Sponsor Area
From the relation,
we find that an extremely minute field bends charged particles in a circle of very large radius. Over a small distance, the deflection due to the circular orbit of such large radius may not be noticeable, but over the gigantic interstellar distances, the deflection can significantly affect the passage of charged particles, e.g., cosmic rays.
Given,
Angle between magnetic moment and magnetic field,
External magnetic field, B= 0.25 T
Torque, = 4.5 x 10–2 J
Using formula
where, m is the magnetic pole strength.
A bar magnet of magnetic moment 1.5 JT–1 lies aligned with the direction of a uniform magnetic field of 0.22 T.
(a) What is the amount of work required by an external torque to turn the magnet so as to align its magnetic moment: (i) normal to the field direction, (ii) opposite to the field direction?
(b) What is the torque on the magnet in cases (i) and (ii)?
Given,
Magnetic moment of bar magnet, M = 1.5 JT–1
Uniform magnetic field, B = 0.22 T
(a) Work done on magnet
(i) Amount of work required to turn the magnet normal to the field direction is,
W1 = – MB [cos 90° – cos 0°]
= + MB
= MB = 1.5 x 0.22
= 0.33 N-m
(ii) Amount of work required to turn the magnet in a direction opposite to the field direction is,
W2 = – MB [ cos 180° - cos 0°]
= 2MB
= – MB [cos θ2 – cos θ1]
= 2 MB = 2 x 0.33
= 0.66 J
(b) Torque acting on the magnet.
(i) τ = MB sin 90°
= MB
= 0.33 J.
It works in the direction that tends to align the magnetic moment vector along B.
(ii) Torque acting on the magnet so as to align it's magnetic moment in a direction opposite to the field direction.
τ = MB sin θ
Here, θ = 180°
τ = 1.5 x 0.22 x sin 180°
= 1.5 x 0.22 x 0
= 0.
Sponsor Area
A closely wound solenoid of 2000 turns and area of cross-section 1.6 x 10–4 m2, carrying a current of 4.0 A, is suspended through its centre allowing it to turn in a horizontal plane.
(a) What is the magnetic moment associated with the solenoid?
(b) What is the force and torque on the solenoid if a uniform horizontal magnetic field of 7.5 x 10–2 T is set up at an angle of 30° with the axis of the solenoid?
Given,
Number of turns in the coil, N = 16
Radius of the coil, r = 10 cm = 0.1 m
Current passing thropugh the coil, I = 0.75 A
External magnetic field,
Frequency of oscillation,
Magnetic moment is given by, M= NIA
Therefore,
M = NI
=
= 0.377 JT-1
Now, using formula
Putting values, we get
Tips: -
Here,
δ
Horizontal component of earth's magnetic field is,
Since,
which, is the required earth's magnetic field.
Given,
Declination,
Angle of dip, δ
Horizontal component of earth's magnetic field is,
Since,
The earth's field lies in a vertical plane 12° west of geographic meridian at an angle of 60° above the horizontal.
Given,
Magnetic moment of the bar magnet, M= 0.48 JT-1
Distance from the centre to the point where magnetic field is produced, r = 10 cm = 0.1 m
(i) Magnetic field at its axis
Putting values, we get
(ii) Magnetic field along the equatorial line
Therefore, putting the values we get,
The induced dipole moment in a diamagnetic sample is always opposite to the magnetising field, no matter what the internal motion of the atoms is. Hence, the alignment of dipole has got nothing to do with the temperature.
Magnetisation of a ferromagnet is not a single-valued function of the magnetising field. Its magnetisation value for a particular field depends both on the field and also on history of magnetisation (i.e., how many cycles of magnetisation it has gone through etc.).
In other words, the value of magnetisation is a record or ‘memory’ of its cycles of magnetisation. If information bits can be made to correspond to these cycles, the system displaying such a hysteresis loop can act as a device for storing information.
Here,
Current, I = 2.5 A.
Earth's magnetic field, B = 0.33G = 0.33 x 10–4 T
Angle of dip, δ = 0°
Horizontal component of earth’s field
H = B cos δ
= 0.33 x 10–4 cos 0°
= 0.33 x 10–4 Tesla
Let the neutral points lie at a distance r from the cable.
Strength of magnetic field on this line due to
current in the cable
At neutral point,
Hence, neutral points lie on a straight line parallel to the cable at a perpendicular distance of 1.5 cm above the plane of the paper.
Given,
Earth’s magnetic field, Be = 0.39 G
Angle of dip, δ = 35°
Number of wires, n = 4
declination = 0
distance , r = 4 cm = 4
Current carried by the cable, I = 1 A
∴ Horizontal component of earth’s magnetic field is given by,
BH = Be cos δ
BH = 0.39 cos 35°
= 0.3195 G
Vertical component of earth's magnetic field, Bv = Be sin δ
Bv = 0.39 sin 35°
= 0.224 G
Magnetic field produced by telephone cable having 4 wires is,
Now,
We will find the resultant field below the cable.
