7 ×10-3 k

Voltage sensitivity of second meter
VM2= ϕR.I       = 37.8 × 10-3k × 10        = 2.7 × 10-3k

Hence, the ratio of voltage sensitivity, VM1VM2 = 1.

Question 10

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In a chamber, a uniform magnetic field of 6.5 G (1 G = 10^–4T) is maintained. An electron is shot into the field with a speed of 4.8 x 10⁶ m s^–1 normal to the field. Explain why the path of the electron is a circle. Determine the radius of the circular orbit. (e = 1.6 x 10^–31 C, m_e = 9.1 x 10^–31 kg)

Solution

Here given,
Uniform magnetic field, B = 6.5 x 10^–4 T
Speed with which the electron is shot, v = 4.8 x 10⁶ m/s
Charge on electron, e = 1.6 x 10^–19 C
Angle made by the electron with the field, θ = 90° Mass of electron, m = 9.1 x 10^–31 kg
Radius of the circular orbit, r =?

(i) Force on the moving electron due to magnetic field will be,

F = evB \sin θ .

The direction of this force is perpendicular to

\vec{v}

and

\vec{B}

therefore, this force will only change the direction of motion of the electron without affecting its velocity i.e., this force will provide the centripetal force to the moving electron and hence, the electron will move on the circular path.

If r is the radius of circular path traced by electron, then

e v B s i n 90 ° = m v^{2} / r \Rightarrow r = \frac{m v}{B e} \Rightarrow r = \frac{(9.1 \times 10^{- 31}) \times (4.8 \times 10^{6})}{(6.5 \times 10^{- 4}) \times (1.6 \times 10^{- 19})} = 4.2 \times 10^{- 2} m = 4.2 cm

Question 11

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In Question 4.11 obtain the frequency of revolution of the electron in its circular orbit. Does the answer depend on the speed of the electron?

Solution

Given ,
B = 6.5 G= 6.5

\times

10^-4 T
v= 4.8

\times

10⁶ m/s

Frequency of the revolution of a charged particle is given by

ν = \frac{Bq}{2 πm} = \frac{6.5 \times 10^{- 4} \times 1.6 \times 10^{- 19}}{2 \times 3.14 \times 9.1 \times 10^{- 31}} = \frac{10.4 \times 10^{- 23}}{51.148 \times 10^{- 31}} = 0.18198 \times 10^{8} v = 18 \times 10^{6} Hz = 18 MHz

The frequency is independent of the velocity.

Question 12

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(a) A circular coil of 30 turns and radius 8.0 cm, carrying a current of 6.0 A is suspended vertically in a uniform horizontal magnetic field of magnitude 1.0 T. The field lines make an angle of 60° with the normal of the coil. Calculate the magnitude of the counter torque that must be applied to prevent the coil from turning.
(b) Would your answer change if the circular coil were replaced by a planar coil of some irregular shape that encloses the same area? (All other particulars are also unaltered).

Solution

Given, a current carrying circular coil.

Number of turns in the coil, N = 30
Current, I = 6.0A
Uniform horizontal magnetic field, B = 1.0 T
Angle made by coil with the magnetic field lines, $α = 60 °$
Radius of the circular coil, r = 8.0 cm = 8 x 10^-2m.

Therefore,
Area of the coil, $A = {πr}^{2}$
$= \frac{22}{7} \times {(8 \times 10^{- 2})}^{2}$
$A = 2.01 \times 10^{- 2} m^{2}$

(a) Now,
Torque that must be applied to prevent the coil from turning
$τ = NBIA \sin α$

$= 30 \times 6.0 \times 1.0 \times (2 \times 10^{- 2}) \times \sin 60 °$

$τ = 30 \times 6 \times 1 \times 2 \times \frac{\sqrt{3}}{2} \times 10^{- 2} = 3.12 Nm .$

(b) If the area of the loop is the same, the torque will remain unchanged as the torque on the planar loop does not depend upon the shape.

Question 13

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Two concentric coils X and Y of radii 16 cm and 10 cm respectively lie in the same vertical plane containing the north to south direction. Coil X has 20 turns and carries a current of 16 A ; coil Y has 25 turns and carries a current of 18 A. The sense of current in X is anti clockwise and in Y, clockwise, for an observer looking at the coil facing west. Give the magnitude and direction of the net magnetic field due to the coils at their centre.

Solution

Consider X axis in East-West directions and YY’ in North-South direction respectively.
Plane of coil is in Y-Z axis plane.
Direction of magnetic field due to coil X is in East, and due to coil Y is in west direction.

Given, for coil X
Radius of coil, r_x = 16 cm = 0.16 m
Number of turns, N_x = 20
Current passing through coil, I_x = 16 A

Magnetic field at the centre of coil X is

B_{x} = \frac{μ_{0} I_{x} N_{x}}{2 r_{x}} = \frac{4 π \times 10^{- 7} \times 16 \times 20}{2 \times 0.16} = 4 π \times 10^{- 4} T

The current in the coil X is anticlockwise therefore, the field B_x is directed towards east.

Given, for coil Y
Radius, r_y = 10 cm = 0.10 m
Number of turns, N_y = 25
Current pasing through, I_y = 18 A

Magnetic field at the centre of coil Y is

$B_{y} = \frac{μ_{0} I_{y} N_{y}}{2 r_{y}} = \frac{4 π \times 10^{- 7} \times 18 \times 25}{2 \times 0.10} = 9 π \times 10^{- 4} T$

The direction of magnetic field induction B_y is towards west.

Hence,

Net magnetic field = B_y – B_x= 9 $π$ x 10^–4 – 4 $π$ x 10^–4= 5 $π$ x 10^–4= 1.6 x 10^–3 T (Towards west).

Question 14

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A magnetic field of 100 G (1 G = 10^–4 T) is required which is uniform in a region of linear dimension about 10 cm and area of cross section about 10^–3 m². The maximum current - carrying capacity of a given coil of wire is 15 A and the number of turns per unit length that can be wound round a core is at most 1000 turns m^–1. Suggest some appropriate design particulars of a solenoid for the required purpose. Assume the core is not ferromagnetic.

Solution

Given,
Magnetic field, $B = 100 G = 100 \times 10^{- 4} T = 10^{- 2} T$
Maximum current carrying capacity, I = 15 A
Number of turns per unit length, $n = 1000 m^{- 1}$

Magnetic field insie a solenoid is

$B = μ_{0} n l$
Therefore,
$nI = \frac{B}{μ_{0}} = \frac{10^{- 2}}{4 π \times 10^{- 7}} = \frac{10^{5}}{4 π} = 7961.5 ~ 8000$

We may have,

$I = 10 A and n = 800$ /m as per the requirement

The solenoid may have length 50 cm and cross section 5 x 10^–3 m² (five times given values) so as to avoid edge effects etc.

Question 15

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For a circular coil of radius R and N turns carrying current I, the magnitude of the magnetic field at a point on its axis at a distance x from its centre is given by,
$B = \frac{μ_{0} {IR}^{2} N}{2 {(x^{2} + R^{2})}^{3 / 2}}$
(a) Show that this reduces to the familiar result for field at the centre of the coil.

(b) Consider two parallel co-axial circular coils of equal radius R, and number of turns N, carrying equal currents in the same direction, and separated by a distance R. Show that the field on the axis around the mid-point between the coils is uniform over a distance that is small as compared to R, and is given by,
$B = 0.72 \frac{μ_{0} NI}{R}, approximately .$
[Such an arrangement to produce a nearly uniform magnetic field over a small region is known as Helmholtz coils.]

Solution

(a) Given, a circular coil of radius r and N turns carrying current I.

Then,
Magnitude of magnetic field at a point on axis at a distance x from centre is given by,
   $B = \frac{μ_{0} {IR}^{2} N}{2 {(x^{2} + R^{2})}^{3 / 2}} (axial line)$

At the centre of the coil x =0

Therefore,

   $B = \frac{μ_{0} {IR}^{2} N}{2 R^{3}}$

i.e.,    $B = \frac{μ_{0} IN}{2 R}$

which is same as the standard result.

(b) In figure, O is a point which is mid-way between the two coils X and Y.

Let, B_x be the magnetic field at Q due to coil X.
Then,
$B_{x} = \frac{μ_{0} {NIR}^{2}}{2 {[{(\frac{R}{2} + d)}^{2} + R^{2}]}^{3 / 2}}$

If, $B_{y}$ is the magnetic field at Q due to coil Y, then
   $B_{y} = \frac{μ_{0} {NIR}^{2}}{2 {[{(\frac{R}{2} - d)}^{2} + R^{2}]}^{3 / 2}}$

The currents in both the coils X and Y are flowing in the same direction.

So, the resultant field is given by

Question 16

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A toroid has a core (non-ferromagnetic material) of inner radius 25 cm and outer radius 26 cm around which 3500 turns of wire are wound. If the current in the wire is 11 A, what is the magnetic field (a) outside the toroid (b) inside the core of the toroid (c) in the empty space surrounded by the toroid?

Solution

Given,
Inner radius of torroid core, r₁ = 0.25 m
Outer radius, r₂ = 0.26 m
Number of turns, N = 3500
Current in the wire, I = 11 A

(i)    The magnetic field is zero outside the toroid because, there is no current flowing on outside of the torroid.

(ii) Magnetic field inside the core of the toroid,
$B = μ_{0} nI$    $B = \frac{μ_{0} NI}{l} (∵ n = \frac{N}{l} \to Number of turns per unit length)$

$I = 2 π (\frac{r_{1} + r_{2}}{2})$
$= π (r_{1} + r_{2}) = π (0.25 + 0.26) = π \times 0.51 m$

Putting the values
$B = \frac{(4 \times 10^{- 7}) \times 3500 \times 11}{π \times 0.51} = 3.02 \times 10^{- 2} T$

(iii) The magnetic field is zero in the empty space surrounded by the toroid also.

Question 17

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Answer the following questions:
A magnetic field that varies in magnitude from point to point but has a constant direction (east to west) is set up in a chamber. A charged particle enters the chamber and travels undeflected along a straight path with constant speed. What can you say about the initial velocity of the particle?

Solution

The force on a charged particle moving in a magnetic field is given by F = qvB sin θ.
If the force acting on a particle is 0 then, the charged particle will travel undeflected along a traight path with constant speed in the presence of magnetic field. And, the force acting is 0 when, θ = 0° or180° . That is, when initial velocity v is either parallel or anti-parallel to $\vec{B .}$

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Question 18

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Answer the following questions:
A charged particle enters an environment of a strong and non-uniform magnetic field varying from point to point both in magnitude and direction, and comes out of it following a complicated trajectory. Would its final speed equal the initial speed if it suffered no collisions with the environment?

Solution

Yes, the final speed would equal the initial speed because, the magnetic force can change the direction of velocity and not its magnitude. The force which is exerted by the magnetic field on the charged particle is always perpendicular to it's motion and hence, does no work.

Question 19

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Answer the following questions:
An electron travelling west to east enters a chamber having a uniform electrostatic field in north to south direction. Specify the direction in which a uniform magnetic field should be set up to prevent the electron from deflecting from its straight line path.

Solution

The electrostatic field is directed towards south. Since the electron is a negatively charged particle, therefore, the electrostatic field shall exert a force directed towards north.
So, if the electron is to be prevented from deflection from straight path, the magnetic force on the electron should be directed towards south.
Magnetic force is given by,

\vec{F} = - e (\vec{v} \times \vec{B}) .

{\vec{F}}_{m}

is towards south,

\vec{v}

is travelling towards east.

Applying Fleming's Left-Hand Rule, we find that magnetic field

\vec{B}

should be in the vertically downward direction.

Question 20

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An electron emitted by a heated cathode and accelerated through a potential difference of 2.0 kV, enters a region with uniform magnetic field of 0.15 T. Determine the trajectory of the electron if the field (a) is transverse to its initial velocity, (b) makes an angle of 30° with the initial velocity.

Solution

Potential difference, V = 2 KV = 2000 volt
Uniform magnetic field, B = 0.15 T

(a) If magnetic field is transverse to initial velocity of electron then, the velocity vector has no component in the direction of magnetic field.
∴ Force on electron = Bev sin 90° = Bev

This force acts as the centripetal force

$∴$ $Bev = \frac{{mv}^{2}}{r}$

   $r = \frac{mv}{eB}$ ...(1)

But,    $\frac{1}{2} {mv}^{2} = eV$ ...(2)

$\Rightarrow$ $v = \sqrt{\frac{2 eV}{m}}$
$∴$    $r = \frac{m}{eB} \sqrt{\frac{2 eV}{m}}$ [From 1, 2]

   $v = \frac{1}{B} \sqrt{\frac{2 mV}{e}} = \frac{1}{0.15} \sqrt{\frac{2 \times 9.1 \times 10^{- 31} \times 2000}{1.6 \times 10^{- 19}}} m = 10^{- 3} m = 1.0 mm .$

The electron would move in a circular trajectory of radius 1.0 mm. The plane of the trajectory is normal to B.

(b) If velocity makes an angle 30° with the direction of magnetic field, the velocity can be resolved into v_⊥ and v_||i.e., v cos 30° and v sin 30° respectively.

Due to v_⊥ the electron will move on a circular path. The resultant path will be a combination of straight line motion and circular motion which is called helical.

Thus, $evB \sin θ = \frac{m (v \sin θ)^{2}}{r_{n}}$

for circular motion of radius $r_{n}$

${ev}_{⊥} \times B = \frac{{mv}_{⊥}^{2}}{r_{n}}$
   $v_{⊥} = v \sin θ$

$r_{n} = \frac{mv \sin θ}{eB}$

$r_{n} = \frac{9.1 \times 10^{- 31} \times 2.65 \times 10^{7} \times \sin 30}{1.6 \times 10^{- 19} \times 0.15}$

$= 0.49 \times 10^{- 3} m = 0.49 mm ≃ 0.5 mm$

The linear velocity is = $v \cos θ$
   $= 2.65 \times 10^{7} \times \cos 30 ° = 2.65 \times 10^{7} \times \frac{\sqrt{3}}{2} = 2.3 \times 10^{7} {ms}^{- 1}$

Thus, the electron moves in a helical path of radius 0.49 mm with a velocity component of 2.3 x 10⁷ ms^–1 in the direction of magnetic field.

Question 21

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A magnetic field set up using Helmholtz coils is uniform in a small region and has a magnitude of 0.75 T. In the same region, a uniform electrostatic field is maintained in a direction normal to the common axis of the coils. A narrow beam of (single-species) charged particles, all accelerated through 15 kV, enters this region in a direction perpendicular to both the axis of the coils and the electrostatic field. If the beam remains undeflected when the electrostatic field is 9.0 x 10⁵ Vm^–1, make a simple guess as to what the beam contains. Why is the answer not unique?

Solution

Given,
Magnetic field set up, B= 0.75 T
Electrostatic field, E = 9.0 $\times$ 10⁵ V/m

Since, the magnetioc field is perpendicular to the initial velocity of electron, therefore the electron will move in a circular path.

Kinetic energy of the electron

$\frac{1}{2} {mv}^{2} = eV$
$\frac{e}{m} = \frac{v^{2}}{2 V}$

$\frac{e}{m} = \frac{1.2 \times 10^{6} \times 1.2 \times 10^{6}}{2 \times 15 \times 10^{3}} \frac{e}{m} = \frac{0.24}{5} \times 10^{9} \frac{e}{m} = 0.048 \times 10^{9} \frac{e}{m} = 4.8 \times 10^{7} coulomb / kg$

This charge to mass ratio is equivalent to charge to mass ratio of proton so the charge particle may be deuterons.
However, the answer is not unique.
This is because ${He}^{+ +} (\frac{2 e}{2 m})$ and ${Li}^{+ +} (\frac{3 e}{3 m})$ have also the same value of $\frac{e}{m}$ .

Question 22

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A straight horizontal conducting rod of length 0.45 m and mass 60 g is suspended by two vertical wires at its ends. A current of 5.0 A is set up in the rod through the wires.
(a) What magnetic field should be set up normal to the conductor in order that the tension in the wires is zero?
(b) What will be the total tension in the wires if the direction of current is reversed keeping the magnetic field same as before?
(Ignore the mass of the wires.) g = 9.8 m s^–2.

Solution

Given,
Length of the rod, l= 0.45 m
mass of the suspension, m = 60 g = 60 x 10^–3 kg
Current carried by the conductor, I = 5.0 A

(a) Force needed to balance the weight of the rod,

$F = mg = 0.06 kg \times 9.8 = 0.588 N$

Using the formula,

$F = BI l$

$B = \frac{F}{I l} = \frac{0.588}{5.0 \times 0.45} = 0.26 T$

If the direction of current is from right to left then the direction of magnetic field is horizontal and normal to the conductor and the force due to magnetic field will be upwards (Fleming's left hand rule).

(b) If the direction of current is reversed then ‘BIl’ and ‘mg’ will act vertically downwards.
Therefore,
Total tension in the wires = BIl + mg
= 0.588 + 0.588
= 1.176 N.

Question 23

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The wires which connect the battery of an automobile to its starting motor carry a current of 300 A (for a short time). What is the force per unit length between the wires if they are 70 cm long and 1.5 cm apart? Is the force attractive or repulsive?

Solution

Given,

$Current on first conductor, I_{1} = 300 A Current on second conductor, I_{2} = 300 A Distance between the conductors, r = 1.5 c m = 1.5 \times 10^{- 2} m Length of the conductor, l = 70 cm$

Using Biot -Savart law,

$F = \frac{μ_{0} I_{1} I_{2} l}{2 πr}$

This implies,

Force per unit length =

$\frac{F}{l} = \frac{4 π \times 10^{- 7} \times 300 \times 300}{2 π \times 0.015}$

= $1.2 {Nm}^{- 1}$

The total force between the wires is $F = f l$

$= 1.2 \times 0.70 N = 0.84 N .$

The force is repulsive since the current will flow in opposite direction in the two wires.

Question 24

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A uniform magnetic field of 1.5 T exists in a cylindrical region of radius 10.0 cm, its direction parallel to the axis along east to west. A wire carrying current of 7.0 A in the north to south direction passes through this region. What is the magnitude and direction of the force on the wire if,
(a) the wire intersects the axis,
(b) the wire is turned from N-S to northeast-northwest direction,
(c) the wire in the N-S direction is lowered from the axis by a distance of 6.0 cm?

Solution

a)
Diameter of cylindrical region = 20 cm = 0.20 m
Clearly,
length of conductor, l = 0.20 m.
Also, θ = 90°
Current carried by conductor, I = 7A
Uniform magnetic field, B= 1.5 T

Force, F = BIl sin θ
= 1.5 x 7 x 0.20 sin 90° N
= 2.1 N

Using Fleming's left-hand rule, we find that the force is directed vertically downwards.

(b) If l₁is the length of the wire in the magnetic field, then,

F₁ = BIl₁ sin 45°

But , l₁ sin 45° = l
∴ F₁ = BIl
= 1.5 x 7 x 0.20 N
= 2.1 N

The force is directed vertically downwards( using Fleming's left hand rule).

(c) When the wire is lowered by 6 cm from axis, the length of the wire in the cylindrical magnetic field is 2x.

Now considering the right angled triangle as shown in fig. above,

$x^{2} = 10^{2} - 6^{2} x = \sqrt{64} = 8 cm$
$∴$ $2 x = 16 cm .$

$F_{2} = BI l_{2} = 1.5 \times 7 \times 0.16 N = 1.68 N$

The force is directed vertically downwards.

The result is true for any angle between current and direction of $\vec{B}$ . This is because I sin θ remains constant i.e., 20 cm.

Question 25

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A uniform magnetic field of 3000 G is established along the positive z-direction. A rectangular loop of sides 10 cm and 5 cm carries a current 12A. What is the torque on the loop in the different cases shown in the figure below. What is the force on each case? Which case corresponds to stable equilibrium?

Solution

(a) Given,
Magnetic field along the positive z-direction = 3000 G = 3000 x 10^–4 T = 0.3 T
Current carried by the loop, I = 12 A
Area of rectangular loop, A = 10 x 5 cm²
= 50 x 10^–4 m²

Torque on the loop is given by,

τ = BIA \cos θ

where, θ is the angle between the plane of loop and direction of magnetic field.
Here, θ = 0° .

Therefore, torque is, τ = 0.3 x 12 x 50 x 10^–4
= 1.8 x 10^–2 Nm.

Using Fleming's left hand rule we can say that the direction of torque is along negative y direction.

(b) The torque acting is the same as in case (a) but, the direction of torque is along side 10 cm.

(c) The magnitude of torque is equal to 1.8 x 10^–2 Nm along - x direction of torque on lower arm of 5 cm towards – y axis.
(d)    This case is similar to (c). Direction of torque is 60°.
(e)    zero. (∵ angle between plane of loop and direction of magnetic field is 90°)
(f)    zero.

Force is zero in each case. Stable equilibrium is corresponded by case (e).

Question 26

Wired Faculty App

A circular coil of 20 turns and radius 10 cm is placed in a uniform magnetic field of 0.10 T normal to the plane of the coil. If the current in the coil is 5.0 A, what is the
(a) total torque on the coil,
(b) total force on the coil,
(c) average force on each electron in the coil due to the magnetic field?
(The coil is made of copper wire of cross-sectional area 10^–5 m², and the free electron density in copper is given to be about 10^–29 m^–3.)

Solution

Given,
Number of turns, N = 20
Radius of the coil, r = 10 cm = 10 x 10^–2 m
Magnetic field, B = 0.10 T
Current carried, I = 5.0 A
θ = 0° (angle between field and normal to the coil)
Area of the coil, A = $π$ r²
= $π$ x (10 x 10^–2)² = $π$ x 10^–2 m²(a) Torque, τ = NIBA sin θ
= 20 x 5.0 x 0.10 x $π$ x 10^–2 sin 0°
= 20 x 5.0 x 0.10 x $π$ x 10^–2 x 0 = 0

(b) Net force on a planer current loop in a magnetic field is always zero, as net force due to couple of force is zero.

(c) If v_d is the drift velocity of electron

F = qv x B
= ev_d. B sin 90°

Force on one electron = Be $v_{d} = Be \frac{I}{neA} = \frac{BI}{nA}$
Here,

$n = 10^{29} m^{- 3}, A = 10^{- 5} m^{2}$

$∴$ Force on one electron = $\frac{0.10 \times 5.0}{10^{29} \times 10^{- 5}} = 5 \times 10^{- 25} N .$

Question 27

Wired Faculty App

A solenoid 60 cm long and of radius 4.0 cm has 3 layers of windings of 300 turns each. A 2.0 cm long wire of mass 2.5 g lies inside the solenoid (near its centre) normal to the axis; both the wire and the axis of the solenoid are in the horizontal plane. The wire is connected through two leads parallel to the axis of solenoid to an external battery which supplies a current of 6.0 A in the wire. What value of current (with appropriate sense of circulation) in the windings of the solenoid can support the weight of the wire? g = 9.8 ms^–2.

Solution

Given a solenoid.
Length of the solenoid,l = 60 cm = 0.60 m
Total number of turns, N = 3 x 300 = 900
Length of the wire, l₁ = 2.0 cm = 0.02 m
mass of the wire lying inside the solenoid,m 2.5 g = 2.5 x 10^–3 kg
Current carried by the wire,I₁ = 6.0 A

Let, current I be passed through the solenoid windings, then,
Magnetic field produced inside the solenoid due to current is

$B = \frac{μ_{0} NI}{l}$

Force acting on wire, $= I_{1} l_{1} B = I_{1} l_{1} \frac{μ_{0} NI}{l}$

The wire can be supported if the force on wire is equal to the weight of wire, i.e.

$I_{1} l_{1} \frac{μ_{0} NI}{l} = mg$

$\Rightarrow$ $I = \frac{mg l}{I_{1} l_{1} μ_{0} N} = \frac{2.5 \times 10^{- 3} \times 9.8 \times 0.6}{(6.0) \times (0.02) \times (4 π \times 10^{- 7})} \times 900$

$= 108.27 A$

Question 28

Wired Faculty App

A galvanometer coil has a resistance of 12 Ω and the metre shows full scale deflection for a current of 3 mA. How will you convert the metre into a voltmeter of range 0 to 18 V?

Solution

Given, a galvanometer coil.
Resistance on the galvanometer, R = 12 Ω
Current on galvanometer, I_g =3 mA = 3 x 10^–3A
Voltage ,V = 18 V

By using formula,

   $V = I_{g} (R + R_{g})$

$\frac{V}{I_{g}} = R + R_{g}$
or,
   $R = \frac{V}{I_{g}} - R_{g} = \frac{18}{3 \times 10^{- 3}} - 12 R = 6 \times 10^{3} - 12 = 5988 Ω$

Question 29

Wired Faculty App

A galvanometer coil has a resistance of 15 Ω and the metre shows full scale deflection for a current of 4 mA. How will you convert the metre into an ammeter of range 0 to 6 A?

Solution

Given, a galvanometer coil.
Resistance on galvanometer, G = 15 Ω
Current pasing through the galvanometer,I_g = 4 mA = 4 x 10^–3 A
Current across the ammeter, I = 6 A

Using formula,

S = \frac{I_{g} . G}{I - I_{g}}

Putting values,

S = \frac{4 \times 10^{- 3} \times 15}{6 - 0.004} = \frac{60 \times 10^{- 3}}{5.996} = 10 \times 10^{- 3} Ω

S = 10 m Ω

Therefore, a resistance of 10mΩ should be connected so as to convert the galvanometer into an ammeter.

Question 30

Wired Faculty App

An electron does not suffer any deflection while passing through a region of uniform magnetic field. What is the direction of the magnetic field?

Solution

Since. the electron goes undeflected the direction of magnetic field

\vec{B}

is parallel or antiparallel to the velocity

\vec{v}

of electron.

As,

F = q (\vec{v} \times \vec{B}) = 0

Since,

\vec{v} | | \vec{B}

θ = 0^{o} or 180^{0}

Question 31

Wired Faculty App

What are the units of magnetic permeability?

Solution

The unit of magnetic permeability is Henry per metre or Tesla metre/ampere (TmA^–1).

Question 32

Wired Faculty App

Magnetic field lines can be entirely confined within the core of a toroid, but not within a straight solenoid. Why?

Solution

The magnetic field is given by B=

μ_{o} N I

where, I is the current flowing through the torroid or solenoid.
In a torroid, the magnetic field due to individual wires cancel out everywhere except for the core.
But, in a solenoid magnetic field lines due to individual wires cancel out longitudnally and add up horizontally. Hence, magnetic field lines exist even outside the core.

Question 33

Wired Faculty App

The coils, in certain galvanometers, have a fixed core made of a non-magnetic matallic materials.
Why does the oscillating coil come to rest so quickly such a core?

Solution

Eddy current acting in the core gives rise to restoring torque. And, this restoring torque tries to bring back the coil back to it's original position. Hence, the oscillating coil comes to rest quickly.

Question 34

Wired Faculty App

Two circular coils made of similar wires but of radii 20 cm and 40 cm are connected in parallel.
What will be the ratio of the magnetic field at their centres?

Solution

Given, two circular coils mad of similar wires.
Radius of first wire, r₁ = 20 cm
Radius of second wire, r₂= 40 cm
Magnetic field at the centre of circular coil of radius r, turns N, and current I passing in coil is

B = \frac{μ_{0} NI}{2 r} B = \frac{μ_{0} N}{2 r} \frac{V}{R} [R = resistance of coil] B = \frac{μ_{0} N}{2 r} \frac{V}{2 πrx} [x is resistance per unit length] B = \frac{μ_{0} NV}{4 {πxr}^{2}}

As coils are is parallel so potential difference ‘V’ are equal in both coils

∴

V \propto \frac{1}{r^{2}}

\frac{B_{1}}{B_{2}} = \frac{{r_{2}}^{2}}{{r_{1}}^{2}}

\frac{B_{1}}{B_{2}} = {(\frac{40}{20})}^{2} \frac{B_{1}}{B_{2}} = 4 B_{1} : B_{2} = 4 : 1

Question 35

Wired Faculty App

What is the direction of the force acting on a charged particle q, moving with a velocity $\vec{v}$ in a uniform magnetic field $\vec{B} ?$

Solution

The force acting on a charged particle q is given by,

\vec{F} = q (\vec{v} \times \vec{B})

Magnetic force is always normal to plane of

\vec{V} and \vec{B} .

Question 36

Wired Faculty App

Magnetic lines of force are endless. Comment.

Solution

This is because magnetic lines of force are continuous closed loops and, monopole is not possible for the magnetic field to exist.

Question 37

Wired Faculty App

An electron is moving along +ve x-axis in the presence of uniform magnetic field along +ve y-axis. What is the direction of force acting on it?

Solution

The direction of the force is along negative z axis. ( Using Fleming's right hand rule )

Question 38

Wired Faculty App

A magnetic dipole is situated in the direction of a magnetic field. What is its potential energy? If it is rotated by 180°, then what amount of work will be done?

Solution

Given, magnetic dipole is situated in the direction of magnetic field. Therefore $θ = 0^{o}$ .
Therefore,
Potential energy of dipole, P.E = – MB cos 0°
= – MB

If the dipole is rotated by an angle of 180°.
Work done = MB (cos 0° – cos 180°)
= MB (1 + 1)
= 2 MB.

Question 39

Wired Faculty App

Define magnetic flux. Give its SI unit.

Solution

The total number of magnetic lines of force crossing the surface A in a magnetic field

\vec{B}

is termed as magnetic flux.

Magnetic flux is given by Φ = BA cos θ .

Its SI unit is Weber.

It is a scalar quantity.

Sponsor Area

Question 40

Wired Faculty App

What is the approximate distance upto which earth's magnetic field extends?

Solution

The magnetic field of earth extends to nearly five times the radius of the earth.
i.e., 5 x (6.4 x 10³) km = 3.2 x 10⁴ km.

where, radius of the earth, r= 6.4 x 10³

Question 41

Wired Faculty App

An electron moving through a magnetic field does not experience any force. Under what condition is this possible?

Solution

The force acting on a charged particle in the presence of a magnetic field is given by,
F = qvB sin θ
So either electron is moving parallel to the direction of the magnetic field or it is at rest.

Question 42

Wired Faculty App

An electron moving with a velocity of 107 m/s enters a uniform magnetic field of 1 T along a direction parallel to the field. What would be its trajectory?

Solution

The electron will move undeflected in a straight line as, the particle is travelling in a direction paralle to the direction of magnetic field.
F = qvB sin θ [ θ = 0 ]

Question 43

Wired Faculty App

Write one condition under which an electric charge does not experience a force in a magnetic field.

Solution

When the electrically charged particle is either at rest (v = 0) or parallel to the direction of magnetic field i.e, when θ =

0^{o}

, it does not experience any force in a magnetic field.

Question 44

Wired Faculty App

What is a shunt? State its SI unit.

Solution

A small resistance connected in parallel with a galvanometer so as to convert it into an ammeter is called a shunt. SI unit of shunt is ohm.

Question 45

Wired Faculty App

Write SI unit of magnetic field $\vec{B} .$

Solution

SI unit of magnetic field is tesla (T).

Question 46

Wired Faculty App

A charge q is moving in a region where both the magnetic field $\vec{B}$ and electric field $\vec{E}$ are simultaneously present. What is the Lorentz force acting on the charge?

Solution

Force acting due to the presence of electric field is $q \vec{E}$ .
Force acting because of the presence of magnetic field is $q (\vec{v} \times \vec{B})$ .
Thus, the net force is,
$\vec{F} = q \vec{E} + q (\vec{v} \times \vec{B})$

$\vec{F} = q (\vec{E} + \vec{v} \times \vec{B})$

which is called as the Lorentz force.

Question 47

Wired Faculty App

In the diagram below is shown a circular loop carrying current I. Show the direction of the magnetic field with the help of lines of force.

Solution

The magnetic force lines of a circular loop carrying current I are shown as follows.

Question 48

Wired Faculty App

The force $\vec{F}$ experienced by a particle of charge ‘q’ moving with a velocity ‘v’ in a magnetic field ‘B’ is given by $\vec{F} = q (\vec{v} \times \vec{B}) .$ Which pairs out of these vectors are always at right angles to each other?

Solution

The force acting on the particle is always perpendicular to the velocity of the particle and also, the magnetic field .

i.e.,

\vec{F} ⊥ \vec{v} and \vec{F} ⊥ \vec{B} .

Question 49

Wired Faculty App

A certain proton moving through a magnetic field region experiences maximum force. When does this occur?

Solution

The force experienced by a particle is maximum when, proton moves perpendicular to the magnetic field.

\vec{v} \times \vec{B}

is maximum(

θ = 90 ° .

)

That is, velocity component is perpendicular to magnetic field.

Question 50

Wired Faculty App

Under what conditions is the force acting on a charge moving through a uniform magnetic field minimum?

Solution

When a charge is moving parallel or antiparallel to the direction of the magnetic field, the force acting on the charged particle is zero or minimum.

Question 51

Wired Faculty App

What is the nature of magnetic field in a moving coil galvanometer?

Solution

In a moving coil galvanometer, the magnetic field is radial in nature.

Question 52

Wired Faculty App

Two wires of equal lengths are bent in the form of two loops. One of the loops is square shaped whereas the other loop is circular. These are suspended in a uniform magnetic field and the same current is passed through them. Which loop will experience greater torque? Give reasons.

Solution

Torque on a current carrying loop is given by,

τ = B I A

where, A is the area of the loop.

For a wire of given length,

Area of circular loop =

{πr}^{2}

and,
Area of square loop = a² .

Circular loop has greater area than the square loop. So the circular loop will experience greater torque in the magnetic field, because torque has direct dependance on area of the loop.

Question 53

Wired Faculty App

Write the SI unit of (i) magnetic pole strength (ii) magnetic dipole moment of a bar magnet.

Solution

(i) The SI unit of magnetic pole strength is Ampere-metre (A-m).

(ii) The SI unit of magnetic dipole strength is Ampere metre²(Am²).

