Sponsor Area
Given,
where, r is the distance between two charges.
Using the formula,
Repulsive in nature since both charges are positive.
The electrostatic force on a small sphere of charge 0.4 μC due to another small sphere of charge –0.8 μc in air is 0.2 N.
(a) What is the distance between the two spheres?
(b) What is the force on the second sphere due to the first?
(a) Given,
Now using the formula,
Hence,
where, r is the distance between two spheres.
(b) Force on the second sphere due to the first is same, i.e., 0.2 N because the charges in action are same and force is attractive as charges are unlike in nature.
Check that the ratio is dimensionless. Look up a table of physical constants and determine the value of this ratio. What does the ratio signify?
and
Now,
It also establishes that the electrostatic force is about 1039 times stronger than the gravitational force.
Law of conservation of charge states that total charge on an isolated system of objects always remain conserved.
When a glass rod is rubbed with silk cloth, glass rod becomes positively charged while silk cloth becomes negatively charged and, the amount of positive charge on the glass rod is apparently found to be exactly the same as the negative charge on silk cloth. Since, the measure of charge is same on both, equal amount of charge with opposite nature will cancel out each other. Hence, the total sum of charge on two bodies is zero. Thus, the system of glass rod and silk cloth, which was neutral before rubbing, still possesses no net charge after rubbing. And law of conservation of charge is justified.
As a consequence of conservation of charge, when two charged conductors of same size and same material carrying charges Q1 and Q2 respectively are brought in contact and separated, the charge on each conductor will be
This condition, however, does not hold true if the conductors are of different sizes or of different material.
In that case the charges on the conductors will be Q1' and Q2’ respectively, where Q1 + Q2 = Q1‘ + Q2’.
Hence, two field lines never cross each other.
Two point charges qA = 3 μC and qB = –3 μC are located 20 cm apart in vacuum.
(a) What is the electric field at the midpoint O of the line AB joining the two charges?
(b) If a negative test charge of magnitude 1.5 x 10–9 C is placed at this point, what is the force experienced by the test charge?
Given,
Distance between two charges-d= 20 cm
Distance of charge A and B from point O - r= 20/2 = 10cm =0.1 m
(a)
Let us assume that a unit positive test charge is placed at 0. qA will repel this test charge while qB will attract. Hence, both are directed towards
(b) As a negative test charge of q0 = – 1.5 x 10–6 C is placed at 0. qA will attract it while qB will repel. 
Therefore, the net force
Sponsor Area
Given,
(a) According to gauss law as
Hence,
or
i.e.,
b) No, a conclusion cannot be made that there was no charge in the box. Perhaps, it can be concluded that the net charge inside the box is zero.
Let us assume that the charge q = + 10 μC = 10–5 C is placed at a distance of 5 cm from the square ABCD of each side 10 cm.
The square ABCD can be considered as one of the six faces of a cubic Gaussian surface of each side 10 cm.
Now, the total electric flux through the faces of the cube as per Gaussian theorem
Therefore, the total electric flux through the square ABCD will be
Given:
Electric flux - Φ = –1.0 x 103 Nm2/C
radius of the gaussian surface - r1 =10 cm=0.1 m,
After doubling, the radius of the spherical gaussian surface - r2 =20 cm=0.2 m
(a) Doubling the radius of Gaussian surface will not affect the electric flux since the charge enclosed is the same in both cases. Thus, the flux will remain the same i.e., –1.0 x 103 Nm2/C
(b)Using gauss theorem,
Given,
As,
Here, since electric field is directed radially inward charge q is negative.
Thus,
(a) To the left of the plates, electric fields are equal and opposite as plates are close to each other electric field is zero as surface charge density in outer side is zero.
(b) To the right of the plates, electric fields are equal and opposite as plates are close to each other electric field is zero.
(c) Electric fields between the plates are in same direction as total E.F. on both sides of plate due to surface charge density =
So, electric field of inner side of plate =
and for both plate
Given,
Since, the droplet is stationary,
Weight of the droplet = force due to the electric field
(a) Figure (a) cannot represent electrostatic field lines since electrostatic field lines start or end only at 90° to the surface of the conductor.
(b) Figure (b) too cannot represent electrostatic field lines as electrostatic field lines do not start from a negative charge.Electric field lines always traverse from a region of positive charge to a region of negative charge.
(c) Electrostatic field lines are represented by figure (c).
(d) Figure (d) cannot represent electrostatic field lines since no two such lines of force can intersect each other.
(e) As electrostatic field lines cannot form closed loop, therefore figure (d) also does not represent electrostatic field lines.
Charge on 'up' quark,
Charge on 'down' quark,
Charge on a proton = e
Charge on a neutron = 0
Let a proton contains x ‘up’ quarks and (3 - x) ‘down’ quarks. Then total charge on a proton is
x = 2
and 3 - x = 3 - 2 = 1
i.e., proton contains 2 ‘up’ quarks and 1 ‘down’ quark. Its quark composition should be uud.