As per the right hand thumb rule, the direction of B’ will be opposite to BH at a point below the cable.
Therefore at a point 4 cm below the cable, resultant horizontal component of earth’s field
RH = BH – B’
= 0.3195 – 0.2
= 0.1195 G
Resultant vertical component of earth’s field
Rv = Bv = 0.224 G (unchanged)
∴ Resultant of earth’s field
Now,
Resultant field above the cable is given as per right hand thumb rule which says that, at a point above the cable, B’ will be in the same direction as BH.
Hence, at a point 4 cm above the cable,
Resultant magnetic field is,
A compass needle free to turn in a horizontal plane is placed at the centre of circular coil of 30 turns and radius 12 cm. The coil is in a vertical plane making an angle of 45° with the magnetic meridian. When the current in the coil is 0.35 A, the needle points west to east.
(a) Determine the horizontal component of the earth's magnetic field at the location.
(b) The current in the coil is reversed, and the coil is rotated about its vertical axis by an angle of 90° in the anticlockwise sense looking from above. Predict the direction of the needle. Take the magnetic declination at the places to be zero.
(a)
Number of turns, N = 30
Current in the coil, I = 0.35 A
Radius of the circular coil, r = 12 cm = 0.12 m.
The magnetic field at the centre of the coil is
It acts in a direction perpendicular to the plane of the coil.
Its component parallel to the magnetic meridian is
The component perpendicular to the magnetic meridian is .
Given that the needle points in the west-east direction.
Horizontal component of earth's magnetic field is given by
(b) When the current in the coil is reversed and coil is turned through 90o anticlockwise, the plane of the coil makes an angle of 45° with the magnetic meridian on the other side. Thus, needle will rotate and will set in east to west direction.
Sponsor Area
Given, a magnetic dipole is under the influence of two magnetic fields.
Angle between the field direction, θ = 60°
Magnitude of one field, B1 = 1.2 x 10–2 T
Angle at which the dipole comes to stable equilibrium, θ1 = 15°
Therefore, θ2 = 60° – 15° = 45°.
In equilibrium, torques due to two fields must balance each other.
i.e., τ1 = τ2
M B1 sin θ1 = MB2 sin θ2
which is the required magnitude of the other field.
Given,
Number of atomic dipoles, n = 2.0 x 1024
Magnetic dipole moment of each dipole = 1.5 x 10–23 J T–1
∴ Total possible magnetic dipole moment of the sample, M = 1.5 x 10–23 x 2.0 x 1024
= 30 J T–1
Magnetic field, B1 = 0.64 T
At temperature of 4.2 K, the magnetic saturation is 15%.
∴ Dipole moment achieved at 4.2 K = 15% of M
Hence,
According to Curie's law,
Here,
and
is the magnetic dipole moment at 0.92 K .
A monoenergetic (18 keV) electron beam initially in the horizontal direction is subjected to a horizontal magnetic field of 0.40 G normal to the initial direction. Estimate the up or down deflection of the beam over a distance of 30 cm. Given mass of electron 9.11 x 10–31 kg and charge on electron = 1.6 x 10–19 C.
[Note: Data in this exercise are so chosen that the answer will give you an idea of the effect of earth’s magnetic field on the motion of electron beam from electron gun to the screen in a T.V. set.]
Here,
Energy of the electron beam, E = 18 keV =
Horizontal magnetic field, B = 0.40 G = 0.40
distnace, 
As,
Energy,
In a magnetic field, electron beam is deflected along a circular arc of radius r, such that
In
By using binomial expansion,
Given,
Horizontal component, BH = 0.22 T
Vertical component, BV = 0.38 T
Therefore,
Resultant magnetic field,
Given,
Since,
Therefore, angle of dip = 45o .
Each piece is a magnet with reduced pole strength and reduced magnetic moment.
(i) Pole strength of each new magnet becomes half of original magnet.
(ii) Magnetic moment (M = M . 2l) becomes halved because, length is also halved.
The magnetic moment vectors μs and μl associated with the intrinsic spin angular momentum S and orbital angular momentum I, respectively, of an electron are predicted by quantum theory (and verified experimentally to a high accuracy) to be given by:
μs = – (e/m) S,
μl = – (e/2m) 1
Which of these relations is in accordance with the result expected classically? Outline the derivation of the classical result.
Sponsor Area
The material used for making permanent magnets should have high coercivity so that the magnetisation is not erased by stray magnetic fields, temperature fluctuations or minor mechanical damage.
Two properties for making permanent magnets are:
i) high coercivity
ii)high retentivity
Steel is a suitable material to make permanent magnet.