Question 54

Wired Faculty App

Give two factors by which the current sensitivity / voltage sensitivity of a moving coil galvanometer can be increased.

Solution

Current sensitivity or voltage sensitivity of a moving coil galvanometer can be increased by :

(i) Increasing the number of turns in the galvanometer coil.
(ii) Decreasing the torsion constant of the suspension fibre.

Question 55

Wired Faculty App

A beam of protons with a velocity 4 x 10⁵ m/s enters a uniform magnetic field of 0.3 T at an angle 60° to the magnetic field. Find the radius of the helical path taken by the proton beam. Also find the pitch of the helix mp = 1.67 x 10^–27 kg.

Solution

Given, the velocity of protons, v = 4 x 10⁵ m/s
Uniform magnetuc field, B = 0.3 T
Angle made by particle with magnetic field = 60^o
The magnetic force will act like centripetal force.
Therefore,

Using the formula,

r = \frac{mv}{qB}

r = \frac{1.67 \times 10^{- 27} \times 4 \times 10^{5} \sin 60 °}{1.6 \times 10^{- 19} \times 0.3} = 1.2 \times 10^{- 2} = 1.2 cm

Pitch is the distance moved along the magnetic field in one rotation.

Therefore,

T = \frac{2 πr}{v \sin θ} = 2.175 \times 10^{- 7} S

∴

P = v \cos θ . T

= 4 \times 10^{5} \times \frac{1}{2} \times 2.175 \times 10^{- 7} = 4.35 cm

Question 56

Wired Faculty App

An electron moves around the nucleus in a hydrogen atom of radius 0.51 Å, with a velocity of 2 x 105 m/s. Calculate the following:
(i) the equivalent current due to orbital motion of electron,
(ii) the magnetic field produced at the centre of the nucleus
(iii) the magnetic moment associated with the electron.

Solution

Given,
Radius, r = 0.51 Å = 0.51 x 10^–10 m
velocity, v = 2 x 10⁵ m/s

(i) Equivalent current due to orital motion of electron is given by,

$I = \frac{e}{t} = \frac{e}{2 πr / v} = \frac{ev}{2 πr}$

$I = \frac{1.6 \times 10^{- 19} \times 2 \times 10^{5}}{2 \times 3.14 \times 0.51 \times 10^{- 10}}$

$I = \frac{3.2 \times 10^{- 4}}{3.2028} = 0.99 \times 10^{- 4} = 10^{- 4} A$

(ii) Magnetic field produced at the centre of the electron is,
$B = μnI$

$B = 4 π \times 10^{- 7} \times 1 \times 10^{- 4}$

$B = 4 \times 3.14 \times 10^{- 11} = 12.56 \times 10^{- 11} = 1.256 \times 10^{- 10} T .$

(iii) Magnetic moment assosciated with the electron

$M = IA = I ({πr}^{2}) = 10^{- 4} \times 3.14 \times (0.51 \times 10^{- 10})^{2} = 3.14 \times 0.2601 \times 10^{- 4} \times 10^{- 20}$

$M = 0.816 \times 10^{- 24} = 8.16 \times 10^{- 25} {Am}^{2}$

Question 57

Wired Faculty App

Three long straight and parallel wires, carrying currents, are arranged as shown in the figure below. Find the force experienced by a 25 cm length of wire C.

Solution

Length of the wire on which force is experienced, l = 25 cm
As seen in the fig. the direction of current in C and D wires are in opposite direction so, force of repulsion F will act on C towards right side. Similarly, due to G it will be in left hand side so resultant force on C due to wires D and G will be F.
Thus,

F = F_{1} - F_{2} = \frac{μ_{0} I_{1} I_{2} l}{2 {πr}_{1}} - \frac{μ_{0} I_{1} I_{3} l}{2 {πr}_{2}}

= \frac{μ_{0} I_{1} l}{2 π} [\frac{I_{2}}{r_{1}} - \frac{I_{3}}{r_{2}}]

= \frac{4 π \times 10^{- 7} \times 10 \times 25}{2 π} [\frac{30}{0.03} - \frac{20}{0.10}] = 5 \times 10^{- 7} \times 80 = 400 \times 10^{- 7}

F_c = 4 x 10^–5 N towards left side.

Question 58

Wired Faculty App

The wire shown in figure below carries a current of 60 A. Find the magnetic field B at P.

Solution

Given, current carried by the wire, I= 60 A .

The field at P arises from 3/4th of the circular loop only because P lies on the straight wires themselves.
Thus,

B = \frac{μ_{0}}{4 π} \frac{2 πl}{r} \frac{θ}{360} θ = 270 ° T h e r e f o r e, B = \frac{3 \times 10^{- 7} \times 2 π \times 60}{4 \times 0.02} = 1.4 \times 10^{- 3} T

Question 59

Wired Faculty App

What is the toroid? Using Ampere's circuital law calculate the magnetic field inside the toroid.

Solution

When a solenoid is in the form of a ring then it is treated as toroid. Toroid is basically, a hollow circular ring on which a large number of turns of wire are closely wound.

Consider a toroid carrying current I and has N turns. The magnetic field is set up inside the turns of the toroid. The magnetic lines of force inside the toroid are concentric circles.
By symmetry the magnitude of the field

\vec{B}

is same at all points on the circle of radius r and is directed tangentially to the circle at any point.

∴

\oint \vec{B} . \vec{dl} = \oint B dl cosθ °

\oint \vec{B} . \vec{dl} = \oint B dl \cos 0 °

\oint \vec{B} . \vec{dl} = B 2 πr

By applying Ampere's circuital law,

\oint \vec{B} . \vec{dl} = μ_{0} \times total current passing through circle of radius r . B 2 πr = μ_{0} \times N \times 2 πrl

∴

B = μ_{0} NI

Question 60

Wired Faculty App

A 50 turn coil as shown in the figure below carries of 2 A in a magnetic field B = 0.25 Wb m ^–2. Find the torque acting on the coil. In what direction will it rotate?

Solution

Number of turns on the coil, N = 50
Current carried by the coil, I= 2A
Magnetic field,B = 0.25 Wb m ^–2Torque acting on the coil, $τ$ = ?

As seen in the fig.
The sides AB and DC are along the field lines hence, the force on each side is zero.
The force on each vertical wire is given as
τ = BINA sin θ
τ = 0.25 x 2 x 50 x 0.12 x 0.1 sin θ
= 0.3 N-m (clockwise direction)

Question 61

Wired Faculty App

Increasing the current sensitivity may not necessarily increase the voltage sensitivity of galvanometer. Justify.

Solution

Let the deflection produced in applying voltage V is

φ

then,

voltage sensitivity = \frac{φ}{v} = \frac{NBA}{kR}

The voltage sensitivity may be increased by
(i) increasing N, B, A and,
(ii) decreasing k.

Current sensitivity is deflection per unit current whic is , given by

\frac{φ}{I}

current sensitivity = \frac{NBA}{k}

It can be increased by
(i) increasing NBA
(ii) decreasing k.

There is a factor of resistance R in voltage sensitivity.
Hence, increasing the current sensitivity may not necessarily increase the voltage sensitivity of a galvanometer.

Question 62

Wired Faculty App

In an exercise to increase current sensitivity of a galvanometer by 25%, its resistance is also increased 1.5 times. How will the voltage sensitivity of the meter be affected ?

Solution

Here given,
Resistance is increased to 1.5 times.
i.e, R' = 1.5 R

Current sensitivity is increased by 25%.

i.e., $I'_{s} = I_{s} + \frac{25}{100} I_{s} = \frac{125}{100} I_{s} = \frac{5}{4} I_{s}$

$Initially, voltage sensitivity, V_{s} = \frac{I_{s}}{R} and, V'_{s} = \frac{I_{s}'}{R^{'}} = \frac{(5 / 4) I_{s}}{1.5 R} = \frac{5}{6} V_{s}$

Therefore,

% increase in voltage sensitivity= $(1 - \frac{V'_{s}}{V_{s}}) \times 100$ = $(1 - \frac{5}{6}) \times 100 = 16.7 %$

Question 63

Wired Faculty App

A charge ‘q’ moving along the -X-axis with a velocity $\vec{v}$ is subjcted to a uniform magnetic field B acting along the Z-axis as it crosses the origin O.

(i) Trace its trajectory.
(ii) Does the charge gain kinetic energy as it enters the magnetic field? Justify your answer.

Solution

Given,

(i) As the charged particle is moving along negative x axis subjected to uniform magnetic field along Z axis it will experience a force when it crosses the origin F=q (

\overset{⇀}{v} \times \overset{⇀}{B}

).
The trajectory of charged particle q moving along (-)ve X-axis will be helical.

(ii) The force which is acting on the charged particle due to it's magnetic field is always perpendicular to the velocity at each point. Therefore, the field does no work done on a moving charged particle. Hence, it will not gain any kinetic energy. Moreover, magnitude of velocity does not change. Only, the direction of velocity changes for the charged particle.

Question 64

Wired Faculty App

Two straight wires A and B of lengths 10 m and 12 m carrying currents of 4.0 A and 6.0 A respectively in opposite directions lie parallel to each other at a distance of 0.03 m. Estimate the force on a 15 cm section of the wire B near its centre.

Solution

Given,
Length of wire A = 10 m
Length of wire B = 12 m
Current carried by A= 4 A
Current carried by B = 6 A
Distance between the two parallel wires, r = 0.03 m

Force per unit length acting on the wire,

\frac{F}{l} = \frac{μ_{0} I_{1} I_{2}}{2 πr} {Nm}^{- 1} = \frac{4 π \times 10^{- 7} \times 4 \times 6}{2 π \times 0.03} = 1.6 \times 10^{- 4} {Nm}^{- 1}

Force on 15 cm section of wire B (near its centre),
F= 1.6 x 10^–4 x 0.15 N
= 2.4 x 10^–5 N

The force is repulsive as the currents are in opposite directions; the direction of force is normal to the wire away from A.

Question 65

Wired Faculty App

Derive an expression for the torque acting on a loop of N turns, area A, carrying current I, when held in a uniform magnetic field.
With the help of circuit, show how a moving coil galvanometer can be converted into an ammeter of given range. Write the necessary mathematical formula.

Solution

Let, I be the current through the loop PQRS.
Given, a and b are the sides of the rectangular loop.
Area of the loop, A = ab
Number of turns in the loop = N

According to Fleming's left-hand rule, the side PQ experiences a normal inward force, F₁ = laB and,
side SR experiences a normal outward force, F₂ = IaB.
These two equal and opposite forces form a couple which exerts a torque given by

τ = Force x Perpendicular distance
= IaB x b sin θ
= IB (ab) sin θ
= IB A sin θ [ ∵ A = ab]
As the coil has N turns, so
τ = NBA sin θ

Conversion of galvanometer into ammeter: A galvanometer can be converted into an ammeter by connecting a low resistance S in parallel with it. This low resistance is called shunt.

Let I_g be the current with which the galvanometer gives full scale deflection.
As galvanometer and shunt are connected in parallel, so
P.D. across the galvanometer = P.D. across the shunt
$I_{g} R_{g} = (I - I_{g}) R_{s}$

$R_{s} = \frac{I}{I - I_{g}} \times R_{g}$
Hence by connecting a shunt of resistance R_s across the galvanometer, we get an ammeter of desired range.

Question 66

Wired Faculty App

A circular coil of 200 turns, radius 5 cm carries a current of 2.5 A. It is suspended vertically in a uniform horizontal magnetic field of 0.25 T, with the plane of the coil making an angle of 60° with the field lines. Calculate the magnitude of the torque that must be applied on it to prevent it from turning.

Solution

Given,
Number of turns in the coil, N = 200
Radius of coil, r = 5 $\times 10^{- 2} m$
Area of coil, $A = {πr}^{2}$
$= \frac{22}{7} \times 5 \times 10^{- 2} \times 5 \times 10^{- 2} = 7.587 \times 10^{- 3} m^{2}$
Current carried by the coil, I = 2.5 A
Magnetic field, B = 0.25 T
Angle made by coil with the field lines = 60^o

Using the formula for torque,
$τ = NBIA \cos θ$
= 200 x 0.25 x 2.5 x 7.857 x 10^–3 x cos 60° Nm
= 0.49 Nm

An opposite and equal torque is required in order to prevent the coil from turning. Thus, the magnitude of the applied torque should be 0.49 m.

Question 67

Wired Faculty App

Draw the field lines of (a) a bar magnet (b) a current carrying finite solenoid, and (c) an electric dipole.
What basic difference do you notice between the magnetic and electric field lines? How do you explain this difference?

Solution

The field lines of (a) a bar magent, (b) a current carrying finite solenoid and (c) electric dipole are shown below.
At large distances, the field lines are very similar. The curves labelled (i) and (ii) are closed Gaussian surfaces.

There is a basic difference between magnetic and electric field lines. In case of the electric field of an electric dipole, the electric lines of force originate from positive charge and end at the negative charge.
In case of a bar magnet, the magnetic field lines are closed loops, i.e., magnetic field lines do not start or end anywhere.

Question 68

Wired Faculty App

Two long parallel straight wires X and Y separated by a distance of 5 cm in air carry currents of 10 A and 5 A respectively in opposite directions. Calculate the magnitude and direction of the force on a 20 cm length of the wire Y.

Solution

Given, two current carrying parallel wires,
Distance between the wire , r = 5 cm
Current carried by first wire = 10 A
Current carried by second wire = 5A
Length of the wire, l = 20 cm

The Force between two wires can be found using the formula,

F = \frac{μ_{0} I_{1} I_{2} l}{2 πr} = \frac{4 π \times 10^{- 7} \times 10 \times 5 \times 0.20}{2 π \times 0.05} = 4 \times 10^{- 5} N .

Question 69

Wired Faculty App

A charge 2Q is spread uniformly over an insulated ring of radius R/2. What is the magnetic moment of the ring if it is rotated with an angular velocity ω with respect to normal axis?

Solution

Given,
Radius of an insulated ring = R/2
Charge= 2Q
Angular velocity with which the ring is rotated with respect to normal axis = ω

Now,

Charge on the element of length dl of the ring is

d_{q} = λ . dl

\Rightarrow dq = \frac{2 Q}{2 π (R / 2)} dl = \frac{2 Q}{πR} dl

Current due to circular motion of this charge is

dI = dq \times v = \frac{2 Q}{πR} dl \times \frac{ω}{2 π}

(∵ ω = 2 πv)

Magnetic moment due to current dl

dM = dI \times π (R / 2)^{2} = \frac{2 Q}{πr} dl \times \frac{ω}{2 π} \times π (R / 2)^{2}

M = \frac{QωR}{4 π} \int dI = \frac{QωR}{4 π} . 2 πR = \frac{1}{2} {QωR}^{2} .

Question 70

Wired Faculty App

A galvanometer with a coil of resistance 120 ohm shows full scale deflection for a current of 2.5 mA. How will you convert the galvanometer into an ammeter of range 0 to 7.5 A? Determine the net resistance of the ammeter. When an ammeter is put in a circuit, does it read slightly less or more than the actual current in the original circuit? Justify your answer.

Solution

Resistance of coil , R_g = 120 ohm
Current on galvanometer, I_g = 2.5 mA = 0.0025 A
Ammeter range, I = 0- 7.5 A

Galvanometer can be converted into an ammeter bu putting a shunt in parallel with galvanometer.

R_{s} = \frac{I_{g}}{I - I_{g}} \times R_{g} = \frac{0.0025}{7.5 - 0.0025} \times 120 = 0.04 Ω

Therefore,

By connecting a shunt of 0.04 Ω across the given galvanometer, we get an ammeter of range of 0 to 7.5 A.

Net resistance of the ammeter

= \frac{120 \times 0.04}{120 + 0.04} = 0.03998 Ω

When an ammeter is put in a circuit, it reads slightly less than the actual current.
An ammeter has a small resistance. When it is connected in the circuit, it decreases the current by a small amount.

Question 71

Wired Faculty App

An electron of 45 eV energy is revolving in a circular path in a magnetic field of intensity 9 x 10^–5 Wb m^–2. Determine (i) the speed of the electron (ii) radius of the circular path.

Solution

Energy of electron, E= 45 eV

That is,

\frac{1}{2} {mv}^{2} = 45 \times 1.6 \times 10^{- 19}

v = \sqrt{\frac{2 \times 45 \times 1.6 \times 10^{- 19}}{9.1 \times 10^{- 31}}} {ms}^{- 1} = 3.98 \times 10^{6} {ms}^{- 1}

which, is the required speed of electron.

ii) Now, since its a circular path the magnetic force is equal to the centripetal force.

Therefore,

Bev = \frac{{mv}^{2}}{r} r = \frac{mv}{Be}

r = \frac{9.1 \times 10^{- 31} \times 3.98 \times 10^{6}}{9 \times 10^{- 5} \times 1.6 \times 10^{- 19}} m = 0.25 m

is the required radius of the circular path.

Question 72

Wired Faculty App

State Biot-Savart law. A current I flows in a conductor placed perpendicular to the plane of the paper. Indicate the direction of the magnetic field due to a small element $\vec{dl}$ at point P situated at a distance $\vec{r}$ from the the element as shown in the figure.

Solution

Biot-Savart law states that, the magnitude of magnetic field induction at a point due to a current element of length dl, carrying current I, at a point r from the element is given by

dB = \frac{μ_{0}}{4 π} . \frac{I dl \sin θ}{r^{2}}

Vector form :

|\vec{dB}| = \frac{μ_{0}}{4 π} . \frac{|dl \times I|}{r^{3}}

The direction of magnetic field

\vec{dB}

is perpendicular to the plane containing

\vec{dl}

and

\vec{r}

and, is directed inward.

Question 73

Wired Faculty App

If electron velocity is $2 \hat{i} + 3 \hat{j}$ and it is subjected to magnetic field of $4 \hat{k},$ then its
path will change
speed will change
both (a) and (b)
none of the above

Solution

path will change

Question 74

Wired Faculty App

An electron having energy 10 eV is circulating in path having magnetic field of 10^–4T. The speed of the electron will be
3.8 x 10⁶ m/s
1.9 x 10¹² m/s
3.8 x 10¹² m/s
1.9 x 10⁶ m/s

Solution

1.9 x 10⁶ m/s

Question 75

Wired Faculty App

A particle having charge 100 times that of an electron is revolving in a circular path by radius 0.8 m with one rotation per second. The magnetic field produced at the centre is
10^–17 μ₀
10¹⁷ μ₀
10^–16 μ₀
10^–15 μ₀

Solution

10^–17 μ₀

Question 76

Wired Faculty App

A particle of mass m and charge q is placed at rest in a uniform electric field E and then released. The kinetic energy attained by the particle after moving a distance y is
qEy²
qE²y
qEy
q²Ey

Solution

qEy

Question 77

Wired Faculty App

A galvanometer of resistance 25 Ω is shunted by a 2.5 Ω wire. The part of total current that flows through the galvanometer is given by
$\frac{I}{I_{0}} = \frac{1}{11}$
$\frac{I}{I_{0}} = \frac{2}{11}$
$\frac{I}{I_{0}} = \frac{3}{11}$
$\frac{I}{I_{0}} = \frac{4}{11}$

Solution

\frac{I}{I_{0}} = \frac{1}{11}

Question 78

Wired Faculty App

The resistance of a galvanometer is 50 $Ω$ and the current required to give full scale deflection is 100 A. In order to convert it into an ammeter for reading up to 10 A, it is necessary to put a resistance of
5 x 10^–32Ω
5 x 10^–4 Ω
5 x 10^–5 Ω
5 x 10^–2 Ω

Solution

5 x 10^–4 Ω

Question 79

Wired Faculty App

The resistance of the coil of ammeter is R. The shunt resistance required to increase its range four fold should have a resistance
R/3
R/4
4R
R/5

Solution

R/3

Sponsor Area

Question 80

Wired Faculty App

Two long parallel wires P and Q are held perpendicular to the plane of the paper with distance of 5 m between them. If P and Q carry current of 2.5 A and 5A respectively in the same direction, then the magnetic field at a point half way between the wire is
$μ_{0} / π$
$\sqrt{3} μ_{0} / π$
μ₀/2 $π$
μ₀/2 $π$

Solution

μ₀/2

π

Question 81

Wired Faculty App

A wire of length 2m carries a current of 1 ampere is bend to form a circle. The magnetic moment of the coil is
$2 π$
$π / 2$
$π / 4$
$1 / π$

Solution

1 / π

Question 82

Wired Faculty App

The magnetic field of given length of wire for single turn coil at its centre is B, then its value for two turns coil for the same wire is
B/4
B/2
4B
2B

Solution

Question 83

Wired Faculty App

When a charged particle enters in a uniform magnetic field, its kinetic energy.
remains constant
increases
decreases
becomes zero

Solution

remains constant

Question 84

Wired Faculty App

The deflection in moving coil galvanometer falls from 50 divisions to 10 divisions, when a shunt of 12 $Ω$ is applied, the resistance of galvanometer coil is
12 Ω
24 Ω
48 $Ω$
50 $Ω$

Solution

Ω

Question 85

Wired Faculty App

If the number of turns, area and current through a coil is given by n, A and I respectively, then its magnetic moment will be
nIA
n²IA
nIA²
$nI / \sqrt{A}$

Solution

nIA

Question 86

Wired Faculty App

The scale of a galvanometer of resistance 100 ohm contains 25 divisions. It gives a deflection of one division on passing a current of 4 x 10^–4 A. The resistance in ohms to be added to it, so that it may become a voltmeter of range 2.5 volt is
100
150
250
300

Solution

150

Question 87

Wired Faculty App

The electric current in a circular coil of two turns produces a magnetic induction of 0.2 T at its centre. The coil is unwound and is rewound into a circular coil of four turns. The magnetic induction at the centre of the coil now is, in tesla (if same current flows in the coil)
0.2
0.4
0.6
0.8

Solution

0.8

Question 88

Wired Faculty App

State Biot-Savart law Using Biot-Savart law, derive an expression for the magnetic field at the centre of a circular coil of number of turns ‘N’, radius ‘r’ carrying a current.

Solution

Biot-savart law states that, the magnetic field induction dB at a point due to current element is

$d B = \frac{μ_{o}}{4 π} \frac{I d l s i n θ}{r^{2}}$
where,
I.dl is the current element,
r is the distance of point from the current elemnt and,
$θ is the angle between \overset{⇀}{dl} and \overset{⇀}{r} .$

Expression for magnetic field at the centre of a circular coil :

Consider a circular coil of radius 'r' with center O. Let, I be the current flowing in the circular coil. Assume, that the current coil is made of a large number of current elements each of length 'dl' .
current element = I.dl
Now, as per Biot Savart law we have,

$d B = \frac{μ_{o}}{4 π} \frac{I d l s i n θ}{r^{2}}$
Since, the angle between $\overset{⇀}{dl} a n d \overset{⇀}{r}$ is 90^0.Therefore,
$d B = \frac{μ_{o}}{4 π} \frac{I d l}{r^{2}}$
Thus,

B= $\int d B = \int \frac{μ_{o}}{4 π} \frac{I d l}{r^{2}} = \frac{μ_{o}}{4 π} \frac{I}{r^{2}} \int d l$
But,
$\int d l = t o t a l l e n g t h o f c i r c u l a r c o i l = c i r c u m f e r e n c e o f c u r r e n t l o o p = 2 π r$

Therefore,
$B = \frac{μ_{O}}{4 π} \frac{2 πnI}{r}$
where,
n is the number of turns in the coil.

Question 89

Wired Faculty App

A proton and an $α$ -Particle of same kinetic energy in turn move through a uniform magnetic field B in a plane normal to the field. Compare the radii of the paths of the two particles.

Solution

Given , a proton and an alpha particle move through a uniform magnetic field b with same kinetic energy.
The magnetic force acting is equal to the centripetal force.
Therefore,
$\frac{m v^{2}}{r} = q v B \Rightarrow r = \frac{m v^{2}}{q v B} = \frac{m v}{q B} = \frac{p}{q B} = \frac{\sqrt{2 m E}}{q B}$

radius of alpha particle is,

$r_{α} = \frac{\sqrt{2 m_{α} E}}{q_{α} B} = \frac{\sqrt{2 \times 4 m_{p} E}}{2 q_{p} B}$

radius of proton is,

$r_{p} = \frac{\sqrt{2 m_{p} E}}{q_{p} B}$

Dividing both equations, we get

$\frac{r_{α}}{r_{p}} = \frac{2}{1}$

That is, radius of alpha particle is twice as that of the radius of proton.

Question 90

Wired Faculty App

Define the SI unit of magnetic field. “A charge moving at right angles to a uniform magnetic field does it undergo change in kinetic energy”. Why?

Solution

The strength of magnetic field at a point is called one Tesla, if a charge of 1 Coulomb when moving with a velocity of 1 metre per second along a direction perpendicular to the direction of magnetic field, experiences a force of 1 Newton.

The magnitude of velocity remains same and hence, the field does no work on the particle. Therefore there is no change in kinetic energy.

Question 91

Wired Faculty App

Give the principle of a cyclotron. Draw a labelled diagram of cyclotron and describe how a positively charged particle is accelerated in it?

Solution

Principle of cyclotron :
The working of the cyclotron is that a positively charged particle can be accelerated to sufficiently high energy using smaller oscillating electric field by making the cyclotron cross the same elctric field time and again by magnetic field.

Working of cyclotron:
1.The positive ion which is to be accelerated is bought at a point in between D₁ and D_2.Suppose, D₁ is at a negative potential and D₂ is at a positive potential. Hence, the ion will be accelerated towards D₁.
2. The ion will be in a free space once it enters D₁ and therefore, it moves with a constant speed say, v.
3. A perpendicular magnetic field B is applied. So, the ion will undergo a circular path of radius r in D₁ .
The circular path is given by,
$q v B = \frac{m v^{2}}{r}$
where, m is the mass of ion and,
q is the charge on ion.

$∴ r = \frac{mv}{qB}$
4. Now, time taken by an ion to describe a semicircular path is given by,

t = $\frac{πr}{v} = \frac{π m}{q B} = \frac{π}{B \frac{(q}{m)}} = c o n s \tan t$

5. If, this time during which the positive ion describes a semi-circular path is the time during which half-cycle of electric oscillator is completed, then as the ion arrives in the gap between the two dees, polarity of the two dees gets reversed. Consequently, positive ion is accelerated towards D₂ and the ion enters D₂ with greater speed which will remain constant in D₂.
6. Ion will describe a semicircular path of greater radius due to perpendicular magnetic field and will again arrive in a gap between the two dees exactly at that instant when, the polarity of two dees is reversed.
7. Thus, positive ion will keep on accelerating everytime it comes into the gap between the dees and will describe a circular path of greater and greater radius henceforth, acquiring sufficiently high kinetic energy.

Question 92

Wired Faculty App

Obtain an expression for the magnetic moment associated with the orbital motion of an electron.

Solution

Consider, a stationary heavy nucleus around which an electron is performing a uniform circular motion.
Charge , q= +Ze
Current constituted , I = e/T .... (1)
where, T is the time period of revolution.

Let, r be the orbital radius of the electron.
v be the orbital speed.
Then,
T= $\frac{2 πr}{v}$ .

Putting this equation in (1) we get,

$I = \frac{ev}{2 πr}$
Magnetic moment is given by,
$μ = IA = I \times {πr}^{2} = \frac{ev}{2 πr} \times {πr}^{2} = \frac{evr}{2}$

Question 93

Wired Faculty App

What is the effect of current flowing in the same direction in two long straight and parallel conductors? Give the definition of ‘ampere’ based on this effect.

Solution

In two long parallel straight conductors, when the current is flowing across the wires in the same direction, they attract each other. i.e, parallel currents attract.

One ampere is the value of that steady current which, when maintained in each of the two very long, straight, parallel conductors of negligible cross-section, and placed one metre apart in vacuum, would produce on each of the conductors a force equal to $2 \times 10^{7}$ newtons per metre of length.

Question 94

Wired Faculty App

Derive the expression for the torque on a rectangular coil of area A, carrying a current I, placed in a magnetic field B. The angle between the direction of B and the vector perpendicular to the plane of the coil is 0.

Solution

Let, a rectangular loop carrying a steady current I, placed in a uniform magnetic field B experiences a torque.
Angle between the field and the normal to the coil be angle $θ$ .

Force on arm BC and DA are equal and opposite, and is acting along the axis of the coil. Being, collinear along the axis they cancel each other resulting in no net force or torque.
Let, the forces on arms AB and CD be F₁ and F₂.
Magnitude of F₁ and F₂ is given by,
F₁= F₂ = BbI .
Since, these forces are equal and opposite net force is 0.
But, there is a torque acting on the loop as a result of these forces which, is given by

$τ = F_{1} \frac{a}{2} \sin θ + F_{2} \frac{a}{2} \sin θ = I ab B \sin θ = I A B \sin θ$

Now, as theta tends to 0, perpendicular distance between the force of the coule also approaches 0. This makes the forces collinear and, net force and torque becomes 0.

Therefore,
$τ = 0 for, θ \to 0$ .

Question 95

Wired Faculty App

What is the magnetic moment associated with a coil of 1 turn, area of cross-section 10^–4m² carrying a current of 2A?

Solution

Number of turns, n = 1
Area of cross section, A = 10^-4 m²
Current , I =2 A
Magnetic moment assosciated with the coil is,

m= nIA
= 1

\times 2 \times 10^{- 4}

= 2

\times 10^{- 4} {Am}^{2}

Question 96

Wired Faculty App

An electron is projected with a speed of 10⁵ ms^–1 at right angles to a magnetic field of 0.019 G. Calculate the radius of the circle described by the electron.

Solution

Given,
Speed of an electron, v = 10⁵ m/s
Magnetic field, B = 0.019 G = 0.019 $\times 10^{- 4} T$

Therefore,
Radius of the circle described by the electron, $r = \frac{mv}{qB}$ .

= $\frac{9.1 \times 10^{- 31} \times 10^{5}}{1.60 \times 10^{- 19} \times 0.019 \times 10^{- 4}}$

= 29.9 cm

is the calculated radius of the circle.

Question 97

Wired Faculty App

Draw a labelled diagram of a moving coil galvanometer, Explain the principle on which it works. Deduce an expression for the torque acting on a rectangular current carrying loop kept in uniform magnetic field. Write two factors on which the current sensitivity of a moving coil galvanometer depend.

Solution

Moving coil galavanometer is an instrument used for the detection and measurement of small currents.

Principle : The working of moving coil galvanometer is that when a current carrying coil is placed in a varying magnetic field, it experiences torque.

Torque acting on a rectangular current carrying coil:

Let, the coil PQRS be freely suspended in a magnetic field.
Let,
l = length of PQ or RS of the coil,
b = breadth of QR or SP of the coil,
n= number of turns in the coil,
B= strength of magnetic field in which the coil is suspended,
I = current passing through the coil in the direction PQRS.
Then,
Area of the coil, A = l $\times$ b

At any instant, $θ$ is the angle which the normal drawn on the plane of the coil makes with the direction of magnetic field.

The rectangular current carrying coil, when placed in a magnetic field experiences a torque whose magnitude is given by,
$τ = n I B A s i n θ$

Current sensitivity of a moving coil galvanometer depends on :
i) Number of turns of the coil
ii) Magnetic field applied.

Question 98

Wired Faculty App

Two identical conducting wires AOB and COD are placed at right angles to each other. The wire AOB carries an electric current I₁ and COD carries a current I₂. What will be the magnetic field on point lying at a distance d from O, in a direction perpendicular to the plane of the wires AOB and COD?

Solution

Given,
Two identical conducting wire AOB and COD are placed at right angles to each other.

Current carried by AOB = I₁
Current carried by COD = I₂

Magnetic field on AOB , B₁= $\frac{μ_{0} I_{1}}{2 πd} (- j)$
Magnetic field on COD, B₂ = $\frac{μ_{o} I_{2}}{2 πd} (+ i)$

Therefore,
Resultant magnetic field , B = $\sqrt{{B_{1}}^{2} + {B_{2}}^{2}}$

= $\frac{μ_{o}}{2 πd} \sqrt{{I_{1}}^{2} + {I_{2}}^{2}}$

Question 99

Wired Faculty App

A charge particle moves through a magnetic field perpendicular to its direction. Which of following quantity will change or remain unchanged (t) kinetic energy (ii) momentum?

Solution

When a charged particle is moving perpendicular to the direction of magnetuic field both, kinetic energy and momentum remains the same. The reason is, only the direction of the velocity component changes whereas, the magnitude of velocity remains the same.

Question 100

Wired Faculty App

A charge particle with charge q enters a region of constant E and B with velocity V perpendicular to both E and B, and comes out without any charge in magnitude or direction of V, then establish a relation between V, E and B.

Solution

A charged particle goes undeflected when,
qE = qvB

$\Rightarrow$ $\Rightarrow$ v = E/B

The force due to electric field and magnetic field are acting on the charged particle and will oppose each other if $\overset{⇀}{v} i s a l o n g \overset{⇀}{E} \times \overset{⇀}{B}$ .
Now, v = $\frac{E B}{B^{2}} = \frac{E B \sin 90^{o}}{B^{2}} = \frac{|\overset{⇀}{E} \times \overset{⇀}{B}|}{B^{2}}$
Therefore,

$\overset{⇀}{v} = \frac{|\overset{⇀}{E} \times \overset{⇀}{B}|}{B^{2}}$
which is the required relation . $Error: timeout waiting for response.$

Question 101

Wired Faculty App

A long straight wire of radius a carries a steady current I. The current is uniformly distributed across its cross-section. Find the ratio of magnetic field at a/2 and 2a.

Solution

Given a long straight wire of radius 'r' carries uniform current.

The current enclosed by first amperian path = $\frac{{Iπr}^{2}_{1}}{{πR}^{2}} = \frac{I {r^{2}}_{1}}{R^{2}}$

Now, using Ampere circuital law,'
$\oint B . d l = μ_{o} i$
$\Rightarrow$ B = $\frac{μ_{o} i}{2 π r_{1}} = \frac{μ_{o}}{2 π r_{1}} \times \frac{I {r_{1}}^{2}}{R^{2}} = \frac{μ_{o} I r_{1}}{2 π R^{2}}$

Magnetic field induction at a distance r₂ is,
B' = $\frac{μ_{o}}{4 π} \frac{2 I}{r_{2}}$

Therefore,
Ratio of magnetic field at a/2 and 2a is,
$\frac{B}{B'} = \frac{r_{1} r_{2}}{R^{2}} = \frac{(\frac{a}{2}) . 2 a}{a^{2}} = 1$

Question 102

Wired Faculty App

A solenoid of length 75 cm, has a radius of 1 cm and has a total of 750 turns wound on it. It carries a current of 4A. Calculate the magnitude of the axial magnetic field inside the solenoid. If a proton were to move with a speed of 10³ ms^–1 along the axis of this current carrying solenoid, what would be the force experienced by this proton?

Solution

Given, a solenoid.
Length of the solenoid, l = 75 cm
Radius of the coil, r = 1 cm
Number of turns in the coil, N = 750
Current carried by the wire, I = 4 A

Magnetic field inside a solenoid, B = $μ_{o} n I$

Number of turns per metre = $\frac{750}{0.75} = 1000 t u r n s / m$

Therefore,
Magnetic field, B = $4 π \times 10^{- 7} \times 1000 \times 4$ $4 π \times 10^{- 7} \times 1000 \times 4$
= 16 $\times$ 3.14 $\times$ 10^-4
= 5.024 $\times$ 10^-3T

Force experienced by the proton, F
F = qvB sin $θ$
Speed of the proton, v = 10³ m/s
Charge on proton, q = +e = 1.6 $\times 10^{- 19} C$

The direction of magnetic field is along the axis.
$θ$ is the angle between velocity and magnetic field which is 0^oin this case .

$∴ \sin 0 = 0$
Hence the force acting on proton = 0.

Question 103

Wired Faculty App

What is direction of force acting on a charged particle q, moving with a velocity u in a uniform magnetic field B?

Solution

The force is acting in a sideways direction, perpendicular to both the velocity and the ,agnetic field.
$\overset{⇀}{F} = q [\overset{⇀}{v} \times \overset{⇀}{B}]$

The direction of force is given by the right hand rule.

Question 104

Wired Faculty App

Why should the spring/suspension wire in a moving coil galvanometer have low torsional constant?

Solution

Low torsional constant for moving coil galvanometer facilitates greater deflection $ϕ$ in coil for given value of current and hence increases the sensitivity of galvanometer.

Question 105

Wired Faculty App

Using Ampere's circuital law, derive an expression for the magnetic field along the axis of current carrying torodial solenoid of N number of turns having radius r.

Solution

Magnetic field due to a toroidal solenoid :

A long solenoid in the the form of a circular ring is known as torroidal solenoid.

Let,
number of turns pr unit length of the solenoid be 'n',
I - Current flowing throught the coil,
B- magnetic field inside the turns of the solenoid.

Magnetic lines of force inside the torroid is in the form of concentric circles. Using the law of symmetry, the magnetic field is along the tangent at every point on the circle and is same at each point of the circle.

For points inside the core of toroid,
Let us consider a circle of radius 'r'.
Length of torid = circumference of circular path = 2 $π$ r
Total no. of turns in torroid = n. (2 $π$ r )
Current flowing across each turn = I
Therefore,
Total current enclosed by circular path = n (2 $π$ r) I
Now, using Ampere's circuital law,
   $\oint B . d l = μ_{o} I$
and    $\oint B . d l \cos 0 = B . 2 π r$

Substituting this in Ampere's law we get,

   $B . 2 πr = μ_{O} (n . 2 πr . I) B = μ_{O} n I$

which is the required magnetic field along the axis of current carrying toroidal solenoid.

Question 106

Wired Faculty App

Magnetic field lines can be entirely confined within the core of a toroid, but not within a straight solenoid. Why?

Solution

Toroid is a circular ring having no ends. Hence, the magnetic field lines form continuous loops and there is no flux leakage. But, in a straight solenoid there is a leakage of flux across the ends.

Question 107

Wired Faculty App

A charge ‘q’ moving along the X-axis with a velocity u is subjected to a uniform magnetic field B acting along z-axis as it crosses the origin.
(i) Trace it trajectory.
(ii) Does the charge gain kinetic energy as it enters the magnetic field? Justify your answer.

Solution

Given,
Charge is moving along the x-axis and,
Magnetic field is moving along z-direction.

Therefore, the magnetic force is such that the particle will follow a circular motion.
$F_{m} = q (\overset{⇀}{v} \times \overset{⇀}{B}) = q (V \overset{⏜}{i} \times B \overset{⏜}{k}) = - qVB \overset{⏜}{k} \overset{⏜}{y}$

Since, the force and velocity are perpendicular to each other, the particle does not gain energy.
W = $\overset{⇀}{F} . \overset{⇀}{v} = F v \cos θ$ [ $θ = 90^{o}$ ]

Hence, the force will not cause any change in velocity.

Question 108

Wired Faculty App

An iron core is inserted into a solenoid 0.5 m long with 400 turns per unit length. The area of cross-section of the solenoid is 0.001 m². (a) Find the permeability of the core when a current of 5 A flows through the solenoid winding. Under these conditions, the magnetic flux through the cross-section of the solenoid is 1.6 x 10^–3 Wb. (b) Find the inductance of the solenoid under these conditions.

Solution

Given,
Length of the solenoid, l = 0.5 m
Number of turns per unit length, n = 400
Area of cross- section of the solenoid, A= 0.001 m²

The magnetic induction on the axis of the solenoid is given by

B = μ [μ_{0} nI] B = μ (\frac{μ_{0}}{4 π}) 4 πni . . . . (1)

where,
μ is the permeability of the medium,
n the number of turns per unit length and,
i is the current passing through the coil.

(a) The normal to the area is along the direction of the field.

Therefore,

Magnetic flux,

ϕ = BA,

Given,

Flux, ϕ = 1.6 \times 10^{- 3} Wb, A r e a, A = 0.001 = 10^{- 3} m^{2}

Therefore,

B = \frac{ϕ}{A} = \frac{1.6 \times 10^{- 3}}{10^{- 3}} = 1.6 Wb m^{- 2}

Since,

No . of turns, n = 400, Current, i = 5 A and, \frac{μ_{0}}{4 π} = 10^{- 7} {Hm}^{- 1}

We have form equation (1),

1.6 = μ \times 10^{- 7} \times 4 π \times 400 \times 5

which gives,

μ = \frac{1.6 \times 10^{7}}{4 π \times 5 \times 400} = 636.7 \approx 637

(b) Total number of turns in the solenoid is given by

N = n \times l = 400 \times 0.5 = 200

Total flux through the solenoid is,

N ϕ = 200 \times 1.6 \times 10^{- 3} = 0.32 Wb

Self inductance,

L = \frac{Nϕ}{i} = \frac{0.32}{5} = 0.064 H = 64 mH

Alternatively,

L = \frac{{μμ}_{0} N^{2} A}{l} = {μμ}_{0} n^{2} lA

= 637 \times (4 π \times 10^{- 7}) \times 400 \times 400 \times 0.5 \times 10^{- 3} = 6.4 \times 10^{- 2} H = 64 mH

Question 109

Wired Faculty App

Using Biot-Savart’s law, derive an expression for the magnetic field at the centre of a circular coil of radius R, number of turns N, carrying current i.

Solution

As per Biot-Savart law, the magnetic field due to a current element

\vec{dl}

at the observation point whose position vector

\vec{r}

is given by

\vec{dB} = \frac{μ_{0} I}{4 π} . \frac{\vec{dl} \times \vec{r}}{r^{3}}

where, μ₀ is the permeability of free space.

Consider a circular loop of wire of radius 'r' carrying a current 'I' and also a current element 'dl' of the loop.

The direction of dl is along the tangent, so dl ⊥ r.

Using Biot-Savart law,

Magnetic field at the centre O due to this current element is

dB = \frac{μ_{0} I}{4 π} \frac{dl \sin 90 °}{r^{3}} = \frac{μ_{0} I}{4 π} \frac{dl}{r^{2}}

The magnetic field due to all such current elements will point into the plane of paper at the centre O.
Hence, the total magnetic field at the centre O is given by

B = \int dB = \int \frac{μ_{0} Idl}{4 {πr}^{2}}

B = \frac{μ_{0} I}{4 {πr}^{2}} \int dl = \frac{μ_{0} I}{4 {πr}^{2}} . l

= \frac{μ_{0} I}{4 {πr}^{2}} . 2 πr B = \frac{μ_{0} I}{2 r}

For a coil of N turns, magnetic field, B =

\frac{μ_{0} NI}{2 r}

Question 110

Wired Faculty App

Two small identical circular coils marked 1, 2 carry equal currents and are placed with their geometric axes perpendicular to each other as shown in the figure. Derive an expression for the resultant magnetic field at 0.

Solution

Given, two identical circular loops which are placed perpendicular to each other carry equal currents.

Now,

Magnetic field at O due to loop 1.

B_{1} = \frac{μ_{0} {iR}^{2}}{2 {(x^{2} + R^{2})}^{3 / 2}}

, acting towards left.

Magnetic field at O due to loop 2.

B_{2} = \frac{μ_{0} {iR}^{2}}{2 {(x^{2} + R^{2})}^{3 / 2}}

,acting vertically upwards .

where, R is the radius of each loop.

Therefore,
Resultant magnetic field at O will be

B = \sqrt{B_{1}^{2} + B_{2}^{2}} = \sqrt{2 B_{1}}

(∵ B_{1} = B_{2})

= \frac{μ_{0}}{\sqrt{2}} \frac{i R^{2}}{{(x^{2} + R^{2})}^{3 / 2}}

This resultant field acts at an angle of 45° with the axis of loop 1.

Question 111

Wired Faculty App

Two parallel co-axial circular coils of equal radius ‘R’ and equal number of turns ‘N’, carry equal currents ‘I’ in the same direction and are separated by a distance ‘2R’. Find the magnitude and direction of the net magnetic field produced at the mid-point of the line joining their centres.

Solution

Given, two parallel co-axial circular coils of equal radius 'R' and equal number of turns 'N' carry equal currents 'I' in the same direction seperated by distance '2R' .

Magnetic field at the mid-point due to loop 1 is given by,

B_{1} = \frac{μ_{0} i R^{2}}{2 {(R^{2} + R^{2})}^{3 / 2}},

acting towards right .

Magnetic field at the mid-point due to loop 2 is,

B_{2} = \frac{μ_{0} i R^{2}}{2 {(R^{2} + R^{2})}^{3 / 2}},

acting towards right.

∴ Net resultant magnetic field at the mid-point is,

B = B_{1} + B_{2}

B = \frac{μ_{0} i R^{2}}{{(2 R^{2})}^{3 / 2}}

which is acting towards right.

Question 112

Wired Faculty App

Two long parallel wires carrying currents 2.5 ampere and I ampere in the same direction (directed into the plane of the paper) are held at P and Q respectively such that they are perpendicular to the plane of paper. The points P and Q are located at a distance of 5 metre and 2 metre respectively from a collinear point R (see figure).

(i) An electron moving with a velocity of 4 x 10⁵ m/s along the positive x-direction experiences a force of magnitude 3.2 x 10^–20 N at the point R. Find the value of I.
(ii) Find all the positions at which a third long parallel wire carrying a current of magnitude 2.5 amperes may be placed so that the magnetic induction at R is zero.

Solution

Given, two long parallel wires.
Current carried by first wire = 2.5 A
Current carried by second wire = 2.5 A
Distance of point P from R = 5 m
Distance of point Q from R = 2 m

(i) Magnetic field B₁ at R due to P is,

B_{1} = \frac{μ_{0} I}{2 π r} = \frac{μ_{0}}{2 π} (\frac{2.5}{5})

Magnetic field

B_{2}

at R due to Q is,

B_{2} = \frac{μ_{0}}{2 πr} \frac{I}{r} = \frac{μ_{0}}{2 π} (\frac{I}{2})

Therefore,
Resultant magnetic field at R is given by,

B = B_{1} + B_{2} = \frac{μ_{0}}{2 π} (\frac{2.5}{5} + \frac{I}{2}) . . . . . . (1)

This Resultant field at R acts downwards and perpendicular to PX.

Now,
Force experienced by electron moving along PX is

\vec{F} = e (\vec{v} \times \vec{B}) [θ = 90 °]

\vec{F}

is perpendicular to both

\vec{v} and \vec{B} .

Because of the negative charge of electron, the force F acts perpendicular to the plane of paper directed upwards.

Now,

F = evB

∴

B = \frac{F}{ev} = \frac{3.2 \times 10^{- 20}}{1.6 \times 10^{- 19} \times 4 \times 10^{5}} = 0.5 \times 10^{- 6}

...(2)

Equating this to (1), we get

\frac{μ_{0}}{2 π} [\frac{2.5}{5} + \frac{I}{2}] = 0.5 \times 10^{- 6}

which gives

I = 2 [\frac{2 π}{μ_{0}} (0.5 \times 10^{- 6}) - \frac{2.5}{5}] = 2 [\frac{2 π \times (0.5 \times 10^{- 6})}{4 π \times 10^{- 7}} - \frac{2.5}{2}]

= 2 [\frac{2.5}{1} - \frac{2.5}{5}] = 4 A

(ii) In this case, we can consider the following alternatives:

(a) If 'r' is the distance of the current 2.5 A which, is directed into the plane of paper from point R then,
we have

B_{3} = (\frac{μ_{0}}{2 π}) (\frac{2.5}{r})

Now,

B_{1} + B_{2} + B_{3} = 0

Therefore,

\frac{μ_{0}}{2 π} (\frac{2.5}{5} + \frac{4}{2} + \frac{2.5}{r}) = 0

which gives, r= -1 m .
Thus the third wire is located at 1 m from R on RX.

(b) When the current 2.5 A is directed out from the plane of paper in an upward direction.

Therefore,
I₃ = – 2.5 A

Hence, we will have

\frac{μ_{0}}{2 π} (\frac{2.5}{5} + \frac{4}{2} + \frac{2.5}{r}) = 0

which gives r = 1 m i.e., the third wire is located 1 m from R on RQ.

Question 113

Wired Faculty App

Depict the magnetic field lines due to two straight, long parallel conductors carrying currents I₁ and I₂ in the same direction. Hence deduce on expression for the force acting per unit length on one conductor due to the other. Is this force attractive or repulsive?
Figure shows a rectangular current carrying loop placed 2 cm away from a long, straight, current carrying conductor. What is the direction and magnitude of the net force acting on the loop?

Solution

Given, two infinitely long conductors AB and CD with currents I₁ and I₂ in same direction respectively, placed parallel to each other, and separated by distance r.

Magnetic field produced by current at any point of CD is $B_{1} = \frac{μ_{o} I_{1}}{2 π r}$

This field acts perpendicular to CD and into the plane of paper. It exerts a force on wire CD carrying current I₂.

Force exerted on unit length of CD is

$F = B_{1} I_{2} I = \frac{μ_{0} I_{1}}{2 πr} \times I_{2} \times 1$

$F = \frac{μ_{0} I_{1} I_{2}}{2 πr} = \frac{μ_{0}}{4 π} . \frac{2 I_{1} I_{2}}{r}$

According to Fleming's left hand rule, this force acts on CD towards AB.

Similarly, conductor CD also exerts an equal force AB towards itself. Hence the two wires get attracted towards each other.

Numerical:

Force on AB, $F_{1} = \frac{μ_{0}}{4 π} . \frac{2 I_{1} I_{2}}{r_{1}} \times length of AB$

Question 114

Wired Faculty App

A circular coil of 25 turns and radius 6.0 cm, carrying a current of 10 A, is suspended vertically in a uniform magnetic field of magnitude 1.2 T. The field lines run horizontally in the plane of the coil. Calculate the force and torque on coil due to the magnetic field. In which direction should a balancing torque be applied to prevent the coil from turning?

Solution

Given, a circular coil.
Number of turns on the coil, n = 25
Radius of the coil, r = 6 cm
Current carried by the coil, I = 10 A
Uniform magnetic field, B = 1.2 T

Consider any element

\vec{dl}

of the wire.
Force on this element is I

(\vec{dl} \times \vec{B}) .

For each element

\vec{dl},

there is another length element

- \vec{dl}

on the closed loop.

Since

\vec{B}

is uniform, therefore, the forces cancel for each pair of such elements.
Therefore, the net force on the coil is zero.

The torque

\vec{τ}

on a plane loop of any shape carrying a current I in a magnetic field B is given by

\vec{τ} = IA \hat{n} \times \vec{B}

where,

\hat{n}

is a unit vector normal to the plane of the loop (direction of motion of a right- handed screw rotating in the direction of current).

For a circular coil of radius r and N turns,

A = N \times {πr}^{2}

In the given problem, the angle between normal vector,

\hat{n} and magnetic field, \vec{B}

is 90°.
Now,
τ = BIA sin α

= 1.2 \times 10 \times 25 \times π {(0.06)}^{2} \sin 90 ° = 3.39 N m

The direction of

\vec{τ}

is vertically upwards. So as to, prevent the coil from turning, an equal and opposite torque must be applied.

Question 115

Wired Faculty App

State Ampere's circuital law and give it's mathematical formulation.

Solution

Ampere's circuital law states that, the line integral of magnetic field induction $\overset{⇀}{B}$ around a closed path in vacuum is equal to $μ_{o}$ times the total current passing through the closed path .

Mathematically,
$\oint B . d l = μ_{o} I$

Question 116

Wired Faculty App

A rectangular coil of sides 8 cm and 6 cm having 2000 turns and carrying a current of 200 mA is placed in a uniform magnetic field of 0.2 T directed along the positive x-axis.
(a) What is the maximum torque the coil can experience? In which orientation does it experience the maximum torque?
(b) For which orientations of the coil is the torque zero? When is this equilibrium stable and unstable.

Solution

We know that a current loop, having n turns, each of area A, carrying current I, when placed in a magnetic field

\vec{B}

, experience a torque whose magnitude is given by
τ = nlAB sin α ...(i)
where

α

is the angle which the normal on the plane of the current loop makes with the direction of magnetic field, i.e., angle between

\vec{A} and \vec{B} .

Here,Number of turns, N = 2000
Current carried by the coil, I = 200 mA = 200 x 10^–3 A
Area of the coil, A = 8 x 6 sq. cm = 48 x 10^–4 m²
Uniform magnetic field, B = 0.2 T.

(a) Torque acting on the coil will be maximum when
sin

α

= 1
or when

α

= 90°
Therefore,
∴ Maximum torque, τ_max = NIAB
= 2000 x (2.0 x 10^–3) x (48 x 10^–4) x 0.2
= 0.384 N/m

In this situation, the plane of the coil is parallel to the direction of magnetic field i.e., the plane of the coil is in the direction of X-axis.

(b) Torque on the coil will be zero, if
sin

α

= 0 or when,

α

= 0° or 180°.
It will be so if plane of the coil is perpendicular to the direction of magnetic field i.e., the plane of the coil is along Y or Z-axis.

The coil will be in stable equilibrium when,

\vec{A}

is parallel to

\vec{B}

and is in unstable equilibrium when

\vec{A}

is antiparallel to

\vec{B} .

Question 117

Wired Faculty App

What is the relationship between the current and the magnetic moment of a current carrying circular loop? Use the expression to derive the relation between the magnetic moment of an electron moving in a circle and its related angular momentum?

Solution

Let us assume that an electron of mass 'm_e' and charge 'e' revolves in a circular orbit of radius 'r' around the positive nucleus in anticlockwise direction.

The angular momentum of the electron due to its orbital motion is given by
L = m_evr    ...(i)

Let the period of orbital motion of the electron be T.
Then, the electron crosses any point on its orbit after every T seconds.

Therefore, orbital motion of electron is equivalent to a current.
   $I = e . (\frac{1}{T})$
The period of revolution of the electron is given by
   $T = \frac{2 πr}{v}$

$∴$ $I = e (\frac{1}{2 πr / v}) = \frac{ev}{2 πr}$

The area of the electron orbit, $A = {πr}^{2}$

The magnetic dipole moment of the atom is

   $M = IA$

$M = \frac{ev}{2 πr} \times {πr}^{2} = \frac{evr}{2}$ ...(ii)
Now, using the equation (i) we have

   $M = (\frac{e}{2 m_{e}}) L$ [From I]

In vector rotation
   $\vec{M} = - (\frac{e}{2 m_{e}}) \vec{L}$

which, tells us that the magnetic dipole moment vector is directed in a direction opposite to that of angular momentum vector.

Question 118

Wired Faculty App

A coil in the shape of an equilateral triangle of side 0.02 m is suspended from a vertex such that it is ranging in a vertical in plane magnetic field of 5 x 10^–2 T. Find the couple acting on the coil when a current of 0.1 ampere is passed through it and the magnetic field is parallel to its plane.

Solution

Given,
Side of equilateral triangle, a = 0.02 m
Magnetic field, B = 5 x 10^–2 T
Current passd through the coil, I = 0.1 A

As the coil is in the form of an equilateral triangle, its area is

A = \frac{\sqrt{3}}{4} \times 0.02 \times 0.02 A = \sqrt{3} \times 10^{- 4} m = \sqrt{3} \times 10^{- 4} m^{2}

Torque on current carring coil in magnetic field is

τ = BINA sinα

where,

α = 90 °

N = 1

τ = BIA

= 5 \times 10^{- 2} \times 0.1 \times \sqrt{3} \times 10^{- 4} N^{- m} = 0.5 \times \sqrt{3} \times 10^{- 6} = 5 \sqrt{3} \times 10^{- 7} N m

Question 119

Wired Faculty App

A solenoid of length 0.4 m and having 500 turns of wire carries a current of 3 A. A thin coil having 10 turns of wire and of radius 0.01 m carries a current of 0.4 A. Calculate the torque required to hold the coil in the middle of the solenoid with its axis perpendicular to the axis of the solenoid.

Solution

The torque acting on a current loop of turn N and current i in a magnetic field of induction B due to a solenoid is given by
τ = (N i A) B sin θ
where,
i is the current flowing in N turns of the loop with face area A and making an angle

θ

with the field.
N i A can be regarded as the magnetic dipole moment M.
The magnetic dipole moment M of the loop lies along the axis of the loop.
Magnetic field B due to a solenoid is given by

B = μ_{0} n i_{0}

where,
n is the number of turns per unit length of the solenoid and,
i₀ is the current in the solenoid.
The direction of B is along its axis.
Therefore, when the small coil is held with its axis perpendicular to the axis of the solenoid, the torque becomes

τ = μ_{0} {ni}_{0} N i A

Here,

N = 10, n = \frac{500}{0.4} = 1250, i_{0} = 3 A, i = 0.4 A

A = {πr}^{2} = π {(0.01)}^{2} = π \times 10^{- 4} m^{2} μ_{0} = 4 π \times 10^{- 7}

Therefore,

τ = 4 π \times 10^{- 7} \times 1250 \times 3 \times 10 \times 0.4 \times π \times 10^{- 4}

= 5.922 \times 10^{- 6} N m

Sponsor Area

Question 120

Wired Faculty App

A current carrying coil of 200 turns has an area of cross-section 1 x 10^–4 m². When suspended freely through its centre, it can turn in a horizontal plane. What is the magnetic moment of the coil for a current of 5 A? Also calculate the net force and torque on coil if a uniform horizontal field of 10 x 10^–2 T is set up at an angle of 30° with axis of coil when it is carrying the same current.

Solution

Here,
Number of turns, N = 200
Area of cross-section. A = 1 x 10^–4 m²Current passing through the coil, I = 5 A
M =?
Magnetic momemnt of the coil, M = N IA
= 200 x 5 x 10^–1
= 10^–1 JT^–1In a uniform horizontal field B = 10 x 10^–2 T, the ends of coil experience equal and opposite forces. Therefore, net force on the coil = 0.
$θ = 30^{o}$

Therefore, torque acting on the coil is,

$Torque, τ = MB \sin θ = 10^{- 1} \times 10 \times 10^{- 2} \sin 30 ° = 10^{- 2} \times \frac{1}{2} = 0.5 \times 10^{- 2} Nm$

Question 121

Wired Faculty App

A galvanometer of resistance 50 Ω gives full scale deflection for a current of 0.05 A. Calculate the length of the shunt wire required to convert the galvanometer into an ammeter of range 0 to 5 A. The diameter of the shunt wire is 2 mm and its resistivity is 5 x 10^–7 Ω m.

Solution

Given,
Resistance of galvanometer, R_g = 50 Ω
Current, I_g = 0.05 A
Ammeter current, I = 5A
Diameter of the shunt wire, d = 2 mm = 2

\times

10^-3 m
Therefore, radius, r = 1

\times 10^{- 3} m

Now,
By using the formula,

S = \frac{{GI}_{g}}{I - I_{g}}

Putting the values,

S = \frac{50 \times 0.05}{5 - 0.05} Ω = \frac{50}{99} Ω

Also,

S = p \frac{1}{{πr}^{2}} or l = \frac{{πr}^{2} S}{p}

Putting the values, we get

l = \frac{22}{7} \times {(1 \times 10^{- 3})}^{2} \times \frac{50}{99} \times \frac{1}{5 \times 10^{- 7}} = 3.175 m .

which, is the required length of the shunt wire.

Question 122

Wired Faculty App

Two parallel wires Q and R carry currents 5 A and 10 A respectively in opposite directions. A plane lamina ABCD intersects the wires at right angles at points Q and R. Find the magnitude of the total magnetic induction at point P located in the lamina as shown in the figure.

Solution

Given,
Current carried by wire Q, I₁ = 5 A
Current carried by wire R, I₂ = 10 A

The magnetic induction at P due to the 10 A current, is

B_{1} = \frac{μ_{0} I_{1}}{2 {πr}_{1}} = \frac{4 π \times 10^{- 7} \times 10}{2 π \times 200 \times 10^{- 3}} T = 10 μT

The magnetic induction at P due to the 5 A current,

B_{2} = \frac{μ_{0} I_{2}}{2 {πr}_{2}}

B_{2} = \frac{4 A \times 10^{- 7} \times 5}{2 A \times 80 \times 10^{- 3}} B_{2} = \frac{5 \times 100 \times 10^{- 7}}{100} = 125 \times 10^{- 1} \times 10^{- 6} = 12.5 \times μT

Hence, Resultant magnetic induction is,

B = \sqrt{B_{1}^{2} + B_{2}^{2} + 2 B_{1} B_{2} \cos (180 ° - θ)}

where,

{QR}^{2} = {PQ}^{2} + {PR}^{2} - 2 PQ . PR \cos θ

\cos θ = \frac{- 150^{2} + 80^{2} + 200^{2}}{2 \times 80 \times 200} = + 0.7469

Therefore,

B = 10^{- 6} \sqrt{10^{2} + 12 . 5^{2} - 2 \times 10 \times 12.5 \times 0.7469} = 8.34 \times 10^{- 6} T = 8.34 μT

Question 123

Wired Faculty App

Two concentric coplanar semi-circular conductors form part of two current loops as shown in the figure. If their radii are 11 cm and 4 cm, calculate the magnetic induction at the centre.

Solution

Given, two concentric coplanar semi-circular conductors.
Radius of inner loop, r₁ = 4 cm
Radius of outer loop, r₂ = 11 cm
Current carried by the loops is 20 A and 40 A respectively.
Magnetic field due to circular loop at centre is given by,

B = \frac{μ_{0} I}{2 r}

Magnetic induction at centre due 4 qudadrant of wire is,

= \frac{1}{4} \frac{μ_{0}}{2} (\frac{40}{r_{1}} - \frac{40}{r_{2}}) - \frac{1}{4} \frac{μ_{0}}{2} (\frac{20}{r_{1}} - \frac{20}{r_{2}}) = \frac{4 π \times 10^{- 7}}{8} [20 (\frac{1}{r_{1}} - \frac{1}{r_{2}})] = \frac{4 π \times 10^{- 7}}{8} [20 (\frac{1}{4 \times 10^{- 2}} - \frac{1}{11 \times 10^{- 2}})] = 5 \times 10^{- 5} weber / m^{2} (inward)

Question 124

Wired Faculty App

Derive an expression for the maximum force experienced by a straight conductor of length I, carrying current I and kept in a uniform magnetic field, B.

Solution

Consider a straight conductor PQ of length l, area of cross section A, carrying current I placed in a uniform magnetic field

\vec{B}

.

Suppose the conductor is placed along x-axis and magnetic field acts along y-axis.
Hence, current I flows from end P to Q and electrons drift from Q to P.

Let,

\vec{v_{d}} = drift velocity of electron

- e = charge on each electron

Therefore,

Magnetic Lorentz force on an electron is given by

\overset{⇀}{f} = - e (\vec{v_{d}} \times \vec{B})

[∵ F = q (\vec{v} \times \vec{B})]

If, n is the number of free electrons per unit volume of the conductor, then
total number of free electrons in the conductor will be N = n (AI) = nAl.

$∴$
Total force on the conductor is

$\vec{F} = N \vec{f} = n A l [- e (\vec{v_{d}} \times \vec{B})] = - n A l e (\vec{v_{d}} \times \vec{B}) . . . (i)$

But, the current through a conductor is related with drift velocity by the relation
$I = n A e {\vec{v}}_{d}$

$∴$ Il = n A e v_dl
We represent I $\vec{l}$ as current element vector and, it acts in the direction of flow of current i.e., along OX.
Then we have I $\vec{l}$ and $\vec{v_{d}}$ in opposite directions. So,

$I \vec{l} = - n Ale \vec{v_{d}}$ ...(ii)

Therefore, from (i) and (ii), we have

$\vec{F} = I (\vec{l} \times \vec{B})$

Magnitude of
F = Il B sin θ
When $π = 90 °, then, F_{\max} = B I l$

The direction of force on a current carrying conductor placed perpendicular to the magnetic field is given by Fleming’s left hand rule. If we stretch the fore finger, central finger and the thumb of our left hand mutually perpendicular to each other such that the fore finger points in the direction of magnetic field, central finger in the direction of current, then the thumb gives the direction of force experienced by the conductor.

Question 125

Wired Faculty App

A straight horizontal conducting rod of length 0.60 m and mass 60 g is suspended by two vertical wires at its ends. A current of 5.0 A is set up in the rod through the wire.
(a) What magnetic field should be set up normal to the conductor in order that the tension in the wire is zero?
(b) What will be total tension in the wires if the direction of current is reversed, keeping the magnetic field same as before (Ignore mass of the wire), g = 10 ms^–2.

Solution

Given,
Length of the conducting rod, l = 0.60 m
Mass, m = 60 g = 60 x 10^–3 kg
Current set up in the rod, I = 5.0 A

(a) Tension in the wire is zero if the force on the wire carrying current due to magnetic field is equal and opposite to the weight of wire,
i.e., BIl = mg

$\Rightarrow$ $B = \frac{mg}{I l} = \frac{(60 \times 10^{- 3}) \times 10}{5.0 \times 0.60} = 0.20 T$

The force on the conductor due to magnetic field will be upwards if the direction of magnetic field is horizontal and normal to the conductor.
(b) When direction of current is reversed, BIl and mg will act vertically downwards, then the effective tension in the wires,
T = BIl + mg
= 0.2 x 5.0 x 0.60 + (60 x 10^–3) x 10
= 1.2 N.

Question 126

Wired Faculty App

An electron beam passes through a magnetic field of 4 x 10^–3 weber/m² and an electric field of 2 x 104 Vm^–1, both acting simultaneously. The path of electron remaining undeviated, calculate the speed of the electrons. If the electric field is removed, what will be the radius of the electron path?

Solution

Given,an electron beam.
Magnetic field, B = 4 x 10^–3 weber/m²Electric field, E = 2 x 10⁴ V/m

The force on the moving electron due to electric field is equal and opposite to the force on moving electron due to magnetic field since the path of moving electron is undeviated,

Therefore,
$eE = evB$
i.e.,
$v = \frac{E}{B} = \frac{2 \times 10^{4}}{4 \times 10^{- 3}} = 5 \times 10^{6} m / s .$

If, the electric field is removed and, when electron is moving perpendicular to magnetic field, the radius r of circular path traced by electron is given by,

$r = \frac{mv}{eB} = \frac{(9.1 \times 10^{- 31}) \times (5 \times 10^{6})}{(1.6 \times 10^{- 19}) \times 4 \times 10^{- 3}}$
$= 7.11 \times 10^{- 3} m = 7.11 mm$

Question 127

Wired Faculty App

A current I flows along the length of an infinitely long, thin walled pipe. Then what will be the magnetic field at a point inside the pipe.

Solution

Given,
Current I is flowing along the length of an infinitely long, thin walled pipe.

According to Ampere's circuital law,
$\int B . d l = μ_{o} I$
Since, no current exists in the medium, inside the pipe,
I=0 $∴$ B = 0.
Hence, the magnetic field at a point inside the pipe = 0

Question 128

Wired Faculty App

A narrow stream of protons and deuterons, having the same momentum values, enter a region of uniform magnetic field directed perpendicular to their common direction of motion. What would be the ratio of the radii of the circular paths, described by the protons and deutrons?

Solution

We are given, stream of protons and deutrons haveing same momentum entering a region of uniform magnetic field perpendicular to the direction of motion.

Charge on deutron is twice as that of proton.

Using the formula,
$r = \frac{mv}{qB}$
$∴$ $\frac{r_{p}}{r_{D}} = \frac{q_{D}}{q_{p}} = \frac{2 q}{q} = 2$
Ratio of the radii of proton to deutron is 2:1 .

Question 129

Wired Faculty App

Find the magnitude of the force on each segment of the wire shown, if a magnetic field of 0.30 T, is applied parallel to AB and DE take the value of current flowing in the wire, as 1 ampere.

Solution

Given,
Magnetic field applied on segment AB and DE, B = 0.30 T
Current flowing in the wire, I = 1A
Therefore,

Force acting on the wire, F = IBl sin θ
For segment AB and DE, θ = 0
∴ force acting on these segments is zero.