Bring both the bars closer to each other and, if in case there is repulsion in between the bars, then both bars are magnetised. If the bars get attracted to each other, then only 1 bar is magnetised.
To identify which bar is bar magnet:
Lay bar A on the table and hold bar B in your hand with it's one end placed in the middle of bar A. If the bars tend to get attracted then we can say that bar B (in hand) is magnetised and is the bar magnet.
Angle of dip is defined as that angle which the direction of total strength of earth's magnetic field makes with a horizontal line in magnetic meridian. It is represented by .
Let,
Be be the Earth's magnetic field,
is the angle of dip.
V is the vertical component of earth's magnetic field.
H is the horizontal component of earth's magnetic field
Now, resolving Be along horizontal and vertical direction we get,
cos =
which, is the required relation.
When horizontal component BH is equal to vertical component BV of Earth's magnetic field then,
where, is the angle of dip.
Properties:
1. Diamagnetic :
i) magnetic susceptibilty is independent of temperature.
ii) magnetic permeability has value less than 1.
iii) repelled by a strong magnet.
2. Paramagnetic:
i) magnetic susceptibility decreases with rise of temperature.
ii) magnetic permeabilty has value slightly greater than 1.
iii) attracted by a strong magnet.
3. Ferromagnetic:
i) magnetic susceptibility decreases with increase in temperature.
ii) magnetic permeabilty is very large.
iii) highly attracted by a strong magnet.
Magnetic elements of earth at a place are the quantities which describe completely in in magnitude as well as direction, the magnetic field of earth at that place.
The three magnetic elements of earth are:
1. Magnetic declination ( )
2. Magnetic inclination or magnetic dip (), and
3. Horizontal component (H). 
Given,
Horizontal component of Earth's magnetic field = 0.16 G
Using the formula,
which is the required magnitude of earth.
Given,
Magnetic moment, M = 1.0 x 104 J/T
Magnetic field,
Using the formula for work done,
Here,
Length of magnet, 2l = 10 cm = 0.1 m
Magnetic moment, m = 12 Am
Distance of point from the mid point of axial line, d = 20 cm =0.2 m
Magnetic field on the axial line,
i.e.,
At the same distance, on equatorial line, we have
If χ stands for the magnetic susceptibility of a given material, identify the class of materials for which:
(i) – 1 ≥ χ < 0
(ii) 0 < χ < ε, (ε stands for a small positive number).
(a) Write the range of relative magnetic permeability of these materials.
(b) Draw the pattern of the magnetic field lines when these materials are placed in an external magnetic field.
(i) For – 1 ≥ χ < 0 i.e., magnetic susceptibilty is negative, the material is diamagnetic.
(ii) For 0 < χ < ε, material is paramagnetic.
(a) Range of relative magnetic permeability of diamagnetic material is 0 ≤ μr < 1.
Range of relative magnetic permeability of paramagnetic material is 0 < μr < 1 + ε.
(b) Fig. (1) shows the behaviour of magnetic field lines when diamagnetic material is placed in an external field.
Fig. (2) shows the behaviour of magnetic field lines when paramagnetic material is placed in an external field.
Here,
Horizontal component,
Angle of dip,
Using the formula,
Putting values, we get
which is the required vertical component of Earth's magnetic field.
Using the below formula we can find the value of resulatant magnetic field.
(i) Magnetic declination at a place may be defined as the angle between its magnetic meridian and the earth geographic meridian at the place.
(ii) Angle of dip at a place is defined as the angle between the direction of intensity of earth’s magnetic field (BE) and the horizontal direction in magnetic meridian at that place.
(i) Identify the two specimens X and Y.
(ii) State the reason for the behaviour of the field lines in X and Y.
(i) X is a diamagnetic substance.
Y is a paramagnetic substance.
(ii) For a diamagnetic substance, the field lines are repelled or expelled and the field inside the material is reduced when a diamagnetic bar is placed in an external field.
For a paramagnetic substance, the field lines get concentrated inside the material and the field is increased when a paramagnetic bar is placed in an external field.
1. Properties of diamagnetic materials:
(i) They are feebly repelled by magnets.
(ii) Susceptibilty is negative and very small.
2. Properties of paramagnetic materials:
(i) They are feebly attracted by magnets
(ii) Susceptibility is positive and small.
(i) Intensity of magnetisation is a small negative value for diamagnetic substance and, large positive value for a ferromagnetic substance.
(ii) In a non-uniform magnetic field, a diamagnetic substance tends to move from stronger to weaker part while the ferromagnetic substance tends to move from weaker to stronger part of the field.
(iii) For a diamagnetic substance, susceptibility is small negative value while, for a ferromagnetic substance, susceptibility is large positive value.
The actual length of a magnet is called the geometric length of the magnet.
The distance between the poles of a magnet is called the magnetic length of the magnet.
The geometric length of the magnet is nearly 8/7 times the magnetic length of the magnet.
(i) Given,
Magnetic field, B = 800 G
Torqua acting, = 0.016 Nm
Since, Torque acting on magnet,
[ 1G = 10-4T]
which, is the required magnetic moment.
(ii) Work done be an external force to move the magnet from stable to unstable position,
(iii) The displacement and the torque due to the magnetic field are in opposite direction.
So work done by the force due to the external magnetic field is WB = – 0.064 J.
Here,
Magnetic moment, M = 1.6 A m2
Neutral point is obtained at d1 = 20 cm = 0.2 m
Horizontal component of earth's magnetic field, H = ?
When north pole of magnet is pointing south, neutral points lie on axial line of magnet.
At neutral point,
Now,
When magnet is reversed i.e., north pole is pointing north, neutral points lie on equatorial line of the magnet.
We have to calculate d2 i.e., the position where is the neutral point is obtained.
As,
Given,
Number of turns in solenoid, n = 500
Current carried, I = 3A
Relative premeabiltiy of iron core,
Therefore,
Magnetic intensity,
Using the relation between susceptibility and permeability,
Also,
Magnetisation,
Magnetic field inside the core is given by,
i.e.,
If δ1 and δ2 be the angles of dip observed in two planes at right angles to each other and δ is the true angle of dip, then prove that
cot2 δ1 + cot2 δ2 = cot2 δ
Let, δ2 be the (apparent) dip in the second plane.
The angle made by this plane with the magnetic meridian will be (90° – θ).
Effective horizontal component in this plane is BH cos (90° – θ) i.e., BH sin θ.
The vertical component will be Bv only.
...(ii)
Squaring and adding equation (i) and (ii), we get
Hence proved.
(a) Draw diagrams to depict the behaviour of magnetic field lines near a ‘bar’ of:
(i) copper
(ii) aluminium
(iii) mercury, cooled to a very low temperature (4.2 K)
(b) The vertical component of the earth's magnetic field at a given place is times its horizontal component. If total intensity of earth's magnetic field at the place is 0.4 G find the value of:
(i) angle of dip
(ii) the horizontal component of earth's magnetic field.
(ii) Horizontal component of earth's magnetic field
BH = B cos δ
H = 0.4 cos 60°
Fig.(a)
(a)
From fig. (a), it is clear that null points are obtained on the normal bisector when the magnet’s north and south poles face magnetic north and south respectively.
Magnetic field on the normal bisector at a distance r from the centre is given by
provided r is much greater than the length of the magnet. [The above equation is strictly true only for a point dipole].
At a null point, this field is balanced by the earth's field.
Earth's magnetic field,B = 0.38 G
Angle of dip = 0o
So,
...(1)
Since, angle of dip = 0, therefore, Bv = 0 and the horizontal component of the earth’s field equals the field itself.
Next, magnetic field due to a magnet on its axis at point distant r from the centre is given by
...(2)
provided r is much greater than the length of the magnet. (The above equation is strictly true only for a point dipole).
From fig. (a), it is clear that on the axis, this field adds up to the earth’s field.
Thus, the total field at a point on the axis has a magnitude equal to Ba + BH
i.e., ...(3)
and direction along (magnetic moment) [which is parallel to the earth's field in case of (a)].
Thus, for the same distance on the axis as the distance of the null point, the total field is found using equations (1) and (3).
That is,
B = 2 x 0.38 x 10–4 + 0.38 x 10–4
= 3 x 0.38 x 10–4
= 1.14 x 10–4 T
This field is directed along
Note: We did not require the given value of 12.5 cm for the nullpoint distance, except in so far that this was assumed to be much greater than the length of the magnet.
(b) When the bar is turned around by 180°, the magnet’s north and south poles face magnetic south and north respectively i.e., in this case, is antiparallel to the earth's field. 
Fig. (b)
From fig. (b), it is clear that the nullpoints now lie on the axis of the magnet at a distance r’ given by Ba’ = BH
...(4)
Comparing equations (4) and (1), we get
Therefore,
(a) Material A is diamagnetic.
Material B is paramagnetic.
Because, susceptibility, .
The slope of the given graph would give us the susceptibility. Slope of A is less than B. Therefore, A is diamagnetic and B is paramagnetic.
(b) Material B is paramagnetic and has a greater value of susceptibility because, paramagnetic substances have a tendency to pull in magnetic field lines when placed in a magnetic field.
Here,
Number of turns, N = 100
Radius of the coil, r = 0.05 m
Current carried by the coil, I = 0.1 A
External magnetic field, B = 1.5 T
is the angle between B and A of the coil.
Magnetic moment is given by, M = NIA
Amount of work required to turn the coil in an external magnetic field,
Each atom of an iron bar (5 cm x 1 cm x 1 cm) has a magnetic moment 1.8 x 10–23Am2.