Force on segment BC = IBl sin 90°
= 1 (B) 8 x 10^–2 = 0.08 B tesla.

Force acting on CD segment, F = 1.B.10 x 10^–2.sin 30°
= 0.05B tesla.

Question 130

Wired Faculty App

Write the relation for current sensitivity and voltage sensitivity of a moving coil galvanometer? Using these relations, explain the fact that increasing the current sensitivity may not necessary increase the voltage sensitivity.

Solution

Current sensitivity, I_s = $\frac{θ}{I}$

Voltage sensitivity, V_s= $\frac{θ}{V}$

$= \frac{θ}{IR}$

$= \frac{Current sensitivity}{Resistance}$

where, $θ$ is the deflection produced.

The above relation implies, if current sensitivity increases as well as the resistance increases in same order, the voltage sensitivity will remain unchanged.

Question 131

Wired Faculty App

A wire AB is carrying a current of 12A and is lying on the table. Another wire CD, carrying current 5A is arranged just above AB at a height of 1 mm. What should be the weight per unit length of this wire so that CD remains suspended at its position? Indicate the direction of current in CD and the nature of force between two wires.

Solution

Current on AB, I₁ = 12 A
Current on CD, I₂ = 5 A
Distance between wires AB and CD, d = 1 mm = 10^-3 m

Weight of the wire will be balanced by the force of repulsion between the wires AB and CD so that the wire CD remains suspended at it's position

Therefore,

\frac{F}{l} = \frac{μ_{0} I_{1} I_{2}}{2 πd}

i.e.,

\frac{mg}{l} = \frac{μ_{0} I_{1} I_{2}}{2 πd}

= \frac{4 π \times 10^{- 7} \times 12 \times 5}{2 π \times 1 \times 10^{- 3}} {Nm}^{- 1} = 120 \times 10^{- 4} = 0.12 {Nm}^{- 1}

mg/l is the weight per unit length.

The direction of current in CD will be opposite to that of in AB.

Question 132

Wired Faculty App

Two circular loops of radii r and 2r have current I and I/2 flowing through them clockwise and anticlockwise sense respectively. If their equivalent magnetic moments are M₁ and M₂ respectively, state the relation between M₁ and M₂.

Solution

Given, two circular loops of radii r and 2r respectively.
Current on two loops are I (clock-wise) and I/2 (anti-clockwise) respectively.

Now, Magnetic moment is given as M=IA

Therefore,

$M_{1} = I ({πr}^{2})$ and,

$M_{2} = \frac{I}{2} [π (2 r)^{2}]$

$∴$ $\frac{M_{1}}{M_{2}} = \frac{{Iπr}^{2}}{Iπ 2 r^{2}} = \frac{1}{2}$

i.e., $M_{1} = \frac{1}{2} M_{2}$

Question 133

Wired Faculty App

A small magnet of magnetic moment M is placed at a distance r from the origin O with its axis parallel to x-axis as shown. A small coil of one turn, is placed on the x-axis, at the same distance from the origin, with the axis of the coil coinciding with x-axis. For what value of current in the coil does a small magnetic needle, kept at origin remain undeflected? What is the direction of current in the coil?

Solution

For a needle to remain undeflected the magnetic field due to magnet and current loop must be equal and opposite.

The magnetic field of magnet at origin

= \frac{μ_{0}}{4 π} . \frac{M}{r^{3}}

The magnetic field of the small coil at origin

= \frac{μ_{0}}{4 π} \frac{{Iπx}^{2}}{(r^{2} + x^{2})^{3 / 2}}

where, x is the radius of the loop.

For, the needle to remain undeflected we have

\frac{μ_{0}}{4 π} . \frac{M}{r^{3}} = \frac{μ_{0}}{4 π} \frac{{Iπx}^{2}}{(r^{2} + x^{2})^{3 / 2}}

i.e.,

I = \frac{M {(r^{2} + x^{2})}^{3 / 2}}{{πx}^{2}}

The direction of the current in the loop must be anticlockwise.

Question 134

Wired Faculty App

A 3.0 cm wire carrying a current of 10 A is placed inside a solenoid perpendicular to its axis. The magnetic field inside the solenoid is given to be 0.27 T. What is the magnetic force on the wire?

Solution

Here,

Length of the wire, l = 3.0 cm = $3 \times 10^{- 2} m$

Current, I = 10 A

Uniform magnetic field, B = 0.27 T

$θ = 90 °$

Magnetic force on the wire, F = ?

Now using the formula,

F = BI l $\sin θ$ = $0.27 \times 10 \times (3 \times 10^{- 2}) \sin 90^{o} = 0.27 \times 10 \times 3 \times 10^{- 2} \times 1 = 8.1 \times 10^{- 2} N$

Question 135

Wired Faculty App

How does the mutual inductance of a pair of coils change when,

(i) distance between the coils is increased and

(ii) number of turns in the coils is increased?

Solution

Using the formula,

$straight ϕ space equals space MI space$
i) Mutual inductance decreases when the distance between the coils is increased.

ii) Now using the formula,

$straight M subscript 21 space equals space straight mu subscript straight o space straight n subscript 1 space straight n subscript 2 space straight A space straight l$
On increasing the number of turns in the coil, mutual inductance also increases.

Question 136

Wired Faculty App

The motion of copper plate is damped when it is allowed to oscillate between the two poles of a magnet. What is the cause of this damping?

Solution

The changing magnetic flux through the plate produces a strong eddy current in the direction, which opposes the cause as the plate oscillates.

Question 137

Wired Faculty App

A wire AB is carrying a steady current of 12 A and is lying on the table. Another wire CD carrying 5A is held directly above AB at a height of 1 mm. Find the mass per unit length of the wire CD so that it remains suspended at its position when left free. Give the direction of the current flowing in CD with respect to that in AB.
[Take the value of g = 10 ms–2]

Solution

Current carried by wire AB= 12 A

Current carried by CD = 5 A

i) The current carrying conductors will repel each other if the direction of current along both the conductor is same.

ii) When current along the conductors is flowing in the same direction, then the wires will attract each other.

When CD remains suspended above AB,
$Error converting from MathML to accessible text.$

where r is the distance between the wires.

Therefore,
$Error converting from MathML to accessible text.$

The direction of current in CD should be opposite to the direction of current in AB.

Question 138

Wired Faculty App

(a) Using Biot-Savart’s law, derive the expression for the magnetic field in the vector form at a point on the axis of a circular current loop.

(b) What does a toroid consist of? Find out the expression for the magnetic field inside a toroid for N turns of the coil having the average radius r and carrying a current I. Show that the magnetic field in the open space inside and exterior to the toroid is zero.

Solution

Biot-Savart law states that the magnetic field strength (dB) produced due to a current element of current I and length dl at a point having position vector to current element is given by,
$stack d B with rightwards harpoon with barb upwards on top space equals space fraction numerator mu subscript o over denominator 4 pi end fraction fraction numerator I space stack d l with rightwards harpoon with barb upwards on top space cross times space r with rightwards harpoon with barb upwards on top over denominator r cubed end fraction$

where, $mu subscript o$ is permeability of free space.

The magnitude of magnetic field is given by,

$d B space equals space fraction numerator mu subscript o over denominator 4 pi end fraction fraction numerator I space d l space sin straight space theta over denominator r squared end fraction$ ; $theta space$ is the angle between the current element and position vector.

Magnetic field at the axis of a circular loop:

Consider a circular loop of radius R carrying current I. Let, P be a point on the axis of the circular loop at a distance x from its centre O. Let, be a small current element at point A.

Magnitude of magnetic induction dB at point P due to this current element is given by,
$delta B space equals space fraction numerator mu subscript o over denominator 4 pi end fraction fraction numerator I space delta l space sin alpha over denominator r squared end fraction$ ... (1)
The direction of $delta B with italic rightwards harpoon with barb upwards on top$ is perpendicular to the plane containing $stack delta l with rightwards harpoon with barb upwards on top space a n d space r with rightwards harpoon with barb upwards on top$ .
Angle between $stack delta l with rightwards harpoon with barb upwards on top space a n d space r with rightwards harpoon with barb upwards on top space i s space 90 to the power of o$
Therefore,
$stack delta B with rightwards harpoon with barb upwards on top space space equals space italic space fraction numerator mu subscript o I over denominator 4 pi end fraction fraction numerator delta l space sin 90 to the power of 0 over denominator r squared end fraction italic space equals space fraction numerator mu subscript o I delta l space over denominator 4 pi r squared end fraction$ ... (2)

The magnetic induction $stack delta B with rightwards harpoon with barb upwards on top$ can also be resolved into two components, PM and PN’ along the axis and perpendicular to the axis respectively. Thus if we consider the magnetic induction produced by the whole of the circular coil, then by symmetry the components of magnetic induction perpendicular to the axis will be cancelled out, while those parallel to the axis will be added up.

Thus, resultant magnetic induction $B with rightwards harpoon with barb upwards on top$ at axial point P is given by,
$T h e space c o m p o n e n t space o f space stack delta B with rightwards harpoon with barb upwards on top space a l o n g space t h e space a x i s comma space delta B subscript x space equals space fraction numerator mu subscript o over denominator 4 pi end fraction fraction numerator I space delta l space space s i n alpha over denominator r squared end fraction italic space$
$Error converting from MathML to accessible text.$
$therefore space delta B subscript x equals space fraction numerator mu subscript o I space delta l space over denominator 4 pi r squared end fraction. R over r equals fraction numerator mu subscript o I R over denominator 4 pi r cubed end fraction delta l space$
$therefore space delta B subscript x equals space fraction numerator mu subscript o I R over denominator 4 pi left parenthesis R squared plus x squared right parenthesis to the power of 3 divided by 2 end exponent end fraction delta l$

Therefore the magnitude of resultant magnetic induction at axial point P due to the whole circular coil is given by,
B = $contour integral fraction numerator mu subscript o I R over denominator 4 pi left parenthesis R squared plus x squared right parenthesis to the power of 3 divided by 2 end exponent end fraction. d l space equals space fraction numerator mu subscript o I R over denominator 4 pi left parenthesis R squared plus x squared right parenthesis to the power of 3 divided by 2 end exponent end fraction contour integral d l$
$contour integral d l space equals l e n g t h space o f space t h e space l o o p space equals space 2 pi R$

Therefore,
$B space equals space fraction numerator mu subscript o I R over denominator 4 pi left parenthesis R squared plus x squared right parenthesis to the power of 3 divided by 2 end exponent end fraction left parenthesis 2 pi R right parenthesis space space space equals space fraction numerator mu subscript o I R squared over denominator 2 left parenthesis R squared plus x squared right parenthesis to the power of 3 divided by 2 end exponent end fraction space T e s l a$

b) A long solenoid on bending in the form of closed ring is called a toroidal solenoid.

i) For points inside the core of a toroid,

As per Ampere’s circuital law,
$Error converting from MathML to accessible text.$
where, I is the current in the solenoid.

So, resultant net current = NI
$therefore space contour integral space straight B with rightwards harpoon with barb upwards on top space. space dl with rightwards harpoon with barb upwards on top space equals space straight mu subscript straight o space NI space rightwards double arrow space open vertical bar straight B close vertical bar 2 straight pi space straight r space equals space straight mu subscript straight o space NI space rightwards double arrow space space open vertical bar straight B close vertical bar space equals space fraction numerator straight mu subscript straight o space NI space over denominator 2 πr end fraction space space space space space space space space space space space space space space equals straight mu subscript straight o space straight n space straight I space space space space space space space open square brackets because space straight n space equals space fraction numerator straight N over denominator 2 πr end fraction close square brackets$

Since no current is flowing through the points in the open space inside the toroid.

Therefore, I = 0.
So,
$text ∮ end text B with rightwards harpoon with barb upwards on top space. space stack d l with rightwards arrow on top space equals space mu subscript o space I space equals space 0 space rightwards double arrow space open vertical bar B close vertical bar subscript i n s i d e end subscript space equals space 0 space$

Question 139

Wired Faculty App

State the principle of working of a galvanometer.
A galvanometer of resistance G is converted into a voltmeter to measure upto V volts by connecting a resistance R₁ in series with the coil. If a resistance R₂ is connected in series with it, then it can measure upto V/2 volts. Find the resistance, in terms of R₁ and R₂, required to be connected to convert it into a voltmeter that can read upto 2 V. Also find the resistance G of the galvanometer in terms of R₁ and R₂.

Solution

Principle:

When a current carrying loop or coil is placed in the uniform magnetic field, moving coil galvanometer experiences a torque.

A high resistance is connected in series with the galvanometer to convert it into a voltmeter.

The value of resistance is given by,
$straight R straight space equals straight space straight V over straight I subscript straight g minus straight space straight G$
where,
V is the potential difference across the terminals of the voltmeter.
$bold italic I subscript bold italic g bold italic space$ is current through galvanometer and

G is the resistance of the galvanometer.
When resistance R₁ is connected in series with the galvanometer,
$straight R subscript 1 straight space end subscript equals straight space straight V over straight I subscript straight g minus straight G space space space space space space space space space space space space space space space space space space space space space space space... space left parenthesis 1 right parenthesis thin space$

When resistance R₂ is connected in series with the galvanometer,
$straight R subscript 2 straight space equals straight space fraction numerator straight V over denominator 2 straight I subscript straight g end fraction minus straight G straight space$ ... (2)

From equations (1) and (2), we get
$fraction numerator straight V over denominator 2 straight I subscript straight g end fraction equals straight space straight R subscript 1 straight space end subscript minus straight R subscript 2 straight space comma straight space and straight G space equals space straight R subscript 1 space minus space 2 straight R subscript 2$

Resistance R₃ required to convert galvanometer into voltmeter of range 0 to 2V is given by,
$straight R subscript 3 straight space equals straight space fraction numerator 2 straight V over denominator straight I subscript straight g end fraction minus straight space straight G space rightwards double arrow straight R subscript 3 straight space end subscript equals straight space 4 left parenthesis straight R subscript 1 straight space end subscript minus straight R subscript 2 right parenthesis minus left parenthesis 2 straight R subscript 1 straight space end subscript minus straight R subscript 2 right parenthesis space space space space space space space space space space space space space equals straight space 3 straight R subscript 1 straight space end subscript minus 2 straight R subscript 2$
Therefore, R₁ - 2R₂is the galvanometer resistance in terms of R₁ and R₂

Question 140

Wired Faculty App

(a) State Ampere’s circuital law. Use this law to obtain the expression for the magnetic field inside an air cored toroid of average radius ‘r’, having ‘n’ turns per unit length and carrying a steady current I.

(b) An observer to the left of a solenoid of N turns each of cross section area ‘A’ observes that a steady current I in it flows in the clockwise direction. Depict the magnetic field lines due to the solenoid specifying its polarity and show that it acts as a bar magnet of magnetic moment m = NIA.

Solution

Ampere’s Law states that the line integral of magnetic field $bold italic B with rightwards harpoon with barb upwards on top$ around any closed path in vacuum is $straight mu subscript straight o$ times the total current through the closed path.

That is,
$contour integral straight B with rightwards harpoon with barb upwards on top. stack dl straight space with rightwards harpoon with barb upwards on top equals straight space straight mu subscript straight o straight I$
Toroid is a hollow circular ring on which a wire of large number of turns is closely wound.

Let’s consider an air-cored toroid with center O.

We have,
$contour integral straight B with rightwards harpoon with barb upwards on top. stack dl straight space with rightwards harpoon with barb upwards on top equals straight space contour integral straight B straight space dl straight space cosθ Angle space straight theta space between space straight B with rightwards harpoon with barb upwards on top and straight space stack dl straight space with rightwards harpoon with barb upwards on top space is space zero. contour integral straight B straight space dl straight space cosθ straight space equals straight space contour integral stack straight B straight space with rightwards harpoon with barb upwards on top stack dl straight space with rightwards harpoon with barb upwards on top straight space equals straight space straight B cross times straight space circumference straight space of straight space circle straight space of straight space radius straight space straight r rightwards double arrow straight space contour integral straight B with rightwards harpoon with barb upwards on top. stack dl straight space with rightwards harpoon with barb upwards on top straight space equals straight space straight B cross times 2 πr space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 1 right parenthesis As space per space ampere apostrophe straight s space circuital space law comma space contour integral straight B with rightwards harpoon with barb upwards on top. stack dl straight space with rightwards harpoon with barb upwards on top equals straight mu subscript straight o straight space cross times straight space net straight space current straight space enclosed straight space by straight space the straight space circle straight space of straight space radius equals straight space straight mu subscript straight o straight space cross times straight space tota straight space number straight space of straight space turns straight space cross times straight I straight space equals straight space straight mu subscript straight o cross times left parenthesis straight n cross times 2 πr right parenthesis straight I space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space... space left parenthesis 2 right parenthesis thin space Now comma space comparing space equations space left parenthesis 1 right parenthesis space and space left parenthesis 2 right parenthesis comma space we space have straight B cross times 2 πr straight space equals straight space straight mu subscript straight o cross times left parenthesis straight n cross times 2 πr right parenthesis straight I rightwards double arrow straight B straight space equals straight mu subscript straight o nI comma space$

which is the magnetic field due to a toroid carrying current.

b) Given, current is flowing in the clockwise direction for an observer who is on the left side of the solenoid. Implies, left face of the solenoid is the South Pole and right face acts as the North Pole. The magnetic field lines are directed from south to north, inside the bar magnet. Hence, the magnetic field lines are directed from left to right in the solenoid. The figure below illustrates the direction of flow of current inside the solenoid.

Magnetic moment of single current carrying loop = IA

Therefore,

Magnetic moment due to the whole solenoid is, m = N(IA)

Where,

N is the number of turns of solenoid,

I is the current flowing through the loop, and

A is the area of the loop.

Question 141

Wired Faculty App

Write the expression, in a vector form, for the Lorentz magnetic force $straight F with rightwards harpoon with barb upwards on top$ due to a charge moving with velocity $straight V with rightwards arrow on top$ in a magnetic field B. What is the direction of the magnetic force?

Solution

Lorentz magnetic force is given by:
$straight F with rightwards harpoon with barb upwards on top space equals space q left parenthesis stack V space with rightwards harpoon with barb upwards on top space cross times space B with rightwards harpoon with barb upwards on top right parenthesis$ ; q is the magnitude of the moving charge.
The direction of the magnetic force is perpendicular to the plane containing the velocity vector and the magnetic field vector.

Question 142

Wired Faculty App

State the underlying principle of a cyclotron. Write briefly how this machine is used to accelerate charged particles to high energies.

Solution

Cyclotron works on the principle that an oscillating electric field can be used to accelerate a charge particle to high energy.

In a cyclotron, the charged particles across the gap between the two D-shaped magnetic field regions are accelerated by an electric field. The magnetic field is perpendicular to the paths of the charged particles that make them follow circular paths within the two Dee’s. An alternating voltage accelerates the charged particles each time they cross the Dee’s. The radius of each particle’s path increases with its speed. So, the accelerated particles spiral toward the outer wall of the cyclotron.

The accelerating electric field reverses just at the time the charge particle finishes its half circle so that it gets accelerated across the gap between the Dee’s. The particle gets accelerated again and again, and its velocity increases. Therefore, it attains high kinetic energy.

Question 143

Wired Faculty App

(a) State Ampere's circuital law, expressing it in the integral form.

(b) Two long coaxial insulated solenoids, S₁ and S₂ of equal lengths are wound one over the other as shown in the figure. A steady current "I" flow thought the inner solenoid S₁ to the other end B, which is connected to the outer solenoid S₂ through which the same current "I" flows in the opposite direction so as to come out at end A. If n₁ and n₂ are the number of turns per unit length, find the magnitude and direction of the net magnetic field at a point
(i) Inside on the axis and (ii) outside the combined system.

Solution

(a) Ampere’s circuital law states that the line integral of magnetic induction around a closed path in vacuum is equal to times the total current I threading the closed path.

In the above illustration, the Ampere’s Circuital Law can be written as follows:
$contour integral B ⃗. d l rightwards arrow equals mu subscript o space I semicolon space w h e r e comma space i equals vertical line i subscript 1 minus i subscript 2 vertical line.$

(b) (i) The magnetic field due to a current carrying solenoid:
$straight B with rightwards harpoon with barb upwards on top equals straight space straight mu subscript straight o straight space end subscript ni$

where, n = number of turns per unit length

i = current through the solenoid.

From the fig, we can say that, the direction of magnetic field due to solenoid S₁ will be in the upward direction and the magnetic field due to S₂ will be in the downward direction using right hand screw rule.

Therefore,
$straight B subscript net equals straight B subscript straight S 1 end subscript minus straight B subscript straight S 2 end subscript rightwards double arrow straight B subscript net equals straight mu subscript 0 straight n subscript 1 straight I minus straight mu subscript 0 straight n subscript 2 straight I space space space space space space space space space space space space equals straight mu subscript 0 straight I left parenthesis straight n subscript 1 minus straight n subscript 2 right parenthesis$

Net magnetic field is in the upward direction.
(ii) Since, there is no current which is flowing outside the solenoid, the magnetic field is zero.

Question 144

Wired Faculty App

Two identical circular wires P and Q each of radius R and carrying current ‘I’ are kept in perpendicular planes such that they have a common center as shown in the figure. Find the magnitude and direction of the net magnetic field at the common center of the two coils.

Solution

Coils P and Q each of radius R have a common centre:

B_P is directed vertically upwards and B_Qis directed horizontally.

Therefore, resultant magnetic field is given by,
$straight B subscript straight R straight space equals straight space square root of straight B subscript straight P squared plus straight B subscript straight Q squared end root$
We have,
$straight B subscript straight P straight space equals straight space straight B subscript straight Q equals fraction numerator straight mu subscript straight o straight I over denominator 2 straight R end fraction rightwards double arrow straight B straight space equals straight space square root of 2 straight B subscript straight P equals square root of 2 fraction numerator straight mu subscript straight o straight I over denominator 2 straight R end fraction rightwards double arrow straight space straight B straight space equals straight space fraction numerator straight mu subscript straight o straight I over denominator square root of 2 straight R end fraction$
This is the required of the resultant magnetic field.

Direction of magnetic field is given by,
$tan straight space straight theta straight space equals straight space straight B subscript straight P over straight B subscript straight Q equals 1 theta equals pi over 4$
The resultant magnetic field is directed at angle of 45° with either of the fields.

Question 145

Wired Faculty App

A rectangular loop of wire of size 4 cm × 10 cm carries a steady current of 2A. A straight long wire carrying 5A current is kept near the loop as shown.

If the loop and the wire are coplanar, find

(i) the torque acting on the loop and

(ii) the magnitude and direction of the force on the loop due to the current carrying wire.

Solution

i) Torque on the current carrying conductor is given by,
$straight tau straight space equals straight space straight M with rightwards harpoon with barb upwards on top straight space cross times straight B with rightwards harpoon with barb upwards on top straight space equals straight space MB straight space sin straight space straight theta$

In this case, M and B have the same direction. Therefore, = 0^o.

So Torque, = 0.

ii) The magnetic forces are equal and opposite on the parallel current carrying wires.

Therefore, force acting on the loop is given by,

$Error converting from MathML to accessible text.$

The force is attractive in nature or is towards the conductor.

Question 146

Wired Faculty App

A plane electromagnetic wave travels in vacuum along z-direction. What can you say about the direction of electric and magnetic field vectors?

Solution

The direction of electric field vectors is along X-axis.

Magnetic field vector is along Y-axis.

Question 147

Wired Faculty App

A current is induced in coil C₁ due to the motion of current carrying coil C₂.

(a) Write any two ways by which a large deflection can be obtained in the galvanometer G.
(b) Suggest an alternative device to demonstrate the induced current in place of galvanometer.

Solution

a) Two ways by which we can obtain deflection in the galvanometer is by:

i) Moving coil C₂ towards C₁ with high speed.

ii) By placing a soft iron laminated core at the centre of coil C₁.

b) By replacing galvanometer with the torch bulb in coil C₁, one can demonstrate the induced current. The bulb begins to glow as a result of the induced current.

Question 148

Wired Faculty App

State Biot-Savart law, giving the mathematical expression for it.

Use this law to derive the expression for the magnetic field due to a circular coil carrying current at a point along its axis.
How does a circular loop carrying current behave as a magnet?

Solution

where, $mu subscript o$ is permeability of free space.

The magnitude of magnetic field is given by,
$d B space equals space fraction numerator mu subscript o over denominator 4 pi end fraction fraction numerator I space d l space sin straight space theta over denominator r squared end fraction$ , $theta space$ is the angle between the current element and position vector.

Magnetic field at the axis of a circular loop:

Consider a circular loop of radius R carrying current I. Let, P be a point on the axis of the circular loop at a distance x from its centre O.
Let, be a small current element at point A.

The direction of $stack delta B with rightwards harpoon with barb upwards on top$ is perpendicular to the plane containing $stack delta l with rightwards harpoon with barb upwards on top space a n d space r with rightwards harpoon with barb upwards on top$ .
Angle between $stack delta l with rightwards harpoon with barb upwards on top space a n d space r with rightwards harpoon with barb upwards on top space i s space 90 to the power of o$

Therefore,
$stack delta B with rightwards harpoon with barb upwards on top space equals space fraction numerator mu subscript o I over denominator 4 pi end fraction fraction numerator delta l space sin 90 to the power of 0 over denominator r squared end fraction equals fraction numerator mu subscript o I delta l space over denominator 4 pi r squared end fraction$ ... (2)
The magnetic induction $stack delta B with rightwards harpoon with barb upwards on top$ can also be resolved into two components, PM and PN’ along the axis and perpendicular to the axis respectively. Thus if we consider the magnetic induction produced by the whole of the circular coil, then by symmetry the components of magnetic induction perpendicular to the axis will be cancelled out, while those parallel to the axis will be added up.
Thus, resultant magnetic induction $B with rightwards harpoon with barb upwards on top$ at axial point P is given by,
$T h e space c o m p o n e n t space o f space stack delta B with rightwards harpoon with barb upwards on top space a l o n g space t h e space a x i s comma space delta B subscript x space equals space fraction numerator mu subscript o over denominator 4 pi end fraction fraction numerator I space delta l space space s i n alpha over denominator r squared end fraction$
$Error converting from MathML to accessible text.$

Therefore the magnitude of resultant magnetic induction at axial point P due to the whole circular coil is given by,

B = $contour integral fraction numerator mu subscript o I R over denominator 4 pi left parenthesis R squared plus x squared right parenthesis to the power of 3 divided by 2 end exponent end fraction. d l space equals space fraction numerator mu subscript o I R over denominator 4 pi left parenthesis R squared plus x squared right parenthesis to the power of 3 divided by 2 end exponent end fraction contour integral d l space$
$contour integral d l space equals l e n g t h space o f space t h e space l o o p space equals space 2 pi R italic space T h e r e f o r e italic comma italic space B space equals space fraction numerator mu subscript o I R over denominator italic 4 pi italic left parenthesis R to the power of italic 2 italic plus x to the power of italic 2 italic right parenthesis to the power of italic 3 italic divided by italic 2 end exponent end fraction left parenthesis 2 pi R right parenthesis space space space equals space fraction numerator mu subscript o I R squared over denominator 2 left parenthesis R squared plus x squared right parenthesis to the power of 3 divided by 2 end exponent end fraction space T e s l a$
If the coil contains N turns, then $B equals space fraction numerator mu subscript o N I R squared over denominator 2 left parenthesis R squared plus x squared right parenthesis to the power of 3 divided by 2 end exponent end fraction space T e s l a$ is the required magnetic field.
A magnetic needle placed at the center and axis of a circular coil shows deflection. This implies that a circular coil behaves as a magnet.

Question 149

Wired Faculty App

With the help of a labelled diagram, state the underlying principle of a cyclotron. Explain clearly how it works to accelerate the charged particles.
Show that cyclotron frequency is independent of energy of the particle. Is there an upper limit on the energy acquired by the particle? Give reason.

Solution

Underlying principle of cyclotron: When a charged particle is kept in a magnetic field it experiences a force and the perpendicular magnetic field causes the particle to rotate.

Working:
Consider the figure which is shown.

A Dee which is at a negative potential accelerates the positive ions which are produced from the source S. Magnetic field which is perpendicular will move the ions in a circular motion inside the Dees. The magnetic field and the frequency of the applied voltages are so chosen that as the ion comes out of a Dee, the Dees change its polarity (positive becoming negative and vice-versa) and the ion is further accelerated.

Now, the ions move with higher velocity along a circular path of greater radius. The phenomenon is continued till the ion reaches at the periphery of the Dees where an auxiliary negative electrode (deflecting plate) deflects the accelerated ion on the target to be bombarded.

Expression for cyclotron frequency:

Let the velocity of the positive ion having charge ‘q’ be ‘v’.

Then,
$q v B space equals space fraction numerator m v squared over denominator r end fraction$
$rightwards double arrow space r space equals space fraction numerator m v over denominator q B end fraction$ ;m is the mass of ion, r the radius of the path of ion in the dee and B is the strength of the magnetic field.

Angular velocity of the ion is given by,

$straight omega equals straight v divided by straight r equals qB divided by straight m space$

Now, time taken by the ion in describing a semi-circle i.e., turning through an angle is given by,
$t space equals space pi over omega equals fraction numerator pi m over denominator q B end fraction$

The applied alternating potential should also have the same semi-periodic time (T/2) as that taken by the ion to cross either Dee.

That is,

$T over 2 equals t space equals space fraction numerator pi m over denominator q B end fraction$
$rightwards double arrow space T equals space fraction numerator 2 pi m over denominator q B end fraction$ , is the expression for period of revolution.
So, Frequency of revolution of particle is given by, $f space equals space 1 over T equals space fraction numerator q B over denominator 2 pi m end fraction$

This frequency is called the cyclotron frequency which is independent of the speed of the particle.

Question 150

Wired Faculty App

(a) An em wave is traveling in a medium with a velocity . Draw a sketch showing the propagation of the em wave, indicating the direction of the oscillating electric and magnetic fields.
(b) How are the magnitudes of the electric and magnetic fields related to the velocity of the em wave?

Solution

E with rightwards harpoon with barb upwards on top cross times B with rightwards harpoon with barb upwards on top

represents the direction of propagation of wave. The figure below shows us the propagation of the em wave.

a) The direction of oscillating field is

straight i with hat on top space equals space j with hat on top space cross times k with hat on top

b) Speed of EM wave is given by |c| =

straight E subscript straight o over straight B subscript straight o

; E_o is the magnitude of electric field and is the magnitude of magnetic field.

Question 151

Wired Faculty App

(a) Derive the expression for the torque on a rectangular current carrying loop suspended in a uniform magnetic field.
(b) A proton and a deuteron having equal momenta enter in a region of uniform magnetic field at right angle to the direction of the field. Depict their trajectories in the field.

OR

(a) A small compass needle of magnetic moment ‘m’ is free to turn about an axis perpendicular to the direction of uniform magnetic field ‘B’. The moment of inertia of the needle about the axis is ‘I’. The needle is slightly disturbed from its stable position and then released. Prove that it executes simple harmonic motion. Hence deduce the expression for its time period.

(b) A compass needle, free to turn in a vertical plane orients itself with its axis vertical at a certain place on the earth. Find out the values of (i) horizontal component of earth’s magnetic field and (ii) angle of dip at the place.

Solution

A rectangular loop ABCD of dimensions l and b, carrying a steady current is placed in uniform magnetic field as shown in fig; such that normal of the plane is at angle $straight theta$ with the magnetic field lines.

The force F_BC and F_AD on arms BC and AD are equal, opposite and along the axis of the coil, so they cancel each other.

The forces F_AB and F_CD are also equal and opposite, but are not collinear, so they constitute a couple, and the magnitude of the torque can be given as,
$tau space equals space F subscript A B end subscript. b over 2 sin theta space plus space F subscript C D end subscript. b over 2 sin space theta$
Since, $straight tau straight space equals straight space straight F subscript AB. straight b over 2 sinθ straight space plus straight space straight F subscript CD. straight b over 2 sin straight space straight theta$
$vertical line straight F subscript AB vertical line straight space equals straight space vertical line straight F subscript CD vertical line straight space equals straight space BIl therefore straight space straight tau straight space equals straight space BIl cross times straight b straight space sin straight space straight theta space space space space space space equals BI straight space left parenthesis straight l straight space straight b right parenthesis straight space sin straight space straight theta space space space space space space equals straight space BIA straight space sin straight space straight theta$

Since, magnetic moment m I |A|

$straight tau space equals space mB space sin space straight theta$ ; which is the expression of torque.

b) If a charge particle enters right angle to the direction of magnetic field, it follows a circular trajectory, and radius can be given as,
$q v B space equals space fraction numerator m v squared over denominator r end fraction rightwards double arrow space r space equals space fraction numerator m v over denominator q B end fraction equals space fraction numerator p over denominator q B end fraction$

Since momentum are equal and they having equal charges.

So, r_p : r_d = 1:1

If magnetic compass of dipole moment is placed at angle q in uniform magnetic field, and released it experiences a restoring torque.

Torque is given by, $stack tau space with rightwards harpoon with barb upwards on top equals space minus m a g n e t i c space f o r c e space cross times p e r p e n d i c u l a r space d i s tan c e$
$Error converting from MathML to accessible text.$

b) If compass needle orients itself with its axis vertical at a place, then

$straight i right parenthesis space straight B subscript straight H space end subscript equals space 0 space because space BV space equals space vertical line straight B vertical line ii right parenthesis space Angle space of space dip space straight delta space equals space 90 to the power of 0 space space space space space space space space space tan space straight delta space equals space straight B subscript straight v over straight B subscript straight H equals infinity space rightwards double arrow space Angle straight space straight delta space space equals space 90 to the power of straight o$

Question 152

Wired Faculty App

How are electromagnetic waves produced? What is the source of energy of these waves? Draw a schematic sketch of the electromagnetic waves propagating along the + x-axis.
Indicate the directions of the electric and magnetic fields. Write the relation between the velocity of propagation and the magnitudes of electric and magnetic fields.