(a) What will be the magnetic moment of the bar in the state of magnetic saturation.
(b) What will be the torque required to place this magnetised bar perpendicular to magnetic field of 15000 gauss?
Density of iron = 7.8 x 103 kg/m3
Atomic weight of iron = 56
Avagadro’s number = 6.023 x 1023 gm/mole
Sponsor Area
Number of turns, N = 2
Radius of the coil, r = 0.10 m
Deflection, = 30o
H = 0.32 10-4 T
Before passing current, compass needle was along the direction of H.
On passing current, the needle shows a deflection of 30° (with H) under the combined effect of H and B.
Magnetic field due to current is in a direction perpendicular to the plane of the coil.
As compass needle is inclined equally to H and B, as in fig.
Therefore,
Magnet is suspended such that it oscillates in horizonal plane.
When, angle of dip is 30o , magnet performs 20 oscillations per minute.
and,
When angle of dip is 60o , it performs 15 oscillations per minute.
Frequency of oscillation is given by,
[ ]
Therefore,
Hence, the ratio of Earth's magnetic field at these two places is
Given,
Using the formula,
This implies,
which is the required value for apparent value of dip.
A wheel with 8 metallic spokes each 50 cm long is rotated with a speed of 120 rev/min in a plane normal to the horizontal component of the Earth’s magnetic field. The Earth’s magnetic field at the plane is 0.4 G and the angle of dip is 60°. Calculate the emf induced between the axle and the rim of the wheel. How will the value of emf be affected if the number of spokes were increased?
Emf induced in a rod of length ‘l’ with angular speed in a uniform magnetic field is given by, 
For earth’s magnetic field,
Horizontal component is given by, 
Vertical component is given by, 
Therefore, emf induced is given by, 
Emf induced is not dependent on the number of spokes.
Out of the two magnetic materials, 'A' has relative permeability slightly greater than unity while 'B' has less than unity. Identify the nature of the materials 'A' and 'B'. Will their susceptibilities be positive or negative?
A is a paramagnetic material because its permeability is greater than unity and its susceptibility is positive. The relative permeability of a paramagnetic material is between 
and susceptibility is 
For a diamagnetic material, the relative permeability lies between 0 ≤ μr < 1 and its susceptibility lies between −1< χ< 0.
Therefore, 'B' is a diamagnetic material and its susceptibility is negative.
This is because its relative permeability is less than unity.
The horizontal component of the earth’s magnetic field at a place is B and angle of dip is 60°.
What is the value of vertical component of earth’s magnetic field at equator?Vertical component of magnetic field at the equator is zero. At the equator, the direction of magnetic field is horizontal.
The susceptibility of a magnetic material is –2.6 × 10–5. Identify the type of magnetic material and state its two properties.
Negative susceptibility is the property of diamagnetic material.
Two properties of diamagnetic material are:
i) This material expels the magnetic field lines and do not obeys Curies Law.
ii) They have the tendency to move from stronger to weaker part of the external magnetic field.
The permeability of a magnetic material is 0.9983. Name the type of magnetic materials it represents.
Here, permeability is less than 1 i.e., 
.
So, magnetic material is diamagnetic.
A magnetic needle free to rotate in a vertical plane parallel to the magnetic meridian has its north tip down at 60° with the horizontal. The horizontal component of the earth's magnetic field at the place is known to be 0.4 G. Determine the magnitude of the earth's magnetic field at the place.
Given,
Angle of dip,
What ate permanent magnets? Give one example.
Substances that can retain their magnetism for a long duration of time at room temperature are called permanent magnets. They have high retentivity and coercivity.
Eg: Steel, cobalt
Show diagrammatically the behaviour of magnetic field lines in the presence of (i) paramagnetic and (ii) diamagnetic substances. How does one explain this distinguishing feature? 
A paramagnetic material tends to move from weaker to stronger regions of the magnetic field and hence increases the number of lines of magnetic field passing through it. Relative permeability is greater than 1. So, magnetic lines of force pass through the substance.
A diamagnetic material tends to move from stronger to weaker regions of the magnetic field and hence, decreases the number of lines of magnetic field passing through it. Relative permeability is 1 and magnetic lines of force do not pass through the substance.
Distinguishing feature is because of the difference in their relative permeability.
In what way is the behaviour of a diamagnetic material different from that of a paramagnetic, when kept in an external magnetic field?
Difference in the behaviour of a diamagnetic material and paramagnetic material:
| Diamagnetic | Paramagnetic |
| 1. Diamagnetic substance would move towards the weaker region of the magnetic field. | 1. Paramagnetic substance move towards the stronger region of the magnetic field. |
| 2. Diamagnetic substance is repelled by a magnet. | 2. Paramagnetic substance moves towards the magnet. |
| 3. Diamagnetic substance get aligned perpendicular to the field. | 3. Paramagnetic substance get aligned along the magnetic field. |
At a place, the horizontal component of earth's magnetic field is B and angle of dip are 60o. What is the value of a horizontal component of the earth's magnetic field at the equator?
The horizontal component of the electric field is given by
BH = Becos(θ)
where θ is the angle of dip at the given place
Be is teh net magnetic field
At θ = 60o
BH = Be cos(60) = Be x (1/2)
Be = 2BH = 2B (given BH =B)
Now horizontal component at equator
BH = Becos (θ)
angle of dip at equator is 0
BH = Becos(0) = Be = 2B
BH = 2B
The susceptibility of a magnetic material is 0.9853. Identify the type of magnetic material. Draw the modification of the field pattern on keeping a piece of this material in a uniform magnetic field.
The susceptibility of this material is between 0 and 1 so its a paramagnetic material.
Needles N1, N2 and N3 are made of a ferromagnetic, a paramagnetic and a diamagnetic substance respectively. A magnet when brought close to them will
attract all three of them
attract N1 and N2 strongly but repel N3
attract N1 strongly, N2 weakly and repel N3 weakly
attract N1 strongly, but repel N2 and N3 weakly
C.
attract N1 strongly, N2 weakly and repel N3 weakly
Two long conductors, separated by a distance d carry current I1 and I2 in the same direction. They exert a force F on each other. Now the current in one of them increased to two times and its direction reversed. The distance is also increased to 3d. The new value of the force between them is
−2F
F/3
−2F/3
-F/3
C.
−2F/3
Force acting between two current carrying conductors
where d = distance between the conductors,
l = length of each conductor
The length of a magnet is large compared to its width and breadth. The time period of its width and breadth. The time period of its oscillation in a vibration magnetometer is 2 s. The magnet is cut along its length into three equal parts and three parts are then placed on each other with their like poles together. The time period of this combination will be
2 s
2/3 s
2√3 s
2/√3 s
B.
2/3 s
The time period of oscillations of magnet
where I = moment of inertia of magnet = mL2/12 (m is being the mass of magnet)
M = pole strength × L When the three equal parts of magnet are placed on one another with their like poles together, then
The materials suitable for making electromagnets should have
high retentivity and high coercivity
low retentivity and low coercivity
high retentivity and low coercivity
low retentivity and high coercivity
B.
low retentivity and low coercivity
Electromagnets are made of soft iron. The soft iron has high retentivity and low coercivity.
The magnetic susceptibility is negative for,
a paramagnetic material only
ferromagnetic material only
paramagnetic and ferromagnetic materials
diamagnetic material only
D.
diamagnetic material only
The relation between the magnetic permeability and susceptibility os material.
i.e., 
... (i)
For diamagnetic susceptibility,
According to equation (i), the magnetic susceptibility (
) of diamagnetic substance will be negative.
In the case of para and ferromagnetic substances, diamagnetic susceptibility is positive.
A wire carrying current I has the shape as shown in the adjoining figure. Linear parts of the wire are very long and parallel to X -axis while semicircular portion of radius R is lying in a Y-Z plane. The magnetic field at point O is.
C.
The magnetic fiedl in the different regions is given by
A compass needle which is allowed to move in a horizontal plane is taken to a geomagnetic pole. it
will become rigid showing no movement
will stay in any position
will stay in the north-south direction only
will stay in the east-west direction only
B.
will stay in any position
It will stay in any position at geomagnetic north and south poles.
Following figures show the arrangement of bar magnets in different configurations. Each magnet has magnetic dipole moments m. Which configuration has highest net magnetic dipole moment?
C.
The direction of magnetic moment is from S to N
mnet = 
Net magnetic moment will be maximum if cos 
is maximum.
cos 
will be maximum when 
will be minimum. So, at 
mnet will be maximum.
Two identical long conducting wire AOB and COD are placed at right angle to each other, with one above other such that O is their common point for the two. The wires carry I1 and I2 currents, respectively. point P is lying at distance d from O along a direction perpendicular to the plane containing the wires. The magnetic field at the point P will be,
D.
A bar magnet of length l and magnetic dipole moment M is bent in the form of an arc as shown in figure, The new magnetic dipole moment will be
M
3M/π
2M/π
M/2
B.
3M/π
The magnetic moment, M=ml
From figure 
There are four light -weight rod samples A, B, C D separately suspended by threads. A bar magnet is slowly brought near each sample and the following observations are noted
(i) A is feebly repelled
(ii) B is feebly attracted
(iii) C is strongly attracted
(iv) D remains unaffected
C is a diamagnetic material
D is of a ferromagnetic material
A is of a non- magnetic material
B is of paramagnetic material
D.