Solution

Electromagnetic waves are produced by accelerating charged particle. When the charge moves with acceleration, both the magnetic and electric fields change continuously. This change produces electromagnetic waves.

Accelerated charge is the source of energy of these waves.

The direction of oscillating field is $straight i with hat on top space equals space j with hat on top space cross times k with hat on top$ .

Relation between velocity of propagation and magnitude of electric and magnetic field is given by |c| = $straight E subscript straight o over straight B subscript straight o$ ; E_o is the magnitude of electric field and is the magnitude of magnetic field.

Question 153

Wired Faculty App

Two infinitely long straight parallel wires, '1' and '2', carrying steady currents I₁and I₂ in the same direction are separated by a distance d. Obtain the expression for the magnetic field due to the wire '1' acting on wire '2'. Hence find out, with the help of a suitable diagram, the magnitude and direction of this force per unit length on wire '2' due to wire '1'. How does the nature of this force changes if the currents are in opposite direction? Use this expression to define the S.I. unit of current.

Solution

Consider a straight conductor XY lying in the plane of paper. Consider a point P at a perpendicular distance from the straight conductor.

Magnetic field induction (B) at a point P due to current I passing through conductor XY is given by,
$straight B space equals fraction numerator straight mu subscript straight o space straight I over denominator 4 πa end fraction open square brackets sin space straight ϕ subscript 1 space plus space sin space straight ϕ subscript 2 close square brackets$
where, $straight ϕ subscript 1 space and space straight ϕ subscript 2$ are the angles made by point X and Y respectively.
At the centre of the infinite long wire, $straight ϕ$ ₁= $straight ϕ$ ₂ = 90^o

So, magnetic field is given by,
$straight B space equals space fraction numerator straight mu subscript straight o over denominator 4 straight pi end fraction space fraction numerator 2 straight I over denominator straight a end fraction$
Magnetic field produced by current I₁ at any point on conductor Rs is given by,
$straight B subscript 1 space equals space fraction numerator μo space straight I space over denominator 2 πd end fraction$

Force acting on length l of the conductor RS will be,
$Error converting from MathML to accessible text.$

An equal force is exerted into the wire PQ by the field of conductor RS which is given by,
$Error converting from MathML to accessible text.$

Thus, the force is attractive when the current is acting along the same direction. When, current flows in opposite direction, the forces between the two conductors are repulsive.

One Ampere is that value of constant current which when flowing through each of the two parallel uniform long conductors placed in free space at a distance of 1m from each other will attract or repel with a force of 2 10^-7 Newton per metre of their length.

Question 154

Wired Faculty App

Using the concept of force between two infinitely long parallel current carrying conductors, define one ampere of current.

Solution

One ampere is the value of steady current which when maintained in each of the two very long, straight, parallel conductors of negligible cross section and placed one metre apart in vacuum, would produce on each of these conductors a force equal of 2 x 10^-7 N/m of its length.

Question 155

Wired Faculty App

Two equal balls having equal positive charge ‘q’ coulombs are suspended by two insulating strings of equal length. What would be the effect on the force when a plastic sheet is inserted between the two?

Solution

On inserting the plastic sheets between the two balls the force will decrease.

Question 156

Wired Faculty App

Considering the case of a parallel plate capacitor being charged, show how one is required to generalize Ampere’s circuital law to include the term due to displacement current.

Solution

Ampere’s circuital law states that,
$contour integral B with rightwards harpoon with barb upwards on top space. space stack d l with rightwards harpoon with barb upwards on top space equals space mu subscript o space I$

Electric flux across a parallel capacitor is given by,
$straight ϕ subscript straight E space space equals space EA space equals space 1 over straight epsilon subscript straight o straight Q over straight A xA space equals space straight Q over straight epsilon subscript straight o$

Current in the plates of the capacitor is given by,
$i space equals space fraction numerator d Q over denominator d t end fraction italic therefore italic space fraction numerator d ϕ subscript E over denominator d t end fraction italic space italic equals italic space fraction numerator d over denominator d t end fraction open parentheses Q over epsilon subscript o close parentheses italic space italic equals italic space italic 1 over epsilon subscript o fraction numerator d Q over denominator d t end fraction italic rightwards double arrow italic space fraction numerator d ϕ subscript E over denominator d t end fraction italic space italic equals italic space fraction numerator d Q over denominator d t end fraction italic equals i$
This is the missing term in the ampere's law.

Therefore, total current in the conductor is the sum of displacement current and conduction current.
$straight i space equals space straight i subscript straight c space plus space straight i subscript straight d space equals space space straight i subscript straight c space plus space straight epsilon subscript straight o space dϕ over dt Putting space the space value space of space straight I space in space Amper apostrophe straight s space Circuital space law comma space we space have contour integral straight B with rightwards harpoon with barb upwards on top. dl with rightwards harpoon with barb upwards on top space equals space straight mu subscript straight o space straight I subscript straight c space plus space straight mu subscript straight o element of subscript straight o space dϕ subscript straight E over dt This space is space the space required space generalized space form space of space Ampere ’ straight s space law.$

Question 157

Wired Faculty App

(a) Deduce an expression for the frequency of revolution of a charged particle in a magnetic field and show that it is independent of velocity or energy of the particle.

(b) Draw a schematic sketch of a cyclotron. Explain, giving the essential details of its construction, how it is used to accelerate the charged particles.

Solution

a)
Let the velocity of the positive ion having charge ‘q’ be ‘v’.

Then,
$q v B space italic equals space fraction numerator m v to the power of italic 2 over denominator r end fraction italic rightwards double arrow space r space italic equals space fraction numerator m v over denominator q B end fraction italic semicolon italic space m italic space i s italic space t h e italic space m a s s italic space o f italic space i o n italic comma italic space r italic space t h e italic space r a d i u s italic space o f italic space t h e italic space p a t h italic space o f italic space i o n i n italic space t h e italic space d e e italic space a n d italic space B italic space i s italic space t h e italic space s t r e n g t h italic space o f italic space t h e italic space m a g n e t i c italic space f i e l d italic. A n g u l a r italic space v e l o c i t y italic space o f italic space t h e italic space i o n italic space i s italic space g i v e n italic space b y italic comma italic space omega italic equals v over r italic equals fraction numerator q B over denominator m end fraction$

Now, time taken by the ion in describing a semi-circle i.e., turning
through an angle is given by,
$t space equals space pi over omega equals fraction numerator pi m over denominator q B end fraction$

The applied alternating potential should also have the same semi-periodic time (T/2) as that taken by the ion to cross either Dee.

That is,
$straight T over 2 equals straight t straight space equals straight space πm over qB$ ; is the expression for period of revolution.

So, Frequency of revolution of particle is given by,
$f space equals space 1 over T equals space fraction numerator q B over denominator 2 pi m end fraction$

This frequency is called the cyclotron frequency which is independent of the speed of the particle.
b)

Schematic sketch of a cyclotron is as shown below:

Principle: When a charged particle is kept in a magnetic field it experiences a force and the perpendicular magnetic field causes the particle to rotate.

Construction: The cyclotron is made up of two hollow semi-circular disc like metal containers, D1 and D2, called dees.

It uses crossed electric and magnetic fields. The electric field is provided by an oscillator of adjustable frequency.

Working: A Dee which is at a negative potential accelerates the positive ions which are produced from the source S. Magnetic field which is perpendicular will move the ions in a circular motion inside the Dees. The magnetic field and the frequency of the applied voltages are so chosen that as the ion comes out of a Dee, the Dees change its polarity (positive becoming negative and vice-versa) and the ion is further accelerated. Now, the ions move with higher velocity along a circular path of greater radius. The phenomenon is continued till the ion reaches at the periphery of the Dees where an auxiliary negative electrode (deflecting plate) deflects the accelerated ion on the target to be bombarded.

Question 158

Wired Faculty App

(a) Draw a labelled diagram of a moving coil galvanometer. Describe briefly its principle and working.

(b) Answer the following:

(i) Why is it necessary to introduce a cylindrical soft iron core inside the coil of a galvanometer?

(ii) Increasing the current sensitivity of a galvanometer may not necessarily increase its voltage sensitivity. Explain, giving reason.

Solution

b) Moving coil galvanometer:

Principle: The underlying principle of moving coil galvanometer is that a current carrying coil, placed in a uniform magnetic field, experiences torque.

Consider a rectangular coil for which no. of turns = N

Are of cross-section is A = lb

Intensity of the uniform magnetic field = B

Current through the coil = I

Therefore,

Deflecting torque is given by,

BIL x b = BIA

For N number of turns,

$straight tau$ = NBIA

Restoring torque in the spring = k $straight theta$

Therefore,
$NBIA space equals space kθ This space implies comma space straight I thin space equals space open parentheses straight k over NBA close parentheses straight theta That space is comma space straight I space proportional to space straight theta$

b) i) The soft iron coil in a galvanometer will make the field radial. Also, it increases the strength of the magnetic field.
ii) Current sensitivity in the galvanometer is given by, $straight theta over straight I space equals space fraction numerator N B A over denominator k end fraction$
Voltage sensitivity in the galvanometer is given by,
$straight theta over straight V space equals space fraction numerator theta over denominator I R end fraction space equals space open parentheses fraction numerator n B A over denominator k end fraction close parentheses. begin inline style 1 over R end style$
The above two equations imply that increasing the current sensitivity may not necessarily increase the voltage sensitivity.

Question 159

Wired Faculty App

What are the directions of electric and magnetic field vectors relative to each other and relative to the direction of propagation of electromagnetic waves?

Solution

Both electric and magnetic field vectors are perpendicular to each other and perpendicular to the direction of propagation.

Question 160

Wired Faculty App

A circular coil of N turns and radius R carries a current I. It is unwound and rewound to make another coil of radius R/2, current I remaining the same. Calculate the ratio of the magnetic moments of the new coil and the original coil.

Solution

Length of the wire remains same,
$straight N subscript 1 space straight x space 2 πR space equals space straight N subscript 2 space straight x space space 2 straight pi space straight R over 2 therefore space straight N subscript 2 space equals space 2 space straight N$

Magnetic moment of a coil, m = NAI

For the coil of radius R, magnetic moment,
$straight m subscript 1 space equals space straight N subscript 1 IA subscript 1 space equals space straight N subscript 1 Iπ space straight R squared$
Magnetic moment for coil of radius R/2,
$straight m subscript 2 space equals space straight N subscript 2 space IA subscript 2 space equals space fraction numerator 2 straight N subscript 1 IπR squared over denominator 4 end fraction equals fraction numerator straight N subscript 1 IπR squared over denominator 2 end fraction$
Therefore,
$straight m subscript 2 over straight m subscript 1 space equals space 1 colon thin space 2$

Question 161

Wired Faculty App

(a)

Write the expression for the force $straight F with rightwards harpoon with barb upwards on top$ , acting on a charged particle of charge ‘q’, moving with a velocity in the presence of both electric field E and magnetic field B. Obtain the condition under which the particle moves un deflected through the fields.

(b)

A rectangular loop of size l × b carrying a steady current I is placed in a uniform magnetic field. Prove that the torque $straight tau$ acting on the loop is given by $straight tau with rightwards harpoon with barb upwards on top straight space equals straight space stack straight m straight space with rightwards harpoon with barb upwards on top cross times straight space straight B with rightwards harpoon with barb upwards on top$ is the magnetic moment of the loop.

Solution

a) Electric field on the particle is given by,
$stack F subscript e with rightwards harpoon with barb upwards on top space equals space q E with rightwards harpoon with barb upwards on top$
Magnetic force on the particle is given by,

Total force will be,

When the charged particle moves perpendicular to both electric and magnetic field, F = 0

b) Consider a loop PQRS of length l, breadth b suspended in a uniform magnetic field.

Length of the loop, PQ= RS = l

Breadth of the loop, QR = SP = b
At any instance, plane of loop makes angle with the direction of magnetic field B

Suppose that the forces on sides PQ, QR, RS and SP are $stack F subscript 1 with rightwards harpoon with barb upwards on top comma space stack F subscript 2 with rightwards harpoon with barb upwards on top comma space stack F subscript 3 with rightwards harpoon with barb upwards on top comma space a n d space stack F subscript 4 with rightwards harpoon with barb upwards on top$ respectively.

The resultant of forces F₂ and F₄ is zero.

The side PQ and RS of current loop are perpendicular to the magnetic field, therefore the magnitude of each of forces F₁ and F₃ is,
F = I l B sin 90^o= I l B
Moment of couple or torque is given by,
$straight tau space equals space$ (Magnitude of one Force F) x perpendicular distance = (BIl). (b sin $straight theta$ ) = I (lb) B sin $straight theta$

But, lb is the area of the loop = A.

In vector form,

Magnetic dipole moment of rectangular current loop is given by, M = NIA

Direction of torque is perpendicular to the direction of magnetic field.

Question 162

Wired Faculty App

(a) Explain, giving reasons, the basic difference in converting a galvanometer into (i) a voltmeter and (ii) an ammeter.
(b) Two long straight parallel conductors carrying steady current I₁and I₂ are separated by a distance ‘d’. Explain briefly, with the help of a suitable diagram, how the magnetic field due to one conductor acts on the other. Hence deduce the expression for the force acting between the conductors. Mention the nature of this force.

Solution

a)
Conversion of galvanometer into ammeter:

A galvanometer may be converted into ammeter by using very small resistance in parallel with the galvanometer coil.
The small resistance connected in parallel is called a shunt. If G is resistance of galvanometer, I_g is current in galvanometer for full-scale deflection, then for conversion of galvanometer into ammeter of range I ampere, the shunt is given by,

Conversion of galvanometer into voltmeter,

A galvanometer may be converted into voltmeter by connecting high resistance (R) in series with the coil of galvanometer.
If V volt is the range of voltmeter formed, then series resistance becomes,

b)
Magnetic field produced by current I₁ at any point on conductor Rs is given by,

Force acting on length l of the conductor RS will be,

An equal force is exerted into the wire PQ by the field of conductor RS which is given by,

Thus, the force is attractive when the current is acting along the same direction.
When, current flows in opposite direction, the forces between the two conductors are repulsive.

Question 163

Wired Faculty App

Write the expression for Lorentz magnetic force on a particle of charge ‘q’ moving with velocity v in a magnetic field B . Show that no work is done by this force on the charged particle.

OR
A steady current (I₁) flows through a long straight wire. Another wire carrying steady current (I₂) in the same direction is kept close and parallel to the first wire. Show with the help of a diagram how the magnetic field due to the current I₁ exerts a magnetic force on the second wire. Write the expression for this force.

Solution

Lorentz magnetic force is given by,

Therefore, work done is,

As,

So, Work done, W = 0
OR

Consider two long straight conductors PQ and RS placed parallel to each other carrying currents I₁ and I₂ respectively. Conductors experience an attractive force when the conductors move in the same direction while, repulsive force is experienced by the conductors when currents move in opposite direction.

PQ and RS which are placed at a distance r, carry currents I₁ and I₂ in the same direction.
Suppose, a current element ‘ab’ of length of wire RS.

The magnetic field produced by current-carrying conductor PQ at the location of other wire RS,

So, total force on the conductor of length L is given by,

Force acting per unit length of conductor is given by,

Question 164

Wired Faculty App

(a) Using Ampere’s circuital law, obtain the expression for the magnetic field due to a long solenoid at a point inside the solenoid on its axis.

(b) In what respect is a toroid different from a solenoid? Draw and compare the pattern of the magnetic field lines in the two cases.

(c) How is the magnetic field inside a given solenoid made strong?

Solution

a) Consider a symmetrical long solenoid having number of turns per unit length equal to n.

Let I be the current flowing in the solenoid, then by right hand rule, the magnetic field is parallel to the axis of the solenoid.

Field inside the solenoid:
Consider a closed path abcd.

Now, using Ampere’s circuital law to this path, we have

Therefore, B = 0.

This implies, magnetic field outside the solenoid is 0.
Field inside the solenoid:
Consider a closed path pqrs.
The line integral of magnetic field is given by,

For path pq, $B with rightwards harpoon with barb upwards on top space$ and $Error converting from MathML to accessible text.$ are along the same direction,

For path rs, B = 0 because outside the solenoid field is zero.

Using these equations, equation (i) gives,

Now, using Ampere’s law,
$contour integral stack B space with rightwards harpoon with barb upwards on top. space stack d l with rightwards harpoon with barb upwards on top space equals space mu subscript o space I$

This implies,

b)
Magnetic lines do not exist outside the body of a toroid. Toroid is closed and solenoid is open on both sides. Magnetic field is uniform inside a toroid whereas, for a solenoid it is different at two ends and centre.

c) The magnetic field is made strong by,

i) passing large current and

ii) using laminated coil of soft iron.

Question 165

Wired Faculty App

Write any two characteristic properties of nuclear force.

Solution

(i) Nuclear forces are short range attractive forces.

(ii) Nuclear forces are charge – independent.

Question 166

Wired Faculty App

(a) State the principle of the working of a moving coil galvanometer, giving its labelled diagram.

(b) “Increasing the current sensitivity of a galvanometer may not necessarily increase its voltage sensitivity.” Justify this statement.

(c) Outline the necessary steps to convert a galvanometer of resistance R_G into an ammeter of a given range.

Solution

a) Principle of moving galvanometer:

A current carrying coil when kept inside a uniform magnetic field, can experience a torque.

When current (I) is passed in the coil, torque acts on the coil given by,
$straight tau space equals space NIAB space sin space straight theta$

where q is the angle between the normal to plane of coil and the magnetic field of strength B,

N is the number of turns in a coil.

When the magnetic field is radial, as in the case of cylindrical pole pieces and soft iron core, then in every position of coil the plane of the coil, is parallel to the magnetic field lines, so that

$straight theta$ = 90° and sin 90° =1

Deflecting Torque is given by,
$straight tau space equals space NIAB$

If C is the torsional rigidity of the wire and q is the twist of suspension strip, then restoring torque =C

For equilibrium, deflecting torque = restoring torque,
$Error converting from MathML to accessible text.$

That is, deflection of coil is directly proportional to current flowing in the coil.

b)

Clearly, the voltage sensitivity is dependent on current sensitivity and the resistance of the galvanometer. If we increase current sensitivity and resistance G is larger, then it is not certain that voltage sensitivity will be increased. Thus, the increase of current sensitivity does not imply the increase of voltage sensitivity.
c)
Conversion of galvanometer into ammeter:

Inorder to convert galvanometer into ammeter a shunt resistance i.e., a small resistance in parallel is connected across the coil of galvanometer.

Let G be the resistance of galvanometer and Ig the current required for full scale deflection.

Suppose this galvanometer is to converted into ammeter of range I ampere and the value of shunt required is S.

If, I_s is the current in the shunt, then
$space space space space space straight I space equals space straight I subscript straight g space plus space straight I subscript straight S space rightwards double arrow space straight I subscript straight S space equals space left parenthesis straight I space minus space straight I subscript straight g right parenthesis Also space potential space difference space across space straight A space and space straight B comma straight V subscript AB space equals space straight I subscript straight S. straight S space equals space straight I subscript straight g. straight G Substituting space value space of space straight I subscript straight S space from space left parenthesis straight i right parenthesis comma space we space get rightwards double arrow space left parenthesis straight I space minus space straight I subscript straight g right parenthesis straight S space equals space space straight I subscript straight g space straight G rightwards double arrow space space IS space minus space straight I subscript straight g straight S space equals space straight I subscript straight g straight G rightwards double arrow IS space equals space space straight I subscript straight g space left parenthesis straight S plus straight G right parenthesis rightwards double arrow space space straight I subscript straight g space equals fraction numerator straight S over denominator straight S space plus space straight G end fraction straight I straight i. straight e. comma space shunt space required comma space straight S space equals space fraction numerator GI subscript straight g over denominator straight I space minus space straight I subscript straight g end fraction$

which is the required expression for conversion of galvanometer into ammeter.

Question 167

Wired Faculty App

Write the underlying principle of a moving coil galvanometer.

Solution

When a current-carrying coil is placed in a magnetic field, it experiences torque. This is the underlying principle of a moving coil galvanometer.

Question 168

Wired Faculty App

a) Write the expression for the magnetic force acting on a charged particle moving with velocity v in the presence of magnetic field B.

b) A neutron, an electron and an alpha particle, moving with equal velocities, enter a uniform magnetic field going into the plane of paper, as shown. Trace their paths in the field and justify your answer.

Solution

a)
Force acting on a charged particle q, which is moving with velocity v in a magnetic field B, is given by,

$straight F with rightwards harpoon with barb upwards on top space equals space q space left parenthesis v with rightwards harpoon with barb upwards on top space x space B with rightwards harpoon with barb upwards on top space right parenthesis$
The right-hand rule gives the direction of this force. The direction of the force is perpendicular to the plane containing velocity v and magnetic field B.
b)
A charged particle experiences a force when it enters the magnetic field. Due to the presence of magnetic field, the charged particle will move in a circular path. This is because the force is perpendicular to the velocity of the charged particle.
Radius of the circular path in which the charged particle is moving is given by,

$mv squared over straight r space equals space q v B space space space space space r space equals space fraction numerator m v over denominator q B end fraction B space a n d space v space a r e space c o n s t a n t. space W e space c a n space w r i t e comma space r space proportional to space m over q$
Since the neutron has no charge, it will move along a straight line.
The electron will follow a circular path which has a radius smaller than that of the alpha particle. This is because the mass to charge ratio of the alpha particle is more than that of the electron.
Therefore, the electron will move in the clockwise direction and the electron will move in the anticlockwise direction as per the Right Hand Rule.

Question 169

Wired Faculty App

Two long, straight, parallel conductors carry steady currents, I₁ and I₂ , separated by a distance . If the currents are flowing in the same direction, show how the magnetic field set up in one produces an attractive force on the other? Obtain the expression for this force. Hence, define one ampere.

Solution

Magnetic field induction at some point P on wire 2 due to current I₁ passing through wire 1 is given by,

straight B subscript 1 space equals space fraction numerator straight mu subscript straight o space 2 space straight I subscript 1 over denominator 4 space straight pi space straight d end fraction

Magnetic field is produced by wire 1 and current carrying wire 2 lies in magnetic field B₁.

The unit length of wire 2 will experience a force, given by

F₂= B₁ I₂ x 1 =

fraction numerator straight mu subscript straight o over denominator 4 straight pi end fraction space fraction numerator 2 space I subscript 1 space I subscript 2 over denominator d end fraction

According to Fleming's left-hand rule, the force on wire 2 acts in the plane of paper perpendicular to wire 2, directed towards wire 1. Similarly, wire 1 also experiences the same force towards wire 2. Thus, both the conducting wires attract each other with the same force F.
One ampere can be defined as the amount of current flowing through two parallel conductors, which are in the same direction or opposite directions, placed at a distance of one metre in free space, and both the wires attract or repel each other with a force of 2 x 10^-7 per metre of their lengths.

Question 170

Wired Faculty App

Use Biot-Savart law to derive the expression for the magnetic field on the axis of a current carrying circular loop of radius R.

Draw the magnetic field lines due to a circular wire carrying current I.

Solution

Here,
I = current in the loop
R= radius of the loop
X= distance between O and P
dl = conducting element of the loop

According to the Biot-Savart law,
Magnetic field at point P is,

dB space equals space fraction numerator straight mu subscript straight o over denominator 4 straight pi end fraction space fraction numerator straight I space vertical line dl space straight x space straight r vertical line over denominator straight r cubed end fraction straight r squared space equals space space straight x squared space plus space straight R squared Since space dl space and space straight r space is space perpendicular comma space vertical line dl space straight x space straight r vertical line space equals space straight r space dl Therefore comma dB space equals space fraction numerator straight mu subscript straight o over denominator 4 straight pi end fraction. space fraction numerator straight I. space dl over denominator left parenthesis straight x squared plus straight R squared right parenthesis end fraction

dB has two components: dB_x and

dB subscript perpendicular

dB subscript perpendicular

is cancelled out and the x-component remains.
Therefore,

dB subscript straight x space equals space dB space cos space straight theta cos space straight theta space equals space fraction numerator straight R over denominator left parenthesis straight x squared plus straight R squared right parenthesis to the power of bevelled 1 half end exponent end fraction therefore space dB subscript straight x space equals space fraction numerator straight mu subscript straight o straight I space dl over denominator 4 straight pi end fraction. space fraction numerator straight R over denominator left parenthesis straight x squared space plus space straight R squared right parenthesis to the power of bevelled 3 over 2 end exponent end fraction

Summation of dl over the loop is given by,
B =

straight B subscript straight x straight i with hat on top space equals space fraction numerator straight mu subscript straight o space straight I space straight R squared over denominator 2 space left parenthesis straight x squared plus straight R squared right parenthesis to the power of bevelled 3 over 2 end exponent end fraction space straight i with hat on top

Magnetic field lines due to a circular current carrying i is,

Question 171

Wired Faculty App

A long straight current carrying wire passes normally through the centre of the circular loop. If the current through the wire increases, will the be an induced emf in the loop? Justify

Solution

No, Magnetic flux does not change with the change of current.

Question 172

Wired Faculty App

Describe the working principle of a moving coil galvanometer. Why is it necessary to use (i) a radial magnetic field and (ii) a cylindrical soft iron core in a galvanometer? Write the expression for the current sensitivity of the galvanometer.

Can a galvanometer as such be used for measuring the current? Explain.

Solution

Moving coil galvanometer is an instrument used for the detection and measurement of small currents.

Principle: The working of moving coil galvanometer is that when a current carrying coil is placed in a varying magnetic field, it experiences torque.

Consider a rectangular coil for which no. of turns = N

Are of cross-section is A = lb

Intensity of the uniform magnetic field = B

Current through the coil = I

Therefore,

Deflecting torque is given by,

BIL x b = BIA

For N number of turns,

$straight tau$ = NBIA

Restoring torque in the spring = k $straight theta$

Therefore,
$NBIA space equals space kθ This space implies straight I space equals space open parentheses straight k over NBA close parentheses straight theta That space is space straight I proportional to straight theta$
i) The soft iron coil in a galvanometer will make the field radial. Also, it increases the strength of the magnetic field.
ii) Current sensitivity in the galvanometer is given by, $straight theta over straight I space equals space fraction numerator N B A over denominator k end fraction$
Voltage sensitivity in the galvanometer is given by,
$straight theta over straight V space equals space fraction numerator theta over denominator I R end fraction space equals space open parentheses fraction numerator n B A over denominator k end fraction close parentheses. begin inline style 1 over R end style$
Yes a galvanometer can be used for measuring the current.

Question 173

Wired Faculty App

(a) Define the term 'self-inductance' and write its S.I. unit.

(b) Obtain the expression for the mutual inductance of two long co-axial solenoids S₁ and S₂ wound one over the other, each of length L and radii r₁ and r₂ and n₁ and n₂ number of turns per unit length when a current I is set up in the outer solenoid S₂.

Solution

a) Mutual inductance of two coils is equal to the e.m.f induced in one coil when the rate of change of current through the other coil is unity.

SI unit of mutual inductance is Henry.

b) Consider two long solenoids S₁ and S₂ of same length ‘l’ such that S₂ surrounds S₁completely.

Let,
n₁ = Number of turns per unit length of S₁

n₂ = Number of turns per unit length of S₂

I₁ = Current passing through solenoid S₁

$straight empty set subscript 21$ = Flux linked with S2 due to current flowing in S1
$straight ϕ subscript 21 straight space proportional to straight space straight I subscript 1$
$rightwards double arrow space space straight ϕ subscript 21 space equals space straight M subscript 21 straight I subscript 1 space semicolon space straight M subscript 21$ is the coefficient of the mutual inductance of two solenoids.

When current is passed through S₁, emf is induced in S₂ .

Magnetic field inside solenoid S₁ is given by, $straight B subscript 1 space equals space straight mu subscript straight o straight n subscript 1 straight I subscript 1$

Magnetic flux linked with each turn of the solenoid = $straight S subscript 2 space equals space straight B subscript 1 straight A$

Total magnetic flux linked with S₂ is given by,
$straight ϕ subscript 21 space equals space straight B subscript 1 straight A space cross times space straight n subscript 2 straight I space equals space straight mu subscript straight o straight n subscript 1 straight I subscript 1 cross times space straight A cross times space straight n subscript 2 straight I rightwards double arrow space straight space straight ϕ subscript 21 space equals space space straight mu subscript straight o straight n subscript 1 space end subscript straight n subscript 2 AI subscript 1 therefore space space space space space space space space space space space space straight M subscript 21 space equals space straight mu subscript straight o straight n subscript 1 space end subscript straight n subscript 2 AI$

Similarly, mutual inductance between two solenoids, when current is passed through S₂ and emf induced in solenoid S₁ is given by,
$straight M subscript 12 equals straight mu subscript straight o straight n subscript 1 straight n subscript 2 AI therefore space space space straight M subscript 12 space equals space straight M subscript 21 equals straight M$

Hence, coefficient of mutual induction between the two long solenoids is given by,
$straight M equals straight mu subscript straight o straight n subscript 1 straight n subscript 2 AI$

Question 174

Wired Faculty App

A bar magnet of magnetic moment 6 J/T is aligned at 60° with a uniform external magnetic field of 0.44 T. Calculate (a) the work done in turning the magnet to align its magnetic moment (i) normal to the magnetic field, (ii) opposite to the magnetic field, and (b) the torque on the magnet in the final orientation in case (ii).

Solution

M = 6J/T

θ = 60^o

B = 0.44T

$τ = mB sinθ τ = 6 x 0.44 \sin 60^{o} = 6 x 0.44 x \frac{\sqrt{3}}{2} = 3 \sqrt{3} x 0.44 = 2.836 dw \int_{θ = 60^{0}}^{90^{0}} τ . dθ = - mB [\cos 90^{0} - \cos 60^{0}] = 6 x 0.44 x [- \frac{1}{2}] = 3 x 0.44 = 1.32 J (ii) dw = \int_{θ = 60^{0}}^{180^{0}} mB \sin θ . dθ = - mB [\cos 180^{0} - \cos 60^{0}] = 6 x 0.44 x [- 1 - \frac{1}{2}] = - 6 x 0.44 [- \frac{3}{2}] = 9 x 0.44 = 39.6 J (b) \vec{τ} = \vec{m} x \vec{B} τ = m Bsin θ = 6 x 0.44 x \sin 180^{0} τ = 0$

Question 175

Wired Faculty App

An iron ring of relative permeability μ_r has windings of insulated copper wire of n turns per metre. When the current in the windings is I, find the expression for the magnetic field in the ring.

Solution

There will be two magnetic field one due to current and another due to magnetics.

$B = B_{0} + B_{m} B = μ_{0} nI + μ_{0} M B = μ_{0} H + μ_{0} χH B = μ_{0} (1 + χ) H B = μ_{0} μ_{r} H$

Question 176

Wired Faculty App

Two identical wires A and B, each of length ‘l’, carry the same current I. Wire A is bent into a circle of radius R and wire B is bent to form a square of side ‘a’. If B_A and B_B are the values of the magnetic field at the centres of the circle and square respectively, then the ratio B_A /B_B is:

$straight pi squared over 8$
$fraction numerator straight pi squared over denominator 16 square root of 2 end fraction$

$fraction numerator straight pi squared over denominator 8 square root of 2 end fraction$

Solution

Magnetic field in case of circle of radius R, we have

$straight B subscript straight A space equals space fraction numerator straight mu subscript straight o straight I over denominator 2 straight R end fraction As comma space 2 πR space equals space straight l space left parenthesis straight l space is space length space of space straight a space wire right parenthesis straight R space equals fraction numerator straight l over denominator 2 straight pi end fraction straight B subscript straight A space equals space fraction numerator straight mu subscript straight o straight I over denominator 2 space straight x space begin display style fraction numerator straight l over denominator 2 straight pi end fraction end style end fraction space equals space fraction numerator straight mu subscript straight o Iπ over denominator straight l end fraction space... space left parenthesis straight i right parenthesis$

Magnetic field in case of square of side we get

Question 177

Wired Faculty App

Hysteresis loops for two magnetic materials A and B are given below:

These materials are used to make magnets for electric generators, transformer core and electromagnet core. Then it is proper to use:

A for electric generators and transformers.

A for electromagnets and B for electric generators

A for transformers and B for electric generators.

B for electromagnets and transformers.

Solution

B for electromagnets and transformers.

Area of the hysteresis loop is proportional to the net energy absorbed per unit volume by the material, as it is taken over a complete cycle of magnetisation.
For electromagnets and transformers, energy loss should be low.
i.e thin hysteresis curves.
Also |B|→0 When H = 0 and |H| should be small when B →0.