B is of paramagnetic material
Paramagnetic materials will be feebly attracted, the magnetic material will be feebly repelled and ferromagnetic material will be strongly attracted.
A thin ring of radius R metre has charge q column uniformly spread on it. The ring rotates about its axis with a constant frequency of f revolution/s. The value of magnetic induction in Wb m-2 at the centre of the ring is
μoq f/ 2π R
μoq / f 2π R
μoq / 2 f R
μoq f / 2 R
D.
μoq f / 2 R
The magentic field at the centre of the circle
= 
Electromagnets are made of soft iron because soft iron has
low retenitivity and high coercive force
high retentivity and high coercive force
low retentivity and low coercive force
high retentivity and low coercive force
A.
low retenitivity and high coercive force
The material suitable for making electromagnets should high retentivity and low coercivity
A vibration magnetometer placed in magnetic meridian has a small bar magnet. The magnet executes oscillations with a time period of 2s in earth's horizontal magnetic field of 24 μT. When a horizontal field of 18 μT is produced opposite to the earth's field by placing a current-carrying wire, the new time period of magnet will be
1s
2s
3s
4s
B.
2s
Time period in vibration magnetometer
The magnetic moment of a diamagnetic atom is
much greater than one
one
equal to zero
equal to zero
D.
equal to zero
Atoms which have paired electrons have the magnetic moment zero. Because diamagnetism is the intrinsic property of every material and it is generated due to the mutual interaction between the applied magnetic field and orbital motion of electrons.
so, the magnetic moment of a diamagnetic atom is equal to zero.
Two identical bar magnets are fixed with their centres at a distance d apart. A stationary charge Q is placed at P in between the gap of the two magnets at a distance D from the centre O as shown in the figure.
The force on the charge Q is
zero
directed along OP
directed along PO
directed perpendicular to the plane of paper
A.
zero
The force of attraction on Q by the north pole of the first magnet is F. The force of attraction on the same charge by n the north pole of another magnet must be opposite to that of the first magent. Thus, they concel the effect of one other and hence force on the charge Q is zero.
A rectangular, a square a circular and an elliptical loop, all in the (x-y) plane, are moving out of a uniform magnetic field with a constant velocity,
The magnetic field is directed along the negative z- axis direction. The induced emf, during the passage of these loops, out the field region, will not remain constant for
the rectangular, circular and elliptical loops
the circular and the elliptical loops
only the elliptical loops
any of the four loops
B.
the circular and the elliptical loops
Area coming out per second from the magnetic field is not constant for elliptical and circular loops, so induced emf, during the passage of these loops, out of the field region will not remain constant for the circular and the elliptical loops.
Curie temperature is the temperature above which
ferromagnetic material becomes paramagnetic material
paramagnetic material becomes diamagnetic material
paramagnetic material becomes ferromagnetic material
ferromagentic material becomes diamagnetic material
A.
ferromagnetic material becomes paramagnetic material
Ferromagnetism decreases with rising in temperature. If we heat a ferromagnetic substance. If we heat a ferromagnetic substance, then at a definite temperature the ferromagnetic property of the substance suddenly disappears and the substance becomes paramagnetic.The temperature above which a ferromagnetic substance becomes paramagnetic is called the Curie temperature of the substance.
Above Curie temperature
a ferromagnetic substance becomes paramagnetic
a paramagnetic substance becomes diamagnetic
a diamagnetic substance becomes paramagnetic
a paramagnetic substance becomes ferromagnetic
A.
a ferromagnetic substance becomes paramagnetic
Ferromagnetism decreases with rise in temperature. If we heat a ferromagnetic substance, then at a definite temperature the ferromagnetic property of the substance suddenly disappears and the substance becomes paramagnetic. The temperature above which a ferromagnetic substance becomes paramagnetic is called the Curie temperature of the substance.
A 250-Turn rectangular coil of length 2.1 cm and width 1.25 cm carries a current of 85 μA and subjected to a magnetic field of strength 0.85 T. Work done for rotating the coil by 180° against the torque is
9.1 μJ
4.55 μJ
2.3 μJ
1.15 μJ
A.
9.1 μJ
W = MB (cosθ1– cosθ2)
When it is rotated by angle 180º then
W = 2MB
W = 2 (NIA)B
= 2 x 250 x 85 x 10-6[1.25 x 2.1 x 10-4] x 85 x 10-2
9.1μJ
Figure shows a circuit contains three identical resistors with resistance R = 9.0 Ω each, two identical inductors with inductance L = 2.0 mH each, and an ideal battery with emf ε= 18 V. The current'i' through the battery just after the switch closed is
2 mA
0 mA
2 A
0 A
C.
2 A
A thin diamagnetic rod is placed vertically between the poles of an electromagnet. When the current in the electromagnet is switched on, then the diamagnetic rod is pushed up, out of the horizontal magnetic field. Hence the rod gains gravitational potential energy. The work required to do this comes from
The current source
The magnetic field
The induced electric field due to the changing magnetic field
The lattice structure of the material of the rod
A.
The current source
The energy of current source will be converted into potential energy of the rod.
In the given figure, what is the magnetic field induction at point O.
C.
Magnetic field due to the straight wire above O is zero
i.e. = B1 = 0 (since, θ =0°)
The magnetic field due to semicircular part
The magnetic field due to lower straight portion
(upward)
Net magnetic field B = B1 + B2 + B3
=
A uniform electric field and a uniform magnetic field 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
It will turn towards the left of the direction of motion.
It will turn towards the right of the direction of motion
Its velocity will increase
Its velocity will decrease
D.
Its velocity will decrease
The magnetic field will not apply any force to the electrons. On the other hand, the electric field will apply a force on the electron in the opposite direction of the motion. As a result, the velocity of the electron will decrease.
If the magnetic dipole moment of an atom of diamagnetic material, paramagnetic material and ferromagnetic material are denoted by respectively, then
C.
..In diamagnetic substances in each pair of electrons, the spin of both the electrons are in opposite directions. Hence, the electrons of each pair completely cancel the magnetic moment of each other. Thus, the net magnetic moment of each atom of such substances is zero i.e., .
The property of paramagnetism is found in those substances whose atoms, or molecules have an excess of electrons spinning in the same direction. Hence, atoms of paramagnetic substances have permanent magnetic moment ie, . The property of ferromagnetism is found in substances which acquire very strong magnetism when placed in an external magnetic field. Like the paramagnetic substances each atom of ferromagnetic substances also has a also has a permanent magnetic moment i.e, .
A frog can be levitated in magnetic field produced by a current in a vertical solenoid placed below the frog. This is possible because the body of the frog behaves as
paramagnetic
diamagnetic
ferromagnetic
anti-ferromagnetic
B.
diamagnetic
Some substances when placed in a magnetic field are feebly magnetized in the opposite direction of the magnetizing field. Thes substances when brought close to a pole of a powerful magnet are repelled away from the magnet. The body of the frog behaves like a diamagnetic substance, hence the smallest magnetic field placed below the body of the frog will be strong enough to lift it.
When diamagnetic substances are placed in a magnetic field, then they are feebly repelled in the field.
A magnetic field exerts no force on
a magnet
an unmagnetised iron bar
a moving charge
stationary charge
D.
stationary charge
Moving charge can develope magnetic field and also experience the magnetic field. But stationary charge do not have to ability to develope or experience magnetic field from other magnetic fields. Because moving charge gives rise to current which is responsible for the magnetic effects.
F = q vB
F = 0 when v = 0
i.e., a stationary charge experiences no force in a magnetic field.
An ideal solenoid having 5000 turns/m has an aluminium core and carries a current of 5A. If χAl = 2.3 × 10-5, then the magnetic field developed at centre will be
0.031 T
0.048 T
0.027 T
0.050 T
A.
0.031 T
Given:-
Number of turns in solenoid n = 5000 turns/m
Current carried i = 5A
Magnetic susceptibility χAl = 2.3 × 10-5
B = ?
As
B = μo ( H + A )
where,
H =
Where H = magnetic intensity
B0 = external magnetic field
μ0 = permitivity constant
=
H = ni
= 5000 × 5
= 2.5 × 104 A/m
and I = χ H
= 2.3 × 10-5 × 2.5 × 104
I = 0.575 A/m
∴ B = μo ( H + I )
= 4 × 10-7 ( 2.5 × 104 + 0.575 ) T
B = 0.031 T
Assertion: In water, value of magnetic field decreases.
Reason: Water is a diamagnetic substance
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.
A.
If both assertion and reason are true and reason is the correct explanation of assertion.
Water is a diamagnetic substance. The relative permeability of water is less than 1.
Assertion: Magnetic field lines are continuous and closed.
Reason: Magnetic monopole does not exist
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
A.
If both assertion and reason are true and reason is the correct explanation of assertion.
Magnetic field lines represent continuous curves that originate at the north pole of the magnet and end at the south pole.
While we can find electric monopoles in the form of charged particles, we have never observed magnetic monopole. Instead, magnets exist only in the form of dipoles with north and south end
The temperature of transition from ferromagnetic property to paramagnetic property is called
Transition temperature
Critical temperature
Curie temperature
Triplet temperature
C.
Curie temperature
The materials are only magnetic below curie temperature above curie temperature loses this property. This means that by increasing the temperature the orientation of the magnetic moments changes due to which the ferro-magnetic materials loses its magnetic property and change to para-magnetic property.
The temperature at which ferro-magneism changes to para-magnetism called curie-temperature.
Sponsor Area
Sponsor Area