Question 178

Wired Faculty App

An arc lamp requires a direct current of 10 A at 80 V to function. If it is connected to a 220 V (rms), 50 Hz AC supply, the series inductor needed for it to work is close to:

80 H

0.08 H

0.044 H

0.065 H

Solution

0.065 H

I = 10 A, V = 80 V
R = V/I = 80/10 = 8Ω and ω = 50 Hz
For AC circuit, we have

$straight I space equals space fraction numerator straight V over denominator square root of 8 squared plus straight X subscript straight L superscript 2 end root end fraction rightwards double arrow space 10 space equals space fraction numerator 220 over denominator square root of 64 plus straight X subscript straight L superscript 2 end root end fraction rightwards double arrow space square root of 64 space plus straight X subscript straight L superscript 2 end root space equals space 22 squaring space on space both space sides comma space we space get 64 space plus space straight X subscript straight L superscript 2 space equals space 484 straight X subscript straight L superscript 2 space equals space 484 minus 64 space equals space 420 straight X subscript straight L space equals space square root of 420 2 straight pi space straight x space ωL space equals space square root of 420 Series space induction space on space an space arc space lamp. straight L space equals space fraction numerator square root of 420 over denominator left parenthesis 2 straight pi space straight x space 50 right parenthesis end fraction space equals space 0.065 space straight H$

Question 179

Wired Faculty App

A galvanometer having a coil resistance of 100 Ω gives a full-scale deflection, when a current of 1 mA is passed through it. The value of the resistance, which can convert this galvanometer into ammeter giving a full-scale deflection for a current of 10 A, is:

0.01 Ω

2 Ω

0.1 Ω

3 Ω

Solution

0.01 Ω

Maximum voltage that can be applied across the galvanometer coil = 100 Ω x 10^-3 A = 0.1

If R_s is the shunt resistance, then
R_s x 10 A = 0.1 V
R_s = 0.01 Ω

Question 180

Wired Faculty App

Two coaxial solenoids of different radii carry current I in the same direction. Let $stack straight F subscript 1 with rightwards arrow on top$ be the magnetic force on the inner solenoid due to the outer one and $stack straight F subscript 2 with rightwards arrow on top$ be the magnetic force on the outer solenoid due to the inner one. Then:

$stack straight F subscript 1 space with rightwards arrow on top equals stack straight F subscript 2 space with rightwards arrow on top equals space 0$
$stack straight F subscript 1 with rightwards arrow on top$ is radially inwards and $stack straight F subscript 2 with rightwards arrow on top$ is radially outwards
$stack straight F subscript 1 with rightwards arrow on top$ is radially inwards and $stack straight F subscript 2 with rightwards arrow on top$ =0
$stack straight F subscript 1 with rightwards arrow on top$ is radially outward and $stack straight F subscript 2 with rightwards arrow on top$ =0

Solution

stack straight F subscript 1 space with rightwards arrow on top equals stack straight F subscript 2 space with rightwards arrow on top equals space 0

Consider the two coaxial solenoids. Due to one of the solenoids magnetic field at the centre of the other can be assumed to be constant

Due to symmetry, forces on upper and lower part of the solenoid will be equal and opposite and hence resultant is zero.
Therefore option (a) is correct.

Question 181

Wired Faculty App

A rectangular loop of sides 10 cm and 5 cm carrying a current I of 12 A is placed in different orientations as shown in the figures below:

If there is a uniform magnetic field of 0.3 T in the positive z direction , in which orientations the loop would be in (i) stable equilibrium and (ii) unstable equilibrium?

(a) and (b) respectively

(a) and (c) respectively

(b) and (d) respectively

(b) and (c) respectively

Solution

(b) and (d) respectively

Since, B is uniform only torque acts on a current carrying loop
As, $straight tau space equals space bold M bold space bold x bold space bold B vertical line straight tau vertical line space equals space vertical line bold M vertical line vertical line bold B vertical line space sin space straight theta For space orientation space shown space in space left parenthesis straight b right parenthesis space straight theta space equals space 0 to the power of 0 comma space straight tau space equals space 0 space stable space equilibrium$

Question 182

Wired Faculty App

Two long parallel wires are at a distance 2d apart. They carry steady equal currents flowing out of the plane of the paper as shown. The variation of the magnetic field B along the line XX' is given by

Solution

The magnetic field in between because of each of the conductors will be in naturally opposite directions.
∴ Net magnetic field
$straight B subscript in space between space space end subscript space equals space fraction numerator straight mu subscript straight o straight i over denominator 2 space straight pi space straight x end fraction space bold j with bold hat on top space plus space fraction numerator straight mu subscript straight o straight i over denominator 2 space straight pi space left parenthesis 2 straight d minus straight x right parenthesis end fraction left parenthesis negative straight j right parenthesis space equals space fraction numerator straight mu subscript straight o straight i over denominator 2 space straight pi space straight x end fraction space open square brackets 1 over straight x minus fraction numerator 1 over denominator 2 straight d minus straight x end fraction close square brackets space left parenthesis space bold j with bold hat on top space right parenthesis$
At x = d, B in between = 0
For x < d, B in between = $bold j with bold hat on top$
For x> d, B in between = $left parenthesis bold minus bold j with bold hat on top right parenthesis$
Towards x, net magnetic field will add up and direction will be $left parenthesis bold minus bold j with bold hat on top right parenthesis$ . Towards' x' b net magnetic field will add up and direction will be ( $bold j with bold hat on top$ )

Question 183

Wired Faculty App

In a coil of resistance 100 Ω, a current is induced by changing the magnetic flux through it as shown in the figure. The magnitude of change in flux through the coil is

250 Wb

275 Wb

200 Wb

225 Wb

Solution

250 Wb

straight q space equals space fraction numerator increment straight ϕ over denominator straight R end fraction increment straight ϕ space equals space change space in space flux straight q space equals space integral Idt equals space Area space of space current space minus time space graph space equals 1 half space straight x space 10 space straight x space 0.5 space equals space 2.5 space coloumb straight q space equals fraction numerator increment straight ϕ over denominator straight R end fraction increment straight ϕ space equals space 2.5 space straight x space 100 space equals space 250 space space wb

Question 184

Wired Faculty App

A current loop ABCD is held fixed on the plane of the paper as shown in the figure. The arcs BC (radius = b) and DA (radius = a) of the loop are joined by two straight wires AB andCD. A steady current I is flowing in the loop. Angle made by AB and CD at the origin O is 30º. Another straight thin wire with steady current I1 flowing out of the plane of the paper is kept at the origin.

The magnitude of the magnetic field (2) due to the loop ABCD at the origin (O) is

zero

$fraction numerator straight mu space left parenthesis straight b minus straight a right parenthesis over denominator 24 ab end fraction$
$fraction numerator straight mu subscript straight o straight I over denominator 4 space straight pi end fraction space open square brackets fraction numerator straight b minus straight a over denominator ab end fraction close square brackets$
$fraction numerator straight mu subscript 0 straight I over denominator 4 straight pi end fraction space open square brackets 2 left parenthesis straight b minus straight a right parenthesis space plus straight pi over 3 left parenthesis straight a plus straight b right parenthesis close square brackets$

Solution

fraction numerator straight mu space left parenthesis straight b minus straight a right parenthesis over denominator 24 ab end fraction

straight B space equals space 1 over 12 space of space fraction numerator straight mu subscript 0 straight I over denominator 2 end fraction space open parentheses 1 over straight a minus 1 over straight b close parentheses space equals space fraction numerator straight mu subscript 0 straight I space left parenthesis straight b minus straight a right parenthesis over denominator 24 space ab end fraction

Question 185

Wired Faculty App

A current loop ABCD is held fixed on the plane of the paper as shown in the figure. The arcs BC (radius = b) and DA (radius = a) of the loop are joined by two straight wires AB andCD. A steady current I is flowing in the loop. Angle made by AB and CD at the origin O is 30º. Another straight thin wire with steady current I1 flowing out of the plane of the paper is kept at the origin.

Due to the presence of the current I₁ at the origin

The magnitude of the magnetic field (2) due to the loop ABCD at the origin (O) is

The forces on AB and DC are zero

The forces on AD and BC are zero

The magnitude of the net force on the loop is given by $fraction numerator straight mu subscript straight o II subscript 1 over denominator 4 straight pi end fraction space open square brackets 2 left parenthesis straight b minus straight a right parenthesis space plus space straight pi over 3 left parenthesis straight a plus straight b right parenthesis close square brackets$

The magnitude of the net force on the loop is given by $fraction numerator straight mu subscript 0 space II subscript 1 over denominator 24 space ab end fraction space left parenthesis straight b minus straight a right parenthesis$

Solution

The forces on AD and BC are zero

The forces on AD and BC are zero because magnetic field due to a straight wire on AD and BC is parallel to elementary length of the loop.

Question 186

Wired Faculty App

Two coaxial solenoids are made by winding thin insulated wire over a pipe of cross sectional area A = 10 cm2 and length = 20 cm. If one of the solenoids has 300 turns and the other 400 turns, their mutual inductance is (µ₀ = 4π × 10^-7 Tm A^-1)

2.4 π × 10^-5 H

4.8 π × 10^-4 H

4.8 π × 10^-5 H

2.4 π × 10^-4 H

Solution

2.4 π × 10^-4 H

fraction numerator straight mu subscript 0 straight N subscript 1 straight N subscript 2 straight A over denominator calligraphic l end fraction space equals space 2.4 straight pi space straight x space 10 to the power of negative 4 end exponent space straight H

Question 187

Wired Faculty App

A horizontal overhead power line is at a height of 4m from the ground and carries a current of 100 A from east to west. The magnetic field directly below it on the ground is (µ₀ = 4π × 10^-7 T m A^-1)

2.5 × 10^-7 T southward

5 × 10^-6 T northward

5 × 10^-6 T southward

2.5 × 10^-7 northward

Solution

5 × 10^-6 T southward

straight B space equals space fraction numerator straight mu subscript 0 space straight i over denominator 2 space πR end fraction space equals space fraction numerator 4 space straight pi space straight x space 10 to the power of negative 7 end exponent over denominator 2 space straight pi end fraction space straight x space 100 over 4 space equals space 5 space straight x space 10 to the power of 6 space straight T space southward

Question 188

Wired Faculty App

Relative permittivity and permeability of a material are ε_r and µ_r, respectively. Which of the following values of these quantities are allowed for a diamagnetic material?

εr = 0.5, µ_r = 1.5

ε_r = 1.5, µ_r = 0.5

ε_r = 0.5, µ_r = 0.5

ε_r = 1.5, µ_r = 1.5

Solution

ε_r = 1.5, µ_r = 0.5

Question 189

Wired Faculty App

The potential at a point x (measured in µm) due to some changes situated on the x-axis is given by V (x) = 20 /x²- 4) volts. The electric field E at x = 4 µm is given by

5/3 Volt/µm and in the –ve x direction

5/3 Volt/µm and in the +ve x direction

10/9 Volt/µm and in the -ve x direction

10/9 Volt/µm and in the +ve x direction

Solution

10/9 Volt/µm and in the +ve x direction

straight V subscript straight x space equals space fraction numerator 20 over denominator straight x squared minus 4 end fraction straight E space equals space minus space dV over dx space equals space fraction numerator 20 over denominator left parenthesis straight x squared minus 4 right parenthesis squared end fraction space left parenthesis 2 straight x minus 0 right parenthesis space equals space 160 over 144 space equals space 10 over 9

Question 190

Wired Faculty App

In a region, steady and uniform electric and magnetic fields are present. These two fields are parallel to each other. A charged particle is released from rest in this region. The path of the particle will be a

circle

helix

straight line

ellipse

Solution

straight line

Question 191

Wired Faculty App

In an AC generator, a coil with N turns, all of the same area A and total resistance R, rotates with frequency ω in a magnetic field B. The maximum value of emf generated in the coil is ‘

N.A.B.ω

N.A.B.R.ω

N.A.B

N.A.B.R

Solution

N.A.B.ω

Question 192

Wired Faculty App

A long solenoid has 200 turns per cm and carries a current i. The magnetic field at its centre is 6.28 × 10⁻² Weber/m². Another long solenoid has 100 turns per cm and it carries a current i/3. The value of the magnetic field at its centre is

1.05 × 10⁻⁴ Weber/m²

1.05 × 10⁻² Weber/m²

1.05 × 10⁻³ Weber/m²

1.05 × 10⁻¹ Weber/m²

Solution

1.05 × 10⁻² Weber/m²

straight B subscript 2 space equals space fraction numerator straight B subscript 1 straight n subscript 2 straight i subscript 2 over denominator straight n subscript 1 straight i subscript 1 end fraction space equals space fraction numerator left parenthesis 6.28 space straight x space 10 to the power of negative 12 end exponent right parenthesis left parenthesis 100 space straight x space straight i divided by 3 right parenthesis over denominator 200 left parenthesis straight i right parenthesis end fraction space equals space 1.05 space straight x space 10 to the power of negative 2 end exponent space straight W divided by straight m squared

Question 193

Wired Faculty App

A moving coil galvanometer has 150 equal divisions. Its current sensitivity is 10 divisions per milliampere and voltage sensitivity is 2 divisions per millivolt. In order that each division reads 1 volt, the resistance in ohms needed to be connected in series with the coil will be

10³

10⁵

99995

9995

Solution

9995

straight I subscript straight g space equals space 15 space mA straight V subscript straight g space equals space 75 space mV straight R space equals space straight V over straight I subscript straight g minus straight V subscript straight g over straight I subscript straight g

Question 194

Wired Faculty App

In the circuit, the galvanometer G shows zero deflection. If the batteries A and B have negligible internal resistance, the value of the resistor R will be

200Ω

100Ω

500 Ω

1000 Ω

Solution

100Ω

The potential drop across the resistance R is 2V. Now, no current flows through the galvanometer .and if V₁ is the potential drop across the resistor R and V is the total potential then
$fraction numerator VR over denominator 500 space plus straight R space end fraction space space equals space straight V subscript 1 fraction numerator 12 space straight R over denominator 500 space plus straight R space end fraction space equals space 2 space equals space 100 space ohm$

Question 195

Wired Faculty App

One conducting U tube can slide inside another as shown in figure, maintaining electrical contacts between the tubes. The magnetic field B is perpendicular to the plane of the figure. if each tube moves towards the other at a constant speed V, then the emf induced in the circuit in terms of B,

B

-B

zero

2B

Solution

2B $open vertical bar dϕ over dt close vertical bar space equals space 2 straight B calligraphic l v$

Question 196

Wired Faculty App

A uniform electric field and a uniform magnetic field are acting along the same direction in a certain region. If an electron is projected along the direction of the fields with a certain velocity then

its velocity will decrease.

its velocity will increase.

it will turn towards right of direction of motion.

it will turn towards left of direction of motion.

Solution

its velocity will decrease.

straight F with rightwards arrow on top space equals space minus straight e space open square brackets straight E with rightwards arrow on top space plus space straight v with rightwards arrow on top straight x space straight B with rightwards arrow on top close square brackets space equals negative straight e straight E with rightwards arrow on top straight a with rightwards arrow on top space equals space minus space fraction numerator straight e straight E with rightwards arrow on top over denominator straight m end fraction straight v left parenthesis straight t right parenthesis space equals space straight v subscript straight o space minus eE over straight m straight t

Question 197

Wired Faculty App

A charged particle of mass m and charge q travels on a circular path of radius r that is perpendicular to a magnetic field B. The time taken by the particle to complete one revolution is

$fraction numerator 2 πmq over denominator straight B end fraction$
$fraction numerator 2 πq squared straight B over denominator straight m end fraction$
$fraction numerator 2 πqB over denominator straight m end fraction$
$fraction numerator 2 πm over denominator qB end fraction$

Solution

fraction numerator 2 πm over denominator qB end fraction

mω squared straight r space equals Bqωr straight omega space equals space Bq divided by straight m straight T space equals fraction numerator 2 πm over denominator qB end fraction

Question 198

Wired Faculty App

In a potentiometer experiment, the balancing with a cell is at length 240 cm. On shunting, the cell with a resistance of 2Ω the balancing length becomes 120 cm. The internal resistance of the cell is

1 Ω

0.5 Ω

3 Ω

2Ω

Solution

2Ω

straight r space equals space straight R open square brackets calligraphic l subscript 1 over calligraphic l subscript 2 minus 1 close square brackets space space equals space 2 open square brackets 240 over 120 minus 1 close square brackets space equals space 2 straight capital omega

Question 199

Wired Faculty App

The resistance of hot tungsten filament is about 10 times the cold resistance. What will be the resistance of 100 W and 200 V lamp when not in use?

40 Ω

20 Ω

400 Ω

200 Ω

Solution

40 Ω

straight R subscript hot space equals space straight V squared over straight P space equals space fraction numerator 200 space straight x space 200 over denominator 100 end fraction space equals space 400 space straight capital omega Cold space resistance space equals space straight R subscript hot divided by 10 space equals space 400 divided by 10 space equals space 40 space straight capital omega space

Question 200

Wired Faculty App

A magnetic needle is kept in a non-uniform magnetic field. It experiences

a torque but not a force

neither a force nor a torque

a force and a torque.

a force but not a torque.

Solution

a force and a torque.

In non uniform magnetic field, dipole experiences both force and torque.

Question 201

Wired Faculty App

A current I ampere flows along an infinitely long straight thin-walled tube, then the magnetic induction at any point inside the tube is

infinite

zero

$fraction numerator straight mu subscript 0 over denominator 4 straight pi end fraction fraction numerator 2 straight i over denominator straight r end fraction space tesla$
$fraction numerator 2 space straight i over denominator straight r end fraction space Tesla$

Solution

zero

Let R be the radius of a long thin cylindrical shell. To calculate the magnetic induction at a distance r (r < R) from the axis of cylinder, a circular shell of radius r is shown:

Since no current is enclosed in the circle so, from Ampere's circuital law, magnetic induction is zero at every point of the circle. Hence, the magnetic induction at any point inside the infinitely long straight thin-walled tube (cylindrical) is zero.

Question 202

Wired Faculty App

A long wire carries a steady current. It is bent into a circle of one turn and the magnetic field at the centre of the coil is B. It is then bent into a circular loop of n turns. The magnetic field at the centre of the coil will be

nB

n²B

2nB

2n²B

Solution

n²B

The magnetic field at the centre of circular coil is
$straight B space equals fraction numerator straight mu subscript 0 space straight i over denominator 2 straight r end fraction where space straight r space equals space radius space of space circle space space equals space fraction numerator 1 over denominator 2 straight pi end fraction space left parenthesis because space straight I space equals 2 πr right parenthesis straight B space equals space fraction numerator straight mu subscript 0 space straight i space over denominator 2 end fraction space straight x space fraction numerator 2 straight pi over denominator straight I end fraction space equals space fraction numerator straight mu subscript 0 space straight i space straight pi over denominator straight I end fraction space.... space left parenthesis straight i right parenthesis$
When wire of length i bents into a circular loops of n turns, then
l = n × 2 π r
⇒ r = 1/ n x 2 π
Thus, new magnetic field
$straight B apostrophe space equals space fraction numerator straight mu subscript 0 ni over denominator 2 straight r apostrophe end fraction space equals space fraction numerator straight mu subscript 0 ni over denominator 2 end fraction space straight x space fraction numerator straight n space straight x space 2 straight pi over denominator straight l end fraction space equals space fraction numerator straight mu subscript 0 space straight i space straight pi over denominator straight l end fraction space straight x space straight n squared space equals space straight n squared straight B$

Question 203

Wired Faculty App

The magnetic field due to a current carrying circular loop of radius 3 cm at a point on the axis at a distance of 4 cm from the centre is 54 µT. What will be its value at the centre of the loop?

250 µT

150 µT

125 µT

75 µT

Solution

250 µT

Using space formula space straight B space equals fraction numerator straight mu subscript 0 space iR squared over denominator 2 left parenthesis straight R squared space plus straight X squared right parenthesis to the power of 3 divided by 2 end exponent end fraction comma we space get 54 space equals fraction numerator straight mu subscript 0 space straight i space left parenthesis 3 right parenthesis squared over denominator 2 left parenthesis straight R squared space plus straight X squared right parenthesis to the power of 3 divided by 2 end exponent end fraction At space the space centre space of space the space coil comma space straight X equals 0 space and space straight B equals fraction numerator straight mu subscript 0 space straight i over denominator 2 left parenthesis 3 right parenthesis end fraction using space equation space left parenthesis straight i right parenthesis straight B space equals fraction numerator 54 space straight x space 5 cubed over denominator left parenthesis 3 right parenthesis squared space straight x space 3 end fraction equals straight B space equals space 250 space μT

Question 204

Wired Faculty App

In a uniform magnetic field of induction B a wire in the form of semicircle of radius r rotates about the diameter of the circle with angular frequency ω. The axis of rotation is perpendicular to the field. If the total resistance of the circuit is R the mean power generated per period of rotation is

$fraction numerator Bπr squared space straight omega over denominator 2 straight R end fraction$
$fraction numerator left parenthesis Bπr squared straight omega right parenthesis squared over denominator 2 straight R end fraction$
$fraction numerator left parenthesis Bπrω squared right parenthesis over denominator 2 straight R end fraction$
$fraction numerator left parenthesis Bπrω squared right parenthesis squared over denominator 8 straight R end fraction$

Solution

fraction numerator left parenthesis Bπr squared straight omega right parenthesis squared over denominator 2 straight R end fraction

Magnetic space flux space equals space BA space cos space straight theta space equals space straight B. πr squared over 2 space cos space ωt therefore space straight P space equals straight epsilon squared subscript ind over straight R space equals space fraction numerator straight B squared straight pi squared straight r to the power of 4 straight omega squared space sin squared space ωt over denominator 4 straight R end fraction Now comma space less than sin squared space ωt greater than space equals space 1 divided by 2 space left parenthesis mean space value right parenthesis therefore space less than straight P greater than space equals space fraction numerator left parenthesis Bπr squared straight omega right parenthesis squared over denominator 8 straight R end fraction

Question 205

Wired Faculty App

A charged oil drop is suspended in a uniform field of 3 × 10⁴ V/m so that it neither falls nor rises. The charge on the drop will be (take the mass of the charge = 9.9 × 10⁻¹⁵ kg and g = 10 m/s² )

3.3 × 10⁻¹⁸ C

3.2 × 10⁻¹⁸ C

1.6 × 10⁻¹⁸ C

4.8 × 10⁻¹⁸ C

Solution

3.3 × 10⁻¹⁸ C

In steady state, electric force on drop = weight of drop
∴ qE = mg

$straight q space equals space mg over straight E space equals space fraction numerator 9.9 space space straight x space 10 to the power of negative 15 end exponent space straight x space 10 over denominator 3 space straight x space 10 to the power of 4 end fraction space equals space 3.3 space straight x space 10 to the power of negative 18 end exponent space straight C$

Question 206

Wired Faculty App

Two sources of equal emf are connected to an external resistance R. The internal resistances of the two sources are R₁ and R₂ (R₂>R₁). If the potential difference across the source having internal resistance R₂ is zero, then

$straight R space equals space fraction numerator straight R subscript 2 space straight x space left parenthesis straight R subscript 1 space plus straight R subscript 2 right parenthesis over denominator left parenthesis straight R subscript 2 minus straight R subscript 1 right parenthesis end fraction$

R = R₂-R₁

$straight R space equals space fraction numerator straight R subscript 1 straight R subscript 2 over denominator left parenthesis straight R subscript 1 plus straight R subscript 2 right parenthesis end fraction$
$straight R space equals space fraction numerator straight R subscript 1 straight R subscript 2 over denominator left parenthesis straight R subscript 1 minus straight R subscript 2 right parenthesis end fraction$

Solution

R = R₂-R₁

The equivalent resistance of the circuit
Req = R₁ + R₂ + R
I = 2E/ R₁ +R₂ +R
according to the questions
-(VB-VA) = E- IR₂
0 = E- IR₂E= IR₂
E = 2ER₂/R₁+R₂+R
R₁+R₂+R =2R₂
R= R₁-R₂

Question 207

Wired Faculty App

An electron, a proton and an alpha particle having the same kinetic energy are moving in circular orbits of radii r_e, r_p, r_∝ respectively in a uniform magnetic field B. The relation between r_e, r_p, r_∝ is:

r_e < r_∝ < r_p

r_e > r_p = r_∝

r_e < r_p = r_∝

r_e < r_p < r_∝

Solution

r_e < r_p = r_∝

The radius of circular path in a magnetic field is given as,

$r = \frac{\sqrt{2 K m}}{q B} F o r e l e c t r o n, r_{e} = \frac{\sqrt{2 K m_{e}}}{e B} . . . . (i) F o r P r o t o n, r_{p} = \frac{\sqrt{2 K m_{p}}}{e B} . . . . (i i) F o r α p a r t i c l e, r_{α} = \frac{\sqrt{2 K m_{a}}}{q_{α} B} = \frac{\sqrt{2 K 4 m_{p}}}{2 e B} = \frac{\sqrt{2 K m_{p}}}{e B} . . . (i i i) ∴ r_{e} < r_{p} = r_{α} (∵ m_{e} < m_{p})$

Question 208

Wired Faculty App

A long straight wire of radius a carries a steady current I. The current is uniformly distributed over its cross-section. The ratio of the magnetic fields B and B' at radial distances $straight a over 2$ and 2a respectively, from the axis of the wire is,

$1 half$

1

4

1/4

Solution

b) Consider two amperian loops of radius a/2 and 2a as shown in the diagram.

Applying Ampere's circuital law for these loops, we get

contour integral B. d L space equals space mu subscript o space I subscript e n c l o s e d end subscript

For the smaller loop,

rightwards double arrow space straight B space straight x space 2 straight pi straight a over 2 equals straight mu subscript straight o straight x straight I over πa squared xπ space open parentheses straight a over 2 close parentheses squared space space space space space space space space space space space space space space space space space space space space space equals space straight mu subscript straight o space straight I space straight x space 1 fourth space equals space fraction numerator straight mu subscript straight o space straight I over denominator 4 end fraction rightwards double arrow straight B subscript straight I space equals space fraction numerator straight mu subscript straight o straight I over denominator 4 πa end fraction

B₁ acts at a distance a/2 from the axis of the wire.
Similarly, for bigger amperian loop,
B' x

2 straight pi

(2a) =

straight mu subscript straight o straight I

Total current enclosed by amperian loop 2.

rightwards double arrow space straight B apostrophe space equals space fraction numerator straight mu subscript straight o straight I over denominator 4 πa end fraction

, at a distance 2a from the axis of the wire.
Therefore,
Ratio of

fraction numerator straight B over denominator straight B apostrophe end fraction space equals space fraction numerator mu subscript o I over denominator 4 pi a end fraction x fraction numerator 4 pi a over denominator mu subscript o I end fraction equals 1

Question 209

Wired Faculty App

A magnetic needle suspended parallel to a magnetic field requires $square root of 3$ J of work to turn it through 60^o. The torque needed to maintain the needle in this position will be

$2 square root of 3 space straight J$

3 J

$square root of 3 space J$
$3 over 2 space straight J$

Solution

3 J

In this case, work done
$straight W space equals space MB space left parenthesis cosθ subscript 1 minus Cosθ subscript 2 right parenthesis equals space MB space left parenthesis cos space 0 to the power of straight o minus cos space 60 to the power of straight o right parenthesis equals space MB space open parentheses 1 minus 1 half close parentheses space equals space open parentheses MB over 2 close parentheses MB space equals space 2 square root of 3 space straight J space space space space space space space left parenthesis therefore space given space straight W space equals space square root of 3 space straight J right parenthesis straight zeta space equals space MB space sin space 60 to the power of straight o space equals space left parenthesis 2 square root of 3 right parenthesis open parentheses fraction numerator square root of 3 over denominator 2 end fraction close parentheses space straight J space equals space 3 space straight J$

Question 210

Wired Faculty App

The ratio of amplitude of magnetic field to the amplitude of electric field for an electromagnetic wave propagating in vacuum is equal to

the speed of light in vacuum

the reciprocal of the speed of light in vacuum

the ratio of magnetic permeability to the electric susceptibility of vacuum

unity

Solution

the reciprocal of the speed of light in vacuum

as E=cB
so the required ratio
$straight E over straight B space equals straight c fraction numerator straight B over denominator straight E space end fraction space equals 1 over straight c$
Therefore, the ratio of the amplitude of magnetic field to the amplitude of electric field for an electromagnetic wave propagating in a vacuum is equal to the reciprocal of the speed of light.

Question 211

Wired Faculty App

Two similar coils of radius R are lying concentrically with their planes at right angles to each other. The currents flowing in them are I and 2I, respectively. The resultant magnetic field induction at the centre will be

$fraction numerator square root of 5 straight mu subscript straight o straight I over denominator 2 straight R end fraction$
$fraction numerator 3 straight mu subscript straight o straight I over denominator 2 straight R end fraction$
$fraction numerator straight mu subscript straight o straight I over denominator 2 straight R end fraction$
$fraction numerator straight mu subscript straight o straight I over denominator straight R end fraction$

Solution

fraction numerator square root of 5 straight mu subscript straight o straight I over denominator 2 straight R end fraction

The magnetic field (B) at the centre of circular current carrying coil of radius R and current I,
$straight B equals space fraction numerator straight mu subscript straight o straight I over denominator 2 straight R end fraction Similarly comma space if space current space equals space 2 straight I comma space then Magnetic space field space equals space fraction numerator straight mu subscript straight o 2 straight I over denominator 2 straight R end fraction space equals space 2 straight B so space resultant space magnetic space field equals space square root of straight B squared space plus left parenthesis 2 straight B right parenthesis squared end root space equals space square root of 5 straight B squared end root space square root of 5 straight B end root space equals space fraction numerator square root of 5 straight mu subscript straight o straight I over denominator 2 straight R end fraction$

Question 212

Wired Faculty App

An electric dipole of moment p is placed in an electric field of intensity E. The dipole acquires a position such that the axis of the dipole makes an angle θ with the direction of the field. Assuming that the potential energy of the dipole to be zero when θ =90^o, the torque and the potential energy of the dipole will respectively be

pE sin θ, pE cos θ

pE sin θ,-2pE cos θ

pE sin θ, 2 pE cos θ

pE cos θ,-pE sin θ

Solution

pE sin θ, pE cos θ

Here
Torque =pE sin θ

potential space energy space of space the space dipole straight U equals negative integral straight tau space dθ equals negative integral subscript straight pi divided by 2 end subscript superscript 0 space pE space sin space straight theta space dθ space equals space pE left square bracket cosθ minus 0 right square bracket equals negative pE space cosθ

Question 213

Wired Faculty App

A wire loop is rotated in a magnetic field. The frequency of change of direction of the induced emf is

one per revolution

twice per revolution

four times per revolution

six times per revolution

Solution

twice per revolution

Question 214

Wired Faculty App

The electric and the magnetic field, associated with an electromagnetic wave, propagating along the +z - axis, can be represented by

$left square bracket bold E space equals space straight E subscript straight o bold k with hat on top comma space bold B space equals space straight B subscript straight o space bold i with hat on top right square bracket$
$left square bracket bold E space equals space straight E subscript straight o bold j with hat on top comma space bold B space equals space straight B subscript straight o space bold j with hat on top right square bracket$
$left square bracket bold E space equals space straight E subscript straight o bold j with hat on top comma space bold B space equals space straight B subscript straight o space bold k with hat on top right square bracket$
$left square bracket bold E space equals space straight E subscript straight o bold i with hat on top comma space bold B space equals space straight B subscript straight o space bold j with hat on top right square bracket$

Solution

left square bracket bold E space equals space straight E subscript straight o bold i with hat on top comma space bold B space equals space straight B subscript straight o space bold j with hat on top right square bracket

straight mu space equals bold space bold E bold space bold X bold space bold B equals space bold E subscript bold o bold i with bold hat on top bold plus bold space bold B subscript bold o bold space bold j with bold hat on top bold space bold equals bold space straight E subscript straight o straight B subscript straight o bold k with bold hat on top bold E bold space bold X bold space bold B space points space in space the space direction space of space wave space propogatgion.

Question 215

Wired Faculty App

A proton and an alpha particle both enter a region of uniform magnetic field B, moving at angles to the field B. If the radius of circular orbits for both the particles is equal and the kinetic energy acquired by proton is 1 MeV, the energy acquired by the alpha particle will be

4 MeV

0.5 Mev

1.5 MeV

1 MeV

Solution

1 MeV

Radius in magnetic field of circular orbit,
$straight R space equals space mV over qB space equals space fraction numerator square root of 2 mE end root over denominator qB end fraction$
and total energy of a moving particle in a circular orbit, $straight E space equals fraction numerator space straight q squared straight B squared straight R squared over denominator 2 straight m end fraction space For space straight a space proton space enter space in space straight a space region space of space magnetic space field comma straight E subscript 1 space equals space fraction numerator straight e squared space straight x space straight B squared space straight x space straight R squared over denominator 2 space straight x space straight m subscript straight P end fraction space space space.... space left parenthesis straight i right parenthesis Where space straight m subscript straight P space is space the space mass space of space proton. Similarly comma space for space straight a space straight alpha space minus space particle space moves space in space straight a space uniform space magnetic space field straight E subscript 2 space equals space fraction numerator left parenthesis 2 straight e right parenthesis squared space straight x space straight B squared space straight x space straight R over denominator 2 space straight x space left parenthesis 4 straight m subscript straight P right parenthesis end fraction space left square bracket space straight m subscript straight alpha equals space 4 straight m subscript straight p right square bracket space.... space left parenthesis ii right parenthesis Dividing space Eq. space left parenthesis ii right parenthesis space by space Eq. space left parenthesis straight i right parenthesis space we space get straight E subscript 2 over straight E subscript 1 equals space fraction numerator left parenthesis 2 straight e right parenthesis squared space straight x space straight B squared space straight x space straight R over denominator 2 space straight x space left parenthesis 4 straight m subscript straight P right parenthesis end fraction space straight x space fraction numerator 2 space straight x space straight m subscript straight P over denominator straight e squared space straight x space straight B squared space straight x space straight R squared end fraction straight E subscript 2 over straight E subscript 1 space equals space 1 space rightwards double arrow space straight E subscript 2 space equals straight E subscript 1 space equals space 1 space MeV$

Question 216

Wired Faculty App

Two metal wires of identical dimensions are connected in series if σ₁a nd σ₂ are the conductivity of the metal wires respectively, the conductivity of the combination is

$fraction numerator 2 straight sigma subscript 1 straight sigma subscript 2 over denominator straight sigma subscript 2 space plus straight sigma subscript 2 end fraction$
$fraction numerator straight sigma subscript 1 space plus straight sigma subscript 2 over denominator 2 space straight sigma subscript 1 straight sigma subscript 2 end fraction$
$fraction numerator straight sigma subscript 1 space plus straight sigma subscript 2 over denominator straight sigma subscript 1 straight sigma subscript 2 end fraction$
$fraction numerator straight sigma subscript 1 straight sigma subscript 2 over denominator straight sigma subscript 1 space plus straight sigma subscript 2 end fraction$

Solution

fraction numerator 2 straight sigma subscript 1 straight sigma subscript 2 over denominator straight sigma subscript 2 space plus straight sigma subscript 2 end fraction

Net resistance of a metal wire having resistivity rho,
we have

$straight R subscript 1 space equals space straight rho subscript 1 straight L over straight A similarly comma straight R subscript 2 space equals space straight rho subscript 2 straight L over straight A Then comma space net space effective space resistance space of space two space metal space wires. straight R subscript eq space equals space straight R subscript 1 space plus straight R subscript 2 rightwards double arrow space straight rho fraction numerator 2 straight L over denominator straight A end fraction space equals space straight rho subscript 1 straight L over straight A space plus straight rho subscript 2 straight L over straight A rightwards double arrow space 2 straight rho space equals space straight rho subscript 1 space plus straight rho subscript 2 As comma space conductivity space straight sigma space equals space 1 over straight rho comma space we space have 2 over straight sigma space equals space 1 over straight sigma subscript 1 space plus space 1 over straight sigma subscript 2 rightwards double arrow 2 over straight rho space equals space fraction numerator straight sigma subscript 1 plus straight sigma subscript 2 over denominator straight sigma subscript 1 straight sigma subscript 2 end fraction rightwards double arrow space Net space effective space conductivity space of space combined space wires comma straight sigma space equals space fraction numerator 2 straight sigma subscript 1 straight sigma subscript 2 over denominator straight sigma subscript 1 plus straight sigma subscript 2 end fraction$

Question 217

Wired Faculty App

An electron moves on a straight line path XY as shown. The abcd is a coil adjacent in the path of the electron. What will be the direction of the current, if any induced in the coil?

abcd

adcb

the current will reverse its direction as the electron goes past the coil

no current induced

Solution

the current will reverse its direction as the electron goes past the coil

First current develops in direction of abcd but when electron moves away then magnetic field inside loop decrease and current changes its direction.

Question 218

Wired Faculty App

A galvanometer of resistance, G is shunted by a resistance S ohm. To keep the main current in the circuit unchanged,the resistance to be put in series with the galvanometer is

$fraction numerator straight S squared over denominator left parenthesis straight S space plus straight G right parenthesis end fraction$
$fraction numerator SG over denominator left parenthesis straight S plus straight G right parenthesis end fraction$
$fraction numerator straight G squared over denominator left parenthesis straight S space plus straight G right parenthesis end fraction$
$fraction numerator straight G over denominator left parenthesis straight S plus straight G right parenthesis end fraction$

Solution

fraction numerator straight G squared over denominator left parenthesis straight S space plus straight G right parenthesis end fraction

Current will be unchanged if resistance remains same so

$straight G space equals space fraction numerator GS over denominator straight G space plus straight S end fraction space plus straight R straight R space equals space straight G space minus space fraction numerator GS over denominator straight G space plus straight S end fraction straight R space equals space fraction numerator straight G squared over denominator straight G space plus straight S end fraction$

Question 219

Wired Faculty App

A thermocouple force towards the conductor produces an emf of 40 μ V/^oC in the linear range of temperature. A galvanometer of resistance 10 Ω whose sensitivity is 1μA/div, is employed with the thermocouple. The smallest value of temperature difference that can be detected by the system will be.

0.5^oC

1^oC

0.1^oC

0.25^oC

Solution

0.25^oC

For minimum deflection of 1 division require current =1 μA
Voltage required = iR = 1 x 10 = 10 μV
40 μV = 1^oC
10 μV = 1/4 = 0.25^oC

Question 220

Wired Faculty App

Charge q is uniformly spread on a thin ring of radius R. The ring rotates about its axis with a uniform frequency f Hz. The magnitude of magnetic induction at the centre of the ring is

μ_oqf / 2R

μ_oq / 2fR

μ_oq / 2πfR

μ_oqf / 2πR

Solution

μ_oqf / 2R

B = μ_oi / 2R
q = it
⇒ i = q /t = qf
B = μ_oqf / 2R

Question 221

Wired Faculty App

A short bar magnet of magnetic moment 0.4 JT^-1 is placed in a uniform magnetic field of 0.16T. The magnet is stable equilibrium when the potential energy is

-0.64 J

zero

-0.082 J

0.064 J

Solution

0.064 J

We know that
U = - MB cos θ
For stable equilibrium θ = 0^o
So, U = - MB

=- (0.4)(.16) = - 0.064 J

Question 222

Wired Faculty App

A potentiometer circuit is set up as shown. The potential gradient across the potentiometer wire is k volt /cm and the ammeter, present in the circuit, reads 1.0 A when the two-way key is switched off. The balance points, when the key between the terminals (i) 1 and 2 (ii) 1 and 3, is plugged in, are found to be at lengths l₁ and l₂ cm respectively. The magnitudes, of the resistor R and X, in ohm, are then, equal respectively, to

k (l₂ -l₁) and kl₂

kl₁ and k (l₂-l₁)

k (l₂ - l₁) and kl₁

kl₁ and kl₂

Solution

kl₁ and k (l₂-l₁)

The balancing length for R (When 1, 2 are connected) be is l₁ and balancing length for R + x (when 1, 3 is connected is l₂)
$iR space equals space kl subscript 1 and space iR space equals space kl subscript 1 Given space straight i space equals space 1 space straight A straight R space equals space kl subscript 1 space.... space left parenthesis straight i right parenthesis straight R space plus space straight X space equals space kl subscript 2 space..... space left parenthesis ii right parenthesis Subtracting space eq comma space left parenthesis straight i right parenthesis space from space eg. space left parenthesis ii right parenthesis space comma space we space get space straight X space equals space straight k space left parenthesis straight l subscript 2 minus space straight l subscript 1 right parenthesis space$

Question 223

Wired Faculty App

A galvanometer has a coil of resistance 100 Ω and gives a full-scale deflection for 30 mA current. If it is to work as a voltmeter of 30 V range, the resistance required to be added will be

900 Ω

1800

500 Ω

1000 Ω

Solution

900 Ω

Required resistance
$straight R space equals space straight V over straight i subscript straight g minus space straight G equals space fraction numerator 30 over denominator 30 space straight x space 10 to the power of negative 3 end exponent end fraction space minus space 100 space equals space 900 space straight capital omega$

Question 224

Wired Faculty App

A square current carrying loop is suspended in a uniform magnetic field acting in the plane of the loop. If the force on one arm of the loop is $bold F with bold rightwards arrow on top$ , the net force on the remaining three arms of the loops is

3 $bold F with bold rightwards arrow on top$

- $bold F with bold rightwards arrow on top$

$negative 3 space bold F with bold rightwards arrow on top$
$bold F with bold rightwards arrow on top$

Solution

- $bold F with bold rightwards arrow on top$

When a current carrying loop is placed in a magnetic field, the coil experiences a torque given by T = NBiA sin θ. Torque is maximum when θ =90, i.e. the plane of the coil is parallel to the field T_max NBiA

Force $stack bold F subscript bold 1 with bold rightwards arrow on top$ and $bold F with bold rightwards arrow on top subscript 2$ acting on the coil are equal in magnitude and opposite in direction. As the forces $stack bold F subscript bold 1 with bold rightwards arrow on top$ and $bold F with bold rightwards arrow on top subscript 2$ have the same line action their reultant effect on the coil is zero.
The two forces $stack bold F subscript bold 3 with bold rightwards arrow on top bold space bold and bold space stack bold F subscript bold 4 with bold rightwards arrow on top$ are equal in magnitude and opposite in direction. As the two forces have different lines of action, they constitute a torque .Thus, if the force on one arc of the loop is $bold F with bold rightwards arrow on top$ , the net force on the remaining three arms of the loop is -F.

Question 225

Wired Faculty App

A particle having a mass of 10^-2 kg carries a charge of 5 x 10^-8 C. The particle is given an initial horizontal velocity of 10⁵ ms^-1 in the presence of electric field and $bold E with bold rightwards arrow on top$ magnetic field. $bold B with rightwards arrow on top$ To keep the particle moving in a horizontal direction. it is necessary that

1) $bold B with rightwards arrow on top$ should be perpendicular to the direction of velocity and $bold E with bold rightwards arrow on top$ should be along the direction of velocity

2) Both $bold B with rightwards arrow on top$ and $bold E with bold rightwards arrow on top$ should be along the direction of velocity.

3) Both $bold B with rightwards arrow on top$ and $bold E with bold rightwards arrow on top$ are mutually perpendicular and perpendicular to the direction of velocity

4) $bold B with rightwards arrow on top$ should be along the direction of velocity and $bold E with bold rightwards arrow on top$ should be perpendicular to the direction of velocity.

(1) and (3)

(3) and (4)

(2) and (3)

(2) and (4)

Solution

(2) and (3)

Both $bold B with rightwards arrow on top$ and $bold E with bold rightwards arrow on top$ should be along the direction of velocity and Both $bold B with rightwards arrow on top$ and $bold E with bold rightwards arrow on top$ are mutually perpendicular and perpendicular to the direction of velocity.

Question 226

Wired Faculty App

A current loop consists of two identical semicircular parts each of radius R, one lying in the x-y plane and the other in x-z plane. if the current in the loop is i. The resultant magnetic field due to the two semicircular parts at their common centre is

$fraction numerator straight mu subscript straight o space straight i over denominator 2 square root of 2 straight R end root end fraction$
$fraction numerator straight mu subscript 0 straight i over denominator 2 straight R end fraction$
$fraction numerator straight mu subscript straight o straight i over denominator 4 straight R end fraction$
$fraction numerator straight mu subscript straight o straight i over denominator square root of 2 straight R end fraction$

Solution

fraction numerator straight mu subscript straight o space straight i over denominator 2 square root of 2 straight R end root end fraction

straight B subscript 1 space equals space straight B subscript 2 space equals space fraction numerator straight mu subscript 0 straight i over denominator 4 straight R end fraction The space magnetic space field space at space their space common space centre bold B with bold rightwards arrow on top bold space bold equals bold space stack bold B subscript bold 1 with bold rightwards arrow on top bold space bold plus bold space stack bold B subscript bold 2 with bold rightwards arrow on top straight B space equals space square root of straight B subscript 1 superscript 2 plus straight B subscript 2 superscript 2 end root space equals square root of open parentheses fraction numerator straight mu subscript straight o straight i over denominator 4 straight R end fraction close parentheses squared plus open parentheses fraction numerator straight mu subscript 0 straight i over denominator 4 straight R end fraction close parentheses squared end root equals space straight B space fraction numerator straight mu subscript 0 straight i over denominator 2 square root of 2 straight R end root end fraction

Question 227

Wired Faculty App

A wire of resistance 12 Ωm^-1 is bent to form a complete circle of radius 10 cm. The resistance between its two diametrically opposite point, A and B as shown in the figure, is

0.6 π Ω

3 π Ω

6 π Ω

6 π Ω

Solution

0.6 π Ω

Circumference space of space circle space equals space 2 space xπ space 10 over 100 space equals space fraction numerator 2 straight pi over denominator 10 end fraction space equals space straight pi over 5 Resistance space of space wire space equals space 12 space straight x straight pi over 5 space equals space fraction numerator 12 space straight pi over denominator 5 end fraction Resistance space of space each space section space equals space fraction numerator 12 straight pi over denominator 10 end fraction space straight capital omega therefore comma space Equivalent space resistance equals space fraction numerator fraction numerator 12 space straight pi over denominator 10 end fraction straight x fraction numerator 12 space straight pi over denominator 10 end fraction over denominator fraction numerator 12 space straight pi over denominator 10 end fraction space plus fraction numerator 12 space straight pi over denominator 10 end fraction end fraction space equals space fraction numerator 6 space straight pi over denominator 10 end fraction space equals space 0.6 space πΩ

Question 228

Wired Faculty App

A student measures the terminal potential difference (V) of a cell (of emf and internal resistance r) as a function of the current (I) flowing through it. The slope and intercept of the graph between V and I, then respectively, equal

$straight epsilon space and space minus straight r$
$negative straight r space and space straight epsilon$
$straight r space and space minus straight epsilon$
$negative straight epsilon space and space straight r$

Solution

negative straight r space and space straight epsilon

According to Ohm's law
$dV over dI space equals space minus straight r and space straight V space equals straight epsilon space if space straight I space equals space 0 space left square bracket space As space straight V space plus Ir space equals space straight epsilon right square bracket therefore comma space Slope space of space the space graph space equals space minus space and space intercept space equals straight epsilon$

Question 229

Wired Faculty App

The magnetic force acting on a charged particle of charge -2 μC in a magnetic field of 2T acting in y direction, when the particle velocity is $left parenthesis 2 straight i with hat on top space plus 3 straight j with hat on top right parenthesis space straight x space 10 to the power of 6 space ms to the power of negative 1 end exponent space is$

8N in -z direction

4N in z direction

8 N in y direction

8 N in z direction

Solution

8N in -z direction

When a charge q moves with the velocity v inside the magnetic field of strength B, then force on the charge is called magnetic Lorentz force. The magnetic Lorentz force is in the direction of vector
$bold v with bold rightwards arrow on top bold space bold x bold space bold B with bold rightwards arrow on top Magnetic space Lorentz space force space straight F with rightwards arrow on top space equals space straight q space left parenthesis bold v with bold rightwards arrow on top bold space bold x bold space bold B with bold rightwards arrow on top right parenthesis equals space minus 2 space straight x space 10 to the power of negative 6 end exponent space left square bracket space 2 space straight x space 2 space straight x space 10 to the power of 6 right square bracket space equals space 8 space straight N space along space negative space straight z space axis$

Question 230

Wired Faculty App

A particle mass m, charge Q and kinetic energy T enter a transverse uniform magnetic field of induction. $bold B with rightwards arrow on top$ After 3 s the kinetic energy of the particle will be

3T

2T

T

4T

Solution

After passing through a magnetic field, the magnitude of its mass and velocity of the particle remain same, so its energy does not change, ie, kinetic energy will remain T.

Question 231

Wired Faculty App

A long solenoid has 500 turns. When a current of 2 A is passed through it, the resulting magnetic flux linked with each turn of the solenoid is 4 x 10^-3 Wb. The self -inductance of the solenoid is

2.5 H

2.0 H

1.0 H

4.0 H

Solution

1.0 H

The inductance of a coil is numerically equal to the emf induced in the coil when the current in the coil changes at the rate of 1 As^-1. If I is the current flowing in the circuit, then flux linked with the circuit is observed to be proportional to I, ie,
$straight ϕ space proportional to space straight I straight ϕ space equals space LI space... space left parenthesis straight i right parenthesis$
where L is called the self-inductance or coefficients of self-inductance or coefficient of self-inductance or simply inductance of the coil.
Net flux through the solenoid.
Φ = 500 x 4 x 10^-3 = 2 Wb
Or 2 = L x 2 [after putting value in eq. or L = 1H]

Question 232

Wired Faculty App

A galvanometer of resistance 50 Ω is connected to a battery of 3 V along with a resistance of 2950 Ω in series. A full-scale deflection of 30 divisions is obtained in the galvanometer. In order to reduce this deflection to 20 divisions, the resistance in series should be

5050 Ω

5550 Ω

6050 Ω

4450 Ω

Solution

4450 Ω

Current through the galvanometer

$space straight I thin space equals space fraction numerator 3 over denominator left parenthesis 50 plus 2950 right parenthesis end fraction space equals space 10 to the power of negative 3 end exponent space straight A Current space for space 30 space divisions space equals space 10 to the power of negative 3 end exponent space straight A Current space for space 20 space divisions space equals space 10 to the power of negative 3 end exponent over 30 space straight x space 20 equals space 2 over 3 space straight x space 10 to the power of negative 3 end exponent space straight A For space the space same space deflection space to space obtain space for space 20 space divisions comma let space resistance space added space be space straight R therefore comma 2 over 3 space straight x space 10 to the power of negative 3 end exponent space equals space fraction numerator 3 over denominator left parenthesis 50 plus 1 straight R right parenthesis end fraction Or space straight R space equals space 4450 space straight capital omega$

Question 233

Wired Faculty App

Two circular coils 1 and 2 are made from the same wire but the radius of the 1st coil is twice that of the 2nd coil. What is the ratio of potential difference applied across them so that the magnetic field at their centres is the same?

3

4

6

2

Solution

Magnetic field at the centre of a circular coil is
$straight B space equals space fraction numerator straight mu subscript 0 over denominator 4 straight pi end fraction cross times fraction numerator 2 πi over denominator straight r end fraction$
where i is current flowing in the coil and r is radius of coil.
At the centre of coil - 1,
   $straight B subscript 1 space equals space fraction numerator straight mu subscript 0 over denominator 4 straight pi end fraction cross times fraction numerator 2 πi subscript 1 over denominator straight r subscript 1 end fraction space space space space space space space space space... left parenthesis straight i right parenthesis$
At the centre of coil-2
$straight B subscript 2 equals space space fraction numerator straight mu subscript 0 over denominator 4 straight pi end fraction cross times fraction numerator 2 πi subscript 2 over denominator straight r subscript 2 end fraction space space space space space space space... left parenthesis ii right parenthesis$
But    $straight B subscript 1 space equals space straight B subscript 2$
$therefore space space space space fraction numerator straight mu subscript 0 over denominator 4 straight pi end fraction space fraction numerator 2 πi subscript 1 over denominator straight r subscript 1 end fraction space equals space fraction numerator straight mu subscript 0 over denominator 4 straight pi end fraction fraction numerator 2 πi subscript 2 over denominator straight r subscript 2 end fraction or space space space space space space space space straight i subscript 1 over straight r subscript 1 space equals space straight i subscript 2 over straight r subscript 2 As space space space space space space space space straight r subscript 1 space equals space 2 straight r subscript 2 therefore space space space space space space fraction numerator straight i subscript 1 over denominator 2 straight r subscript 2 end fraction space equals space straight i subscript 2 over straight r subscript 2 or space space space space space space space space straight i subscript 1 space equals space 2 straight i subscript 2 space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis iii right parenthesis space space space space space space$
Now, export of potential differences
   $straight V subscript 2 over straight V subscript 1 space equals space fraction numerator straight i subscript 2 cross times straight r subscript 2 over denominator straight i subscript 1 cross times straight r subscript 1 end fraction space equals space fraction numerator straight i subscript 2 cross times straight r subscript 2 over denominator 2 straight i subscript 2 cross times 2 straight r subscript 2 end fraction space equals space 1 fourth$
$therefore space space space space space straight V subscript 1 over straight V subscript 2 space equals space 4 over 1$

Question 234

Wired Faculty App

A long solenoid of diameter 0.1 m has 2 × 104 turns per meter. At the centre of the solenoid, a coil of 100 turns and radius 0.01 m is placed with its axis coinciding with the solenoid axis. The current in the solenoid reduces at a constant rate to 0 A from 4 A in 0.05 s. If the resistance of the coil is 10π2 Ω, the total charge flowing through the coil during this time is

32 πμC

16 μC

32 μC

16 πμC

Solution

32 μC

straight epsilon space equals space minus space straight N dϕ over dt open vertical bar straight epsilon over straight R close vertical bar space equals space straight N over straight R dϕ over dt dq space equals straight N over straight R space dϕ increment straight Q space equals space fraction numerator straight N left parenthesis increment straight ϕ right parenthesis over denominator straight R end fraction increment straight Q space equals space fraction numerator increment straight ϕ subscript total over denominator straight R end fraction open parentheses NBA over straight R close parentheses space equals space fraction numerator straight mu subscript 0 niπr squared over denominator straight R end fraction Putting space values fraction numerator 4 straight pi space straight x space 10 to the power of negative 7 end exponent space straight x space 100 space straight x 4 space xπ space straight x space left parenthesis 0.01 right parenthesis squared over denominator space 10 straight pi squared end fraction space increment straight Q space equals space 32 μC

Question 235

Wired Faculty App

Current sensitivity of a moving coil galvanometer is 5 div/mA and its voltage sensitivity (angular deflection per unit voltage applied) is 20 div/V. The resistance of the galvanometer is

40 Ω

25 Ω

500 Ω

250 Ω

Solution

250 Ω

Current sensitivity of moving coil galvanometer

$I_{s} = \frac{N B A}{C} . . . . (i)$

Voltage sensitivity of moving coil galvanometer,

$V_{s} = \frac{N B A}{C R_{G}} . . . . . . (i i)$

Dividing equation (i) and (ii)
Resistance of galvanometer

$R_{G} = \frac{I_{s}}{V_{s}} = \frac{5 x 1}{20 x 10^{- 3}} = \frac{5000}{20} = 250 Ω$

Question 236

Wired Faculty App

A metallic rod of mass per unit length 0.5 kg m^–1 is lying horizontally on a smooth inclined plane which makes an angle of 30° with the horizontal. The rod is not allowed to slide down by flowing a current through it when a magnetic field of induction 0.25 T is acting on it in the vertical direction. The current flowing in the rod to keep it stationary is

7.14 A

5.98 A

11.32 A

14.76 A

Solution

11.32 A

For equilibrium,

$m g \sin 30^{o} = I / B \cos 30^{o} \Rightarrow I = \frac{m g}{l B} \tan 30^{o} = \frac{0.5 x 9.8}{0.25 x \sqrt{3}} = 11.32 A$

Question 237

Wired Faculty App

The turns of a solenoid, designed to provide a given magnetic flux density along its axis, are wound to fill the space between two concentric cylinders of fixed radii. How should the diameter d of the wire used be chosen so as to minimize the heat dissipated in the windings?

Wire should be multiple of 5d

Wire should be multiple of d/3

the wire is independent of d

Can't say

Solution

the wire is independent of d

The cross-sectional area of the space to be filled is fixed. whilst, that of the wire varies as d². Thus, n ∝ d^-2. The resistance of one turn is inversely proportional to the cross-sectional area of the wire, i.e varies as d^-2 and hence the resistance per unit length of the solenoid is R ∝nd^-2 ∝ d^-4. The flux density B is ∝ nI and therefore the required current I ∝ n^-1 ∝ d². The heat dissipated per unit length is RI², which ∝ d^-4(d²)², i.e., independent of d. Thus, within limits, it does not matter what diameter wire is chosen so far as the heating effect is concerned.

Question 238

Wired Faculty App

A long straight wire is carrying current I in the +z direction. The x-y plane contains a closed circular loop carrying current I₂ and not encircling the straight wire. The force on the loop will be

$\frac{μ_{0} I_{1} I_{0}}{2 π}$

$\frac{μ_{0} I_{1} I_{0}}{4 π}$

zero

Depends on the distance of the centre of the loop from the wire.

Solution

Depends on the distance of the centre of the loop from the wire.

The force depends on the distance of the centre of the loop from the wire.

Question 239

Wired Faculty App

To reduce the range of voltmeter, its resistance needs to be reduced. A voltmeter has resistance Ro and ranges V. Which of the following resistances when connected in parallel will convert it into a voltmeter of range V/n?

nR₀

(n+1)R₀

(n-1)R

None of these

Solution

None of these

When a resistance is connected in parallel, it reduces the resistance of combination but the range is not reduced. For the purpose of reducing the range the resistance in series with galvanometer need to be reduced.

This cannot be achieved by connecting a resistance in parallel with a voltmeter. It should be noted that the range of voltmeter can only be increased but cannot be decreased.

Question 240

Wired Faculty App

A coil having N turns carry a current as shown in the figure. The magnetic field intensity at point P

$\frac{μ_{0} {NiR}^{2}}{2 {(R^{2} + x^{2})}^{\frac{3}{2}}}$

$\frac{μ_{0} Ni}{2 R}$

$\frac{μ_{0} {NiR}^{2}}{{(R + x)}^{2}}$

Zero

Solution

$\frac{μ_{0} {NiR}^{2}}{2 {(R^{2} + x^{2})}^{\frac{3}{2}}}$

From the formula, magnetic field intensity at point P, shown in given figure while carrying current 'i' and having 'N' turns is given by Magneti field at P due to entire circular loop is

$\Rightarrow \frac{μ_{0} {iR}^{2}}{2 {(x^{2} + R^{2})}^{\frac{3}{2}}}$

And hence for N turns is,

$\Rightarrow \frac{μ_{0} N i R^{2}}{2 {(R^{2} + x^{2})}^{\frac{3}{2}}}$

Question 241

Wired Faculty App

An electron moves at right angle to a magnetic field of 1.5×10^-2 T with a speed of 6 x 10⁷ m / s. If the specific charge of the electron is 1.7x 10¹¹ C/ kg. The radius of the circular path will be

2.9 cm

3.9 cm

2.35 cm

2 cm

Solution

2.35 cm

The formula for radius of circular path is

$r = \frac{m ν}{e B} = \frac{ν}{(\frac{e}{m}) B} Suppose the cut off wavelength is represented by λ_{o} so, \frac{h c}{λ_{o}} = work function = W_{o} \frac{h c}{λ_{o}} = 4.2 \times 1.6 \times 10^{- 19} λ_{o} = \frac{6.6 \times 10^{- 34} \times 3 \times 10^{8}}{4.2 \times 1.6 \times 10^{- 19}} = 2946 \times 10^{- 10} \approx 2950 A^{o}$

Question 242

Wired Faculty App

If a charge particle enters perpendicularly in the uniform magnetic field, then:

energy and momentum both remains constant

energy remains constant but momentum changes

both energy and momentum changes

energy changes but momentum remains constant

Solution

energy remains constant but momentum changes

The magnetic field does not work, so kinetic energy and speed of a charged particle in a magnetic field remains constant. The magnetic force, acting perpendicular to the velocity of the particle, will cause circular motion.

Question 243

Wired Faculty App

A coil in the shape of an equilateral triangle of side l is suspended between the pole pieces of a
permanent magnet such that B is in plane of the coil. If due to current i in the triangle a torque T acts on it, the side l of the triangle is

$\frac{2}{\sqrt{3}} {(\frac{τ}{Bi})}^{\frac{1}{2}}$

$\frac{2}{\sqrt{3}} (\frac{τ}{Bi})$

$2 {(\frac{τ}{\sqrt{3} Bi})}^{\frac{1}{2}}$

$\frac{1}{\sqrt{3}} \frac{τ}{Bi}$

Solution

$2 {(\frac{τ}{\sqrt{3} Bi})}^{\frac{1}{2}}$

Torque acting on equilateral triangle in a magnetic field $\vec{B}$ is

$τ = i AB sinθ$

Area of triangle LMN

A = 1/2 × base × height

A = 1/2 ×l × sin 60^o

(since an equilateral triangle has an angle of 60^o each)

$A = \frac{\sqrt{3}}{2} l^{2} and θ = 90^{o}$

Substituting the given value in expression for torque, we have

$τ = i \times \frac{\sqrt{3}}{4} l^{2} B \sin 90^{o} τ = \frac{\sqrt{3}}{4} {il}^{2} B (∵ \sin 90^{o} = 1) Hence, l = 2 {(\frac{τ}{\sqrt{3} Bi})}^{\frac{1}{2}}$

Question 244

Wired Faculty App

An electron moves in a circular orbit with a uniform speed v. It produces a magnetic field B at the centre of the circle. The radius of the circle is proportional, to

$\frac{B}{v}$

$\frac{v}{B}$

$\sqrt{\frac{v}{B}}$

$\sqrt{\frac{B}{v}}$

Solution

$\sqrt{\frac{v}{B}}$

The time period of electron moving in angular orbit

$T = \frac{circumference of circular path}{speed} T = \frac{2 πr}{ν} and equivalent current due to electron flow I = \frac{e}{T} = \frac{e}{(\frac{2 πr}{ν})} I = \frac{eν}{2 πr}$

Magnetic field at centre of circle

$B = \frac{μ_{o} I}{2 r} = \frac{μ_{o} e ν}{4 π r^{2}} \Rightarrow r \propto \sqrt{\frac{ν}{B}}$

Question 245

Wired Faculty App

In a mass spectrometer used for measuring the masses of ions, the ions are initially accelerated by an electric potential V and then made to describe semicircular paths of radius R using a magnetic field B. If V and B are kept constant the ratio $(\frac{charge on the ion}{mass of the ion})$ will be proportional to

$\frac{1}{R}$

$\frac{1}{R^{2}}$

R²

R

Solution

$\frac{1}{R^{2}}$

The radius of the orbit in which ions moving is determined by the relation as given below

$\frac{{mv}^{2}}{R} = qvB$

where m is the mass, v is velocity, q is charge of ion and B is the flux density of the magnetic field, so that qvB is the magnetic force acting on the ion, and $\frac{{mv}^{2}}{R}$ is the centripetal force on the ion moving in a curved path of radius R.

The angular frequency of rotation of the ions about the vertical field B is given by

$ω = \frac{v}{R} = \frac{qB}{m} ω = 2 πv$

where ν is the frequency.

Energy of ion is given by

$E = \frac{1}{2} {mν}^{2} = \frac{1}{2} m ({Rω}^{2}) \Rightarrow E = \frac{1}{2} m R^{2} B^{2} \frac{q^{2}}{m^{2}} \Rightarrow E = \frac{1}{2} \frac{R^{2} B^{2} q^{2}}{m} . . . . (i)$

If ions are accelerated by electric potential V, then energy attained by ions

E = qV .... (ii)

from equation (i) and (ii) we get

$q V = \frac{1}{2} \frac{R^{2} B^{2} q^{2}}{m} \Rightarrow \frac{q}{m} = \frac{2 V}{R^{2} B^{2}}$

If V and B are kept constant

$\frac{q}{m} \propto \frac{1}{R^{2}}$

Question 246

Wired Faculty App

The north pole of a long horizontal bar magnet is being brought closer to a vertical conducting plane along the perpendicular direction. The direction of the induced current in the conducting plane will be

horizontal

vertical

clockwise

Anticlockwise

Solution

clockwise

The induced emf will oppose the motion of the magnet. Applying the right-hand rule, the direction of induced current will be clockwise.

Above image shows the direction of the magnetic force acting on a charged particle. The force on a positively charged particle with velocity v and making an angle θ with the magnetic field B is given by the right-hand rule.

Question 247

Wired Faculty App

An electric current passes through a long straight copper wire. At a distance 5 cm from the straight wire, the magnetic field is B. The magnetic field at 20 cm from the straight wire would be

$\frac{B}{6}$

$\frac{B}{4}$

$\frac{B}{3}$

$\frac{B}{2}$

Solution

$\frac{B}{4}$

The magnetic field at a distance r from the straight wire

B = $\frac{μ_{o} i}{2 πr}$

or B ∝ $\frac{1}{r}$

or $\frac{B_{2}}{B_{1}} = \frac{r_{1}}{r_{2}}$

∴ $\frac{B_{2}}{B} = \frac{5}{20}$

= $\frac{1}{4}$

⇒ $B_{2} = \frac{B}{4}$

Question 248

Wired Faculty App

An electron of mass m and charge q is travelling with a speed v along a circular path of radius r at right angles to a uniform magnetic field B. If speed of the electron is doubled and the magnetic field is halved, then resulting path would have a radius of

$\frac{r}{4}$

$\frac{r}{2}$

2r

4r

Solution

In a perpendicular magnetic field

Magnetic force = centripetal force

Bqv = $\frac{{mv}^{2}}{r}$

⇒ r = $\frac{mv}{Bq}$

∴ $\frac{r_{1}}{r_{2}} = \frac{v_{1}}{v_{2}} \times \frac{B_{2}}{B_{1}}$

$\frac{r_{1}}{r_{2}} = \frac{1}{2} \times \frac{1}{2}$

$r_{2} = 4 r_{1}$

⇒ $r_{2} = 4 r$

Question 249

Wired Faculty App

A galvanometer acting as a voltmeter should have

low resistance in series with its coil

low resistance in parallel with its coil

high resistance in series with its coil

high resistance in parallel with its coil

Solution

high resistance in series with its coil

Voltmeter has high resistance and is always connected in parallel with the circuit. So, to convert a galvanometer into voltmeter, a high resistance must be connected in series with it so that it draws negligible current from the circuit.

Question 250

Wired Faculty App

A galvanometer can be changed into an ammeter by using

low resistance shunt in series

low resistance shunt in parallel

high resistance shunt in series

high resistance shunt in parallel

Solution

low resistance shunt in parallel

A galvanometer can be converted to ammeter by connecting a low resistance ( called shunt resistance) in parallel to the galvanometer. If resistance is connected in parallel, then some of the current will flow through resistance.

Question 251

Wired Faculty App

A proton moving vertically downward enters a magnetic field pointing towards north. In which direction proton will deflect?

East

West

North

South

Solution

East

Proton will represent the direction of current. So, the direction of current is vertically downward.

According to Fleming's left hand rule, if middle finger represents the direction of current and fore-finger represents the direction of magnetic field then thumb will represent the direction of Lorentz force acting on the proton which deflects the proton in east direction.

Question 252

Wired Faculty App

A coil of n number of turns is wound tightly in the form of a spiral with inner and outer radii a and b respectively. When a current of strength I is passed through the coil, the magnetic field at its centre is

$\frac{μ_{o} nI}{(b - a)} \log_{e} \frac{a}{b}$

$\frac{μ_{o} n I}{2 (b - a)}$

$\frac{2 μ_{o} nI}{b}$

$\frac{μ_{o} nI}{2 (b - a)} \log_{e} \frac{b}{a}$

Solution

$\frac{μ_{o} nI}{2 (b - a)} \log_{e} \frac{b}{a}$

Consider an element of thickness (dr) at a distance 'r' from the centre of spiral-coil.

Number of turns in coil = n

Number of turns per unit length = $\frac{n}{b - a}$

a is the inner radii and b is the outer radii of spiral winding.

Number of turns in element dr = dn

Number of turns per unit length in element

dr = $\frac{n dr}{b - a}$

⇒ dn = $\frac{n dr}{b - a}$

Magnetic field at its centre due to element dr is

dB = $\frac{μ_{o} I dn}{2 r}$

= $\frac{μ_{o} I}{2} \frac{n}{(b - a)} \frac{dr}{r}$

By integrating on both side

∴ B = $\int_{a}^{b} \frac{μ_{o} I n dr}{2 (b - a) r}$

= $\frac{μ_{o} I n}{2 (b - a)} \int_{a}^{b} \frac{dr}{r}$

= $\frac{μ_{o} I n}{2 (b - a)} {[\log_{e} r]}_{a}^{b}$ $(∵ \int_{a}^{b} \frac{1}{r} = {[\log_{e} r]}_{a}^{b})$

= $\frac{μ_{o} I n}{2 (b - a)} \log_{e} (\frac{b}{a})$

Question 253

Wired Faculty App

A galvanometer acting as a voltmeter should have

low resistance in series with its coil

low resistance in parallel with its coil

high resistance in series with its coil

high resistance in parallel with its coil

Solution

high resistance in series with its coil

Voltmeter has high resistance and is always connected in parallel with the circuit. So to convert a galvanometer into voltmeter, a high resistance must be connected in series with it so that is draws negligible current from the circuit.

Question 254

Wired Faculty App

Magnetic field at point O will be

$\frac{μ_{o} I}{2 R}$ interior

$\frac{μ_{o} I}{2 R}$ exterior

$\frac{μ_{o} I}{2 R} (1 - \frac{l}{π})$ interior

$\frac{μ_{o} I}{2 R} (1 + \frac{I}{π})$ exterior

Solution

$\frac{μ_{o} I}{2 R}$ interior

Magnetic field at the centre of circular loop

B = $\frac{μ_{o} I}{2 R}$ interior

B is the magnetic field

I is the current flowing through loop

Question 255

Wired Faculty App

If the electric flux entering and leaving an enclosed surface respectively are $ϕ_{1} and ϕ_{2}$ , the electric charge inside the surface will be

$\frac{ϕ_{2} - ϕ_{1}}{ε_{o}}$

$\frac{ϕ_{2} + ϕ_{1}}{ε_{o}}$

$\frac{ϕ_{1} - ϕ_{2}}{ε_{0}}$

$ε_{0} (ϕ_{1} + ϕ_{2})$

Solution

$ε_{0} (ϕ_{1} + ϕ_{2})$

According to Gauss theorem "the net electric flux through any closed surface is equal to the net charge inside the surface divided by ε_o''.

Therefore

$ϕ \propto \frac{q}{ε_{o}}$

Let-q₁ be the charge, due to which flux, is entering the surface,

$ϕ_{1} = \frac{- q_{1}}{ε_{o}}$

$- q_{1} = ε_{o} ϕ_{1}$

Let +q₂ be the charge, due to which flux $ϕ_{2}$ is entering the surface.

$ϕ_{2} = \frac{q_{2}}{ε_{0}}$

⇒ q₂ = $ε_{o} ϕ_{2}$

So, electric charge inside the surface

= $q_{2} - q_{1}$

= ${ε_{0}}_{} ϕ_{2} + ε_{0} ϕ_{1}$

= $ε_{0} (ϕ_{2} - ϕ_{1})$

Question 256

Wired Faculty App

Two long straight wires, each carrying an electric current of 5 A, are kept parallel to each other at a separation of 2.5 cm. Find the magnitude of the magnetic force experiment by 10 cm of a wire.

4.0 × 10^-4 N

3.5 × 10^-6 N

2.0 × 10^-5 N

2.0 × 10^-9 N

Solution

2.0 × 10^-5 N

The field at the site of one wire due to the other is

B = $\frac{μ_{o} i}{2 πd}$

Where μ_o - permeability of free space

i - current

B = $\frac{2 \times 10^{7} \times 5.0}{2.5 \times 10^{- 2}}$

B = 4.0 × 10^-5 T

The force experienced by 10 cm of this wire due to other is

F = il B

= (5.0 ) × ( 10 × 10^-2 ) × ( 4.0 × 10^-5)

F = 2.0 × 10^-5 N

Question 257

Wired Faculty App

The parts of two concentric circular arcs joined by two radial lines and carries current i. The arcs subtend an angle 0 at the centre of the circle. The magnetic field at the centre O, is

$\frac{μ_{0} i (b - a) θ}{4 π ab}$

$\frac{μ_{0} i (b - a)}{(π - θ)}$

$\frac{μ_{0} i (b - a) θ}{πab}$

$\frac{μ_{0} i (a - b)}{2 πab}$

Solution

$\frac{μ_{0} i (b - a) θ}{4 π ab}$

The field at the centre of a circular loop is given by

B = $\frac{μ_{0} i}{2 a}$

The field due to circular arc of inner circle is

B₁ = $(\frac{θ}{2 π}) \frac{μ_{0} i}{2 a}$

The field due to circular arc of outer circle is

B₂ = $(\frac{θ}{2 π}) (\frac{μ_{0} i}{2 b})$

B = B₁ - B₂

B = $\frac{μ_{0} i (b - a)}{4 πab}$

Question 258

Wired Faculty App

A square loop is made by a uniform conductor wire as shown in figure

The net magnetic field at the centre of the loop if side length of the square is a

$\frac{μ_{0} i}{2 a}$

zero

$\frac{μ_{0} i^{2}}{a^{2}}$

None of these

Solution

zero

The current will be equally divided at junction P. The field at the centre due to wires PQ and SR will be equal in magnitude but opposite in the direction, so its effective field will be zero. Similarly, net field due to wires PS and QR is zero. Therefore, the net field at the centre of the loop is zero.

Question 259

Wired Faculty App

A thin bar magnet of length 2 L is bent at the mid-point so that the angle between them is 60°. The new length of the magnet is

$\frac{L}{2 \sqrt{3}}$

$\frac{\sqrt{3} L}{2}$

L

$\frac{2 L}{3}$

Solution

On bending the magnet, the angle between two parts of the magnet is expressed in the diagram below

Now, the minimum distance between the ends P and R

= L sin 30^o + L sin30^o

= 2L sin30^o

= 2L × $\frac{1}{2}$ ....( sin30^o = $\frac{1}{2}$ )

= L

Question 260

Wired Faculty App

A circular current carrying coil has a radius R. The distance from the centre of the coil, on the axis, where B will be $\frac{1}{8}$ of its value at the centre of the coil is

$\frac{R}{\sqrt{3}}$

$\sqrt{3} R$

2 $\sqrt{3}$ R

$\frac{2 R}{\sqrt{3}}$

Solution

$\sqrt{3} R$

B = $\frac{μ_{0} Ni R^{2}}{2 {(R^{2} + x^{2})}^{\frac{3}{2}}}$

= $\frac{μ_{0} N i}{2 R} \frac{1}{{[1 + \frac{x^{2}}{R^{2}}]}^{\frac{3}{2}}}$

Let B_C = $\frac{μ_{0} Ni}{2 R}$

B = B_C $\frac{1}{{[1 + \frac{x^{2}}{R^{2}}]}^{\frac{3}{2}}}$

⇒ $\frac{B_{C}}{8} = \frac{B_{C}}{{(1 + \frac{x^{2}}{R^{2}})}^{\frac{3}{2}}}$ $[∵ B = \frac{B_{C}}{8}]$

⇒ ${(1 + \frac{x^{2}}{R^{2}})}^{\frac{3}{2}}$ = 8

⇒ 1 + $\frac{x^{2}}{R^{2}}$ = 4

⇒ $\frac{x^{2}}{R^{2}}$ = 3

⇒ x = R $\sqrt{3}$

Question 261

Wired Faculty App

A current carrying loop is placed in a uniform magnetic field in four different orientations I, II, III and IV as shown in figure. Arrange them in decreasing order of potential energy

l > lll> ll > lV

l > ll > lll > lV

l > lV > ll > lll

lll > lV > l > ll

Solution

l > lV > ll > lll

As we know that, potential energy of a magnet in a magnetic field.

U = - m .B

where, m = magnetic dipole moment of the magnet

B = magnetic field

Case l :- θ = 180^o

U₁ = $-$ mB. cos 180^o

= mB [ $∵$ cos 180^o = -1 ]

Case ll :- θ = 90^o

U₂ = 0 [ $∵$ cos 90^o = 0 ]

Case lll :- θ is acute angle

θ ∈ ( 0^o, 90^o )

∴ cosθ = positive

Thus U₃= negative

Case lV:- θ is obtuse

θ ∈ ( 90^o, 180^o )

∴ cos ∈ ( 0, -1 )

Thus, U₄ = positive

Therefore, decreasing order of PE is

l > lV > ll > lll

Question 262

Wired Faculty App

A current I is flowing through the loop. The direction of the current and the shape of the loop are as shown in the figure. The magnetic field at the centre of the loop is $\frac{μ_{0} I}{R}$ times ( MA = R, MB = 2R, ∠DMA = 90^o )

$\frac{5}{16}$ , but out of the plane of the paper.

$\frac{5}{16}$ , but into the plane of the paper.

$\frac{7}{16}$ , but out of the plane of the paper.

$\frac{7}{16}$ , but into the plane of the paper.

Solution

$\frac{7}{16}$ , but into the plane of the paper.

Magnetic field at the centre M due to current through the curved portion DA is

$\vec{B_{1}} = \frac{μ_{0} I}{4 πR} \times (\frac{3 π}{2})$

= $\frac{μ_{0} I}{16 R}$

Magnetic field at the centre M due to current through the straight portion CD is B,= 0, since
point M lies on the axis of the straight portion CD.

The resultant magnetic field at the point M is

$\vec{B} = \vec{B_{1}} + \vec{B_{2}} + \vec{B_{3}} + \vec{B_{4}}$

= $(\frac{3 μ_{0} I}{8 R} + 0 + \frac{μ_{0} I}{16 R} + 0)$

⇒ $\vec{B} = \frac{7 μ_{0} I}{16 R}$

Question 263

Wired Faculty App

Assertion: Acceleration of charged particle in non-uniform electric field does not depend on velocity of charged particle.
Reason: Charge is an invariant quantity. That is amount of charge on particle does not depend on frame of reference.

If both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true but reason is not the correct explanation of assertion.

If assertion is true but reason is false.

If both assertion and reason are false.

Solution

If both assertion and reason are true but reason is not the correct explanation of assertion.

If both assertion and reason are false.

$\bar{α} = \frac{q \bar{E}}{m}$

Where $α$ = acceleration ( m/s² )

q = charge ( coulomb )

m = mass ( kg)

E = electric field ( volt × m^-1 )

As $\vec{E}$ varies so does $\bar{α}$ , it does not depend on its velocity.

Question 264

Wired Faculty App

The magnetic field at the centre O of the arc shown in the figure is

2 $I (\sqrt{2} + π) \times \frac{10^{- 7}}{r}$

2 $I (\sqrt{2} + \frac{π}{4}) \times \frac{10^{- 7}}{r}$

$I (\sqrt{2} + π) \times \frac{10^{- 7}}{r}$

$I (\sqrt{2} + \frac{π}{4}) \times \frac{10^{- 7}}{r}$

Solution

2 $I (\sqrt{2} + \frac{π}{4}) \times \frac{10^{- 7}}{r}$

Here, a = $\frac{r}{\sqrt{2}}$

Magnetic field at point O due to AB is

B₁ = $\frac{μ_{0}}{4 π} \frac{I}{a}$

= $\frac{μ_{0} I}{4 π (\frac{r}{\sqrt{2}})}$

Magnetic field at point O due to BCD is

B₂ = $\frac{μ_{0} I}{4 π a} (\frac{π}{2})$

Magnetic fieid at point O due to DE is

B₃ = $\frac{μ_{0} I}{4 π α}$

B₃ = $\frac{μ_{0} I}{4 π (\frac{r}{\sqrt{2}})}$

Resultant magnetic field at point O is

B = B₁ + B₂+ B₃

= $\frac{μ_{0} I}{4 π (\frac{r}{\sqrt{2}})} + \frac{μ_{0} I}{4 πr} (\frac{π}{2}) + \frac{μ_{0} I}{4 π (\frac{r}{\sqrt{2}})}$

= $\frac{μ_{0} I}{4 πr} (\sqrt{2} + \frac{π}{2} + \sqrt{2})$

B = $\frac{μ_{0} I}{4 πr} (2 \sqrt{2} + \frac{π}{2})$

= $\frac{μ_{0} 2 I}{4 πr} (\sqrt{2} + \frac{π}{4})$

B = $\frac{10^{- 7} \times 2 I}{r} (\sqrt{2} + \frac{π}{4})$ .... $[∵ \frac{μ_{0}}{4 π} = 10^{- 7}]$

Question 265

Wired Faculty App

Assertion: Magnetic field is useful in producing parallel beam of charged particle.
Reason: Magnetic field inhibits the motion of
charged particle moving across it.

If both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true but reason is not the correct explanation of assertion.

If assertion is true but reason is false.

If both assertion and reason are false.

Solution

If both assertion and reason are true and reason is the correct explanation of assertion.

Charged particles move freely along magnetic field lines, but are inhibited by the magnetic force from moving across field lines.

Question 266

Wired Faculty App

What is the magnetic field on the axis of a coil of radius r carrying current I at a distance R from the origin?

$\frac{μ_{0} I r^{2}}{2 {(r^{2} + R^{2})}^{\frac{3}{2}}}$

$\frac{μ_{0} {IR}^{2}}{2 {(r^{2} + R^{2})}^{\frac{3}{2}}}$

$\frac{μ_{0} I}{2 (r + R)}$

$\frac{μ_{0} IR}{2 {(r^{2} + R^{2})}^{\frac{1}{2}}}$

Solution

$\frac{μ_{0} I r^{2}}{2 {(r^{2} + R^{2})}^{\frac{3}{2}}}$

Consider a circular coil of radius r, carrying a current I. Consider a point P, which is at a distance X from the centre of the coil. We can consider that the loop is made up of a large number of short elements, generating small magnetic fields. So the total field at P will be the sum of the contributions from all these elements. At the centre of the coil the field will be uniform. As the location of the point increases from the centre of the coil, the field decreases.

By Biot-Savart's law, the field dB due to a small element 'dl' of the circle, centred at A is given by

dB = $\frac{μ_{0} I}{2} \frac{r^{2}}{{(x^{2} + r^{2})}^{\frac{3}{2}}}$

This can be resolved into two components, one along the axis OP, ad other PS, which is perpendicular to OP. Ps get cancelled with PS'. So the magnetic field at a distance x away from the axis of circular coil of radius r is given by

B_x = $\frac{μ_{0} n I}{2} \frac{r^{2}}{{(x^{2} + r^{2})}^{\frac{3}{2}}}$

Question 267

Wired Faculty App

Assertion: When a magnetic dipole is placed in a non uniform magnetic field, only a torque acts on the dipole.
Reason: Force would act on dipole if magnetic field is uniform.

If both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true but reason is not the correct explanation of assertion

If assertion is true but reason is false.

If both assertion and reason are false.

Solution

If both assertion and reason are false.

When dipole is placed uniformly electric field and dipole vector direction is not parallel to field direction, each charge of dipole experiences a force. The magnitude of the two forces are equal but opposite in direction. These equal and unlike give rise to couple. This couple gives a torque to dipole, which rotates the dipole and makes it align to the direction of a field. Once the dipole is aligned in the direction of field, torque vanishes ( r × F will be zero, if r and F are in same direction). Once the dipole is aligned to electric field, the net force will be zero because they are in opposite direction. Hence in uniform field dipole experiences only torque.

When a dipole is placed in non-uniform field, there will be torque as explained above. But once the dipole is aligned to field direction, the forces acting on the charges are not same. Hence there will be a net force acting on the dipole in the direction of magnetic field. Hence in non-uniform field, electric dipole experiences both torque and force.

Physics Part I Chapter 4 Moving Charges And Magnetism

NCERT Solution For Class 12 Physics Physics Part I

A circular coil of wire consisting of 100 turns, each of radius 8.0 cm carries a current of 0.40 A. What is the magnitude of the magnetic field B at the centre of the coil?

A long straight wire carries a current of 35 A. What is the magnitude of the field B at a point 20 cm from the wire?

A long straight wire in the horizontal plane carries a current of 50 A in north to south direction. Give the magnitude and direction of B at a point 2.5 m east of the wire.

A horizontal overhead power line carries a current of 90 A in east to west direction. What is the magnitude and direction of the magnetic field due to the current 1.5 m below the line?

What is the magnitude of magnetic force per unit length on a wire carrying a current of 8 A and making an angle of 30° with the direction of a uniform magnetic field of 0.15 T?

Two long and parallel straight wires A and B carrying currents of 8.0 A and 5.0 A in the same direction are separated by a distance of 4.0 cm. Estimate the force on a 10 cm section of wire A.

A closely wound solenoid 80 cm long has 5 layers of windings of 400 turns each. The diameter of the solenoid is 1.8 cm. If the current carried is 8.0 A, estimate the magnitude of B inside the solenoid near its centre.

In Question 4.11 obtain the frequency of revolution of the electron in its circular orbit. Does the answer depend on the speed of the electron?

Answer the following questions:An electron travelling west to east enters a chamber having a uniform electrostatic field in north to south direction. Specify the direction in which a uniform magnetic field should be set up to prevent the electron from deflecting from its straight line path.

The wires which connect the battery of an automobile to its starting motor carry a current of 300 A (for a short time). What is the force per unit length between the wires if they are 70 cm long and 1.5 cm apart? Is the force attractive or repulsive?

A galvanometer coil has a resistance of 12 Ω and the metre shows full scale deflection for a current of 3 mA. How will you convert the metre into a voltmeter of range 0 to 18 V?

A galvanometer coil has a resistance of 15 Ω and the metre shows full scale deflection for a current of 4 mA. How will you convert the metre into an ammeter of range 0 to 6 A?

An electron does not suffer any deflection while passing through a region of uniform magnetic field. What is the direction of the magnetic field?

What are the units of magnetic permeability?

Magnetic field lines can be entirely confined within the core of a toroid, but not within a straight solenoid. Why?

The coils, in certain galvanometers, have a fixed core made of a non-magnetic matallic materials.Why does the oscillating coil come to rest so quickly such a core?

Two circular coils made of similar wires but of radii 20 cm and 40 cm are connected in parallel.What will be the ratio of the magnetic field at their centres?

What is the direction of the force acting on a charged particle q, moving with a velocity v→ in a uniform magnetic field B→?

Magnetic lines of force are endless. Comment.

An electron is moving along +ve x-axis in the presence of uniform magnetic field along +ve y-axis. What is the direction of force acting on it?

A magnetic dipole is situated in the direction of a magnetic field. What is its potential energy? If it is rotated by 180°, then what amount of work will be done?

Define magnetic flux. Give its SI unit.

What is the approximate distance upto which earth's magnetic field extends?

An electron moving through a magnetic field does not experience any force. Under what condition is this possible?

An electron moving with a velocity of 107 m/s enters a uniform magnetic field of 1 T along a direction parallel to the field. What would be its trajectory?

Write one condition under which an electric charge does not experience a force in a magnetic field.

What is a shunt? State its SI unit.

Write SI unit of magnetic field B→.

A charge q is moving in a region where both the magnetic field B→ and electric field E→ are simultaneously present. What is the Lorentz force acting on the charge?

In the diagram below is shown a circular loop carrying current I. Show the direction of the magnetic field with the help of lines of force.

The force F→experienced by a particle of charge ‘q’ moving with a velocity ‘v’ in a magnetic field ‘B’ is given by F→ = qv→ × B→. Which pairs out of these vectors are always at right angles to each other?

A certain proton moving through a magnetic field region experiences maximum force. When does this occur?

Under what conditions is the force acting on a charge moving through a uniform magnetic field minimum?

What is the nature of magnetic field in a moving coil galvanometer?

Two wires of equal lengths are bent in the form of two loops. One of the loops is square shaped whereas the other loop is circular. These are suspended in a uniform magnetic field and the same current is passed through them. Which loop will experience greater torque? Give reasons.

Write the SI unit of (i) magnetic pole strength (ii) magnetic dipole moment of a bar magnet.

Give two factors by which the current sensitivity / voltage sensitivity of a moving coil galvanometer can be increased.

A beam of protons with a velocity 4 x 105 m/s enters a uniform magnetic field of 0.3 T at an angle 60° to the magnetic field. Find the radius of the helical path taken by the proton beam. Also find the pitch of the helix mp = 1.67 x 10–27 kg.

Three long straight and parallel wires, carrying currents, are arranged as shown in the figure below. Find the force experienced by a 25 cm length of wire C.

The wire shown in figure below carries a current of 60 A. Find the magnetic field B at P.

What is the toroid? Using Ampere's circuital law calculate the magnetic field inside the toroid.

A 50 turn coil as shown in the figure below carries of 2 A in a magnetic field B = 0.25 Wb m –2. Find the torque acting on the coil. In what direction will it rotate?

Increasing the current sensitivity may not necessarily increase the voltage sensitivity of galvanometer. Justify.

In an exercise to increase current sensitivity of a galvanometer by 25%, its resistance is also increased 1.5 times. How will the voltage sensitivity of the meter be affected ?

A charge ‘q’ moving along the -X-axis with a velocity v→ is subjcted to a uniform magnetic field B acting along the Z-axis as it crosses the origin O. (i) Trace its trajectory.(ii) Does the charge gain kinetic energy as it enters the magnetic field? Justify your answer.

Two straight wires A and B of lengths 10 m and 12 m carrying currents of 4.0 A and 6.0 A respectively in opposite directions lie parallel to each other at a distance of 0.03 m. Estimate the force on a 15 cm section of the wire B near its centre.

Derive an expression for the torque acting on a loop of N turns, area A, carrying current I, when held in a uniform magnetic field.With the help of circuit, show how a moving coil galvanometer can be converted into an ammeter of given range. Write the necessary mathematical formula.

Draw the field lines of (a) a bar magnet (b) a current carrying finite solenoid, and (c) an electric dipole.What basic difference do you notice between the magnetic and electric field lines? How do you explain this difference?

Two long parallel straight wires X and Y separated by a distance of 5 cm in air carry currents of 10 A and 5 A respectively in opposite directions. Calculate the magnitude and direction of the force on a 20 cm length of the wire Y.

A charge 2Q is spread uniformly over an insulated ring of radius R/2. What is the magnetic moment of the ring if it is rotated with an angular velocity ω with respect to normal axis?

An electron of 45 eV energy is revolving in a circular path in a magnetic field of intensity 9 x 10–5 Wb m–2. Determine (i) the speed of the electron (ii) radius of the circular path.

State Biot-Savart law. A current I flows in a conductor placed perpendicular to the plane of the paper. Indicate the direction of the magnetic field due to a small element dl→ at point P situated at a distance r→ from the the element as shown in the figure.

If electron velocity is 2i^+3j^ and it is subjected to magnetic field of 4k^, then its path will changespeed will changeboth (a) and (b) none of the above

An electron having energy 10 eV is circulating in path having magnetic field of 10–4T. The speed of the electron will be 3.8 x 106 m/s 1.9 x 1012 m/s3.8 x 1012 m/s1.9 x 106 m/s

A particle having charge 100 times that of an electron is revolving in a circular path by radius 0.8 m with one rotation per second. The magnetic field produced at the centre is 10–17 μ0 1017 μ010–16 μ010–15 μ0

A particle of mass m and charge q is placed at rest in a uniform electric field E and then released. The kinetic energy attained by the particle after moving a distance y is qEy2 qE2yqEyq2Ey

A galvanometer of resistance 25 Ω is shunted by a 2.5 Ω wire. The part of total current that flows through the galvanometer is given by II0 = 111II0 = 211II0 = 311II0 = 411

The resistance of a galvanometer is 50 Ω and the current required to give full scale deflection is 100 A. In order to convert it into an ammeter for reading up to 10 A, it is necessary to put a resistance of 5 x 10–32Ω5 x 10–4 Ω5 x 10–5 Ω5 x 10–2 Ω

The resistance of the coil of ammeter is R. The shunt resistance required to increase its range four fold should have a resistance R/3R/44RR/5

Two long parallel wires P and Q are held perpendicular to the plane of the paper with distance of 5 m between them. If P and Q carry current of 2.5 A and 5A respectively in the same direction, then the magnetic field at a point half way between the wire is μ0/π3μ0/πμ0/2πμ0/2π

A wire of length 2m carries a current of 1 ampere is bend to form a circle. The magnetic moment of the coil is 2ππ/2π/41/π

The magnetic field of given length of wire for single turn coil at its centre is B, then its value for two turns coil for the same wire is B/4 B/24B2B

When a charged particle enters in a uniform magnetic field, its kinetic energy. remains constantincreasesdecreasesbecomes zero

The deflection in moving coil galvanometer falls from 50 divisions to 10 divisions, when a shunt of 12Ω is applied, the resistance of galvanometer coil is 12 Ω 24 Ω48 Ω50 Ω

If the number of turns, area and current through a coil is given by n, A and I respectively, then its magnetic moment will be nIA n2IAnIA2nI/A

The scale of a galvanometer of resistance 100 ohm contains 25 divisions. It gives a deflection of one division on passing a current of 4 x 10–4 A. The resistance in ohms to be added to it, so that it may become a voltmeter of range 2.5 volt is 100150250300

The electric current in a circular coil of two turns produces a magnetic induction of 0.2 T at its centre. The coil is unwound and is rewound into a circular coil of four turns. The magnetic induction at the centre of the coil now is, in tesla (if same current flows in the coil) 0.20.40.60.8

State Biot-Savart law Using Biot-Savart law, derive an expression for the magnetic field at the centre of a circular coil of number of turns ‘N’, radius ‘r’ carrying a current.

A proton and an α-Particle of same kinetic energy in turn move through a uniform magnetic field B in a plane normal to the field. Compare the radii of the paths of the two particles.

Define the SI unit of magnetic field. “A charge moving at right angles to a uniform magnetic field does it undergo change in kinetic energy”. Why?

Give the principle of a cyclotron. Draw a labelled diagram of cyclotron and describe how a positively charged particle is accelerated in it?

Obtain an expression for the magnetic moment associated with the orbital motion of an electron.

What is the effect of current flowing in the same direction in two long straight and parallel conductors? Give the definition of ‘ampere’ based on this effect.

Derive the expression for the torque on a rectangular coil of area A, carrying a current I, placed in a magnetic field B. The angle between the direction of B and the vector perpendicular to the plane of the coil is 0.

What is the magnetic moment associated with a coil of 1 turn, area of cross-section 10–4m2 carrying a current of 2A?

An electron is projected with a speed of 105 ms–1 at right angles to a magnetic field of 0.019 G. Calculate the radius of the circle described by the electron.

Draw a labelled diagram of a moving coil galvanometer, Explain the principle on which it works. Deduce an expression for the torque acting on a rectangular current carrying loop kept in uniform magnetic field. Write two factors on which the current sensitivity of a moving coil galvanometer depend.

Two identical conducting wires AOB and COD are placed at right angles to each other. The wire AOB carries an electric current I1 and COD carries a current I2. What will be the magnetic field on point lying at a distance d from O, in a direction perpendicular to the plane of the wires AOB and COD?

Answer the following questions:
An electron travelling west to east enters a chamber having a uniform electrostatic field in north to south direction. Specify the direction in which a uniform magnetic field should be set up to prevent the electron from deflecting from its straight line path.

The coils, in certain galvanometers, have a fixed core made of a non-magnetic matallic materials.
Why does the oscillating coil come to rest so quickly such a core?

Two circular coils made of similar wires but of radii 20 cm and 40 cm are connected in parallel.
What will be the ratio of the magnetic field at their centres?

What is the direction of the force acting on a charged particle q, moving with a velocity $\vec{v}$ in a uniform magnetic field $\vec{B} ?$

Write SI unit of magnetic field $\vec{B} .$

A charge q is moving in a region where both the magnetic field $\vec{B}$ and electric field $\vec{E}$ are simultaneously present. What is the Lorentz force acting on the charge?

The force $\vec{F}$ experienced by a particle of charge ‘q’ moving with a velocity ‘v’ in a magnetic field ‘B’ is given by $\vec{F} = q (\vec{v} \times \vec{B}) .$ Which pairs out of these vectors are always at right angles to each other?

A beam of protons with a velocity 4 x 10⁵ m/s enters a uniform magnetic field of 0.3 T at an angle 60° to the magnetic field. Find the radius of the helical path taken by the proton beam. Also find the pitch of the helix mp = 1.67 x 10^–27 kg.

A 50 turn coil as shown in the figure below carries of 2 A in a magnetic field B = 0.25 Wb m ^–2. Find the torque acting on the coil. In what direction will it rotate?

A charge ‘q’ moving along the -X-axis with a velocity $\vec{v}$ is subjcted to a uniform magnetic field B acting along the Z-axis as it crosses the origin O.

(i) Trace its trajectory.
(ii) Does the charge gain kinetic energy as it enters the magnetic field? Justify your answer.

Derive an expression for the torque acting on a loop of N turns, area A, carrying current I, when held in a uniform magnetic field.
With the help of circuit, show how a moving coil galvanometer can be converted into an ammeter of given range. Write the necessary mathematical formula.

Draw the field lines of (a) a bar magnet (b) a current carrying finite solenoid, and (c) an electric dipole.
What basic difference do you notice between the magnetic and electric field lines? How do you explain this difference?

An electron of 45 eV energy is revolving in a circular path in a magnetic field of intensity 9 x 10^–5 Wb m^–2. Determine (i) the speed of the electron (ii) radius of the circular path.

State Biot-Savart law. A current I flows in a conductor placed perpendicular to the plane of the paper. Indicate the direction of the magnetic field due to a small element $\vec{dl}$ at point P situated at a distance $\vec{r}$ from the the element as shown in the figure.

If electron velocity is $2 \hat{i} + 3 \hat{j}$ and it is subjected to magnetic field of $4 \hat{k},$ then its
path will change
speed will change
both (a) and (b)
none of the above

An electron having energy 10 eV is circulating in path having magnetic field of 10^–4T. The speed of the electron will be
3.8 x 10⁶ m/s
1.9 x 10¹² m/s
3.8 x 10¹² m/s
1.9 x 10⁶ m/s

A particle having charge 100 times that of an electron is revolving in a circular path by radius 0.8 m with one rotation per second. The magnetic field produced at the centre is
10^–17 μ₀
10¹⁷ μ₀
10^–16 μ₀
10^–15 μ₀

A particle of mass m and charge q is placed at rest in a uniform electric field E and then released. The kinetic energy attained by the particle after moving a distance y is
qEy²
qE²y
qEy
q²Ey

A galvanometer of resistance 25 Ω is shunted by a 2.5 Ω wire. The part of total current that flows through the galvanometer is given by
$\frac{I}{I_{0}} = \frac{1}{11}$
$\frac{I}{I_{0}} = \frac{2}{11}$
$\frac{I}{I_{0}} = \frac{3}{11}$
$\frac{I}{I_{0}} = \frac{4}{11}$

The resistance of a galvanometer is 50 $Ω$ and the current required to give full scale deflection is 100 A. In order to convert it into an ammeter for reading up to 10 A, it is necessary to put a resistance of
5 x 10^–32Ω
5 x 10^–4 Ω
5 x 10^–5 Ω
5 x 10^–2 Ω

The resistance of the coil of ammeter is R. The shunt resistance required to increase its range four fold should have a resistance
R/3
R/4
4R
R/5

A wire of length 2m carries a current of 1 ampere is bend to form a circle. The magnetic moment of the coil is
$2 π$
$π / 2$
$π / 4$
$1 / π$

The magnetic field of given length of wire for single turn coil at its centre is B, then its value for two turns coil for the same wire is
B/4
B/2
4B
2B

When a charged particle enters in a uniform magnetic field, its kinetic energy.
remains constant
increases
decreases
becomes zero

The deflection in moving coil galvanometer falls from 50 divisions to 10 divisions, when a shunt of 12 $Ω$ is applied, the resistance of galvanometer coil is
12 Ω
24 Ω
48 $Ω$
50 $Ω$

If the number of turns, area and current through a coil is given by n, A and I respectively, then its magnetic moment will be
nIA
n²IA
nIA²
$nI / \sqrt{A}$

The scale of a galvanometer of resistance 100 ohm contains 25 divisions. It gives a deflection of one division on passing a current of 4 x 10^–4 A. The resistance in ohms to be added to it, so that it may become a voltmeter of range 2.5 volt is
100
150
250
300

The electric current in a circular coil of two turns produces a magnetic induction of 0.2 T at its centre. The coil is unwound and is rewound into a circular coil of four turns. The magnetic induction at the centre of the coil now is, in tesla (if same current flows in the coil)
0.2
0.4
0.6
0.8

A proton and an $α$ -Particle of same kinetic energy in turn move through a uniform magnetic field B in a plane normal to the field. Compare the radii of the paths of the two particles.

What is the magnetic moment associated with a coil of 1 turn, area of cross-section 10^–4m² carrying a current of 2A?

An electron is projected with a speed of 10⁵ ms^–1 at right angles to a magnetic field of 0.019 G. Calculate the radius of the circle described by the electron.

A charge ‘q’ moving along the X-axis with a velocity u is subjected to a uniform magnetic field B acting along z-axis as it crosses the origin.
(i) Trace it trajectory.
(ii) Does the charge gain kinetic energy as it enters the magnetic field? Justify your answer.

A galvanometer of resistance 50 Ω gives full scale deflection for a current of 0.05 A. Calculate the length of the shunt wire required to convert the galvanometer into an ammeter of range 0 to 5 A. The diameter of the shunt wire is 2 mm and its resistivity is 5 x 10^–7 Ω m.

Two circular loops of radii r and 2r have current I and I/2 flowing through them clockwise and anticlockwise sense respectively. If their equivalent magnetic moments are M₁ and M₂ respectively, state the relation between M₁ and M₂.

How does the mutual inductance of a pair of coils change when,

(i) distance between the coils is increased and

(ii) number of turns in the coils is increased?

Write the expression, in a vector form, for the Lorentz magnetic force $straight F with rightwards harpoon with barb upwards on top$ due to a charge moving with velocity $straight V with rightwards arrow on top$ in a magnetic field B. What is the direction of the magnetic force?

A current is induced in coil C₁ due to the motion of current carrying coil C₂.

(a) Write any two ways by which a large deflection can be obtained in the galvanometer G.
(b) Suggest an alternative device to demonstrate the induced current in place of galvanometer.

State Biot-Savart law, giving the mathematical expression for it.

Use this law to derive the expression for the magnetic field due to a circular coil carrying current at a point along its axis.
How does a circular loop carrying current behave as a magnet?