Let a neutron contains y ‘up’ quarks and (3 – y) ‘down’ quarks. Then total charge on a neutron is
and 3 – y = 3 – 1 = 2
i.e., neutrons contain 1 ‘up’ quark and 2 ‘down’ quarks. Its quark composition should be udd.
(a) Consider an arbitrary electrostatic field configuration. A small test charge is placed at a null point (i.e., where E = 0) of the configuration. Show that the equilibrium of the test charge is necessarily unstable.
(b) Verify this result for the simple configuration of two charges of the same magnitude and sign placed a certain distance apart.
(a) It can be proved by contradiction. Assume that the test charge placed at null point be in stable equilibrium. The test charge displaced slightly in any direction will experience a restoring force towards the null-point as the stable equilibrium requires restoring force in all directions. That is, all field lines near the null point should be directed inwards towards the null point. This indicates that there is a net inward flux of electric field through a closed surface around the null point. But, according to Gauss law, the flux of electric field through a surface enclosing no charge must be zero. This contradicts our assumption. Therefore, the test charge placed at null point must be necessarily in unstable equilibrium.
(b) On the mid-point of the line joining the two charges, the null point lies. The test charge will experience a restoring force if it is displaced slightly on either side of the null point along this line. While the net force takes it away from the null point if it is displaced normal to this line. That is no restoring force acts in the normal direction. But restoring force in all directions is demanded by stable equilibrium, therefore, test charge placed at null point will not be in stable equilibrium.
Given,
An electron is projected with velocity vx= 2.0.
Distance between the plates = 0.5 cm
Electric field between plates - =
Acceleration,
Using formula
We get,
Simplifying for value of t, we get
The electron covers a vertical distance which is given by
Using the formula for quantisation of electric charge,
.
Therefore, electrons make 1 Coulomb of charge.
Sponsor Area
Field lines are the path of small positive test charge. The charge is moving continuously from point to point rather than jumping from one point to another and experiences continuous force in the electrostatic field. The force experienced or the path followed by charge cannot be discontinuous and hence the lines are not broken. Also, electrostatic field lines represent the electric field strenth and the strength of field is never broken.
(i) SI unit of electric field intensity is NC–1 (Newton per Coulomb).
(ii) SI unit of electric dipole moment is C-m (Coulomb-metre).
An electric line of force is an imaginary straight or curved path along which a small positive test charge is supposed to move when free to do so.
The tangent at a point on an electric line of force gives the direction of the resultant electric field at that point.
The relative closeness of electric lines of force in a certain region provides us an estimation of the electric field strength in that region.
In a satellite, there is condition of weightlessness.
Therefore, mg = 0.
On account of electrostatic force of repulsion between the balls, the strings would become horizontal.
Therefore, angle between the strings = 180°.
Also, tension in each string = force of repulsion
Sponsor Area
(i) Positive charge will be produced on glass rod and negative charge will be produced on silk. The electrons are less tightly bound in glass rod as compared to silk.
(ii) Negative charge will be produced on ebonite rod and positive charge will be produced on wool.
Here given the particles have an equal charge joined by a string of length 1 m.
Tension in the string is the force of repulsion (F) between the two charges.
According to Coulomb's law,
Let two equal charges Q each, be held at A and B, where AB = 2x.
C is the centre of AB, where charge q is held, figure below.
Net force acting on q is zero. So q is already in equilibrium.
For the three charges to be in equilibrium, net force on each charge must be zero.
Now, total force on Q at B due to q and Q at A is
Hence proved.
The force experienced by a unit positive charge placed at a point is termed as the electric field intensity at that point.
It is vector quantity and it’s direction is in the direction of force acting on +ve test charge.
The SI unit of electric field intensity is NC–1.
Electric field at an axial point of electric dipole:
Assume point P is located at distance r from the centre of an electric dipole as shown in the figure.
If σ is the surface charge density,
∴
∴ Electric field lines when the charged density of the sphere:
(i) Positive (ii) Negative
(b)
Here given,
diameterof charged conducting sphere = 2.5 m
Charge density is given by
(i)charge on the sphere -
(ii) Total electric flux
(a) Define electric flux. Write its SI units.
(b) The electric field components due to a charge inside the cube of side 0.1 m are as shown:
Ex = x, where = 500 N / C–m
Ey = 0, Ez = 0.
Calculate:
(i) the flux through the cube and
(ii) the charge inside the cube.
(a) Electric flux:
The total number of electric field lines crossing an area placed normal to the electric field is termed as electric flux.
It is denoted by ΦE.
Electric flux is a scalar quantity, its SI unit is Nm2 C–1.
(b) Here, Ex = x where = 500 N/C–m,
Ey = 0, Ez = 0
side of a cube a = 0.1 m
As the electric field has only X-component then,
for each of four faces of cube 1 of Y-axis and Z-axis.
∴ Electric flux is only for left and right face along X-axis of cube.
(i) Electric field at the left face,
(ii) According to Gaussian law
Using the above fig.
Also,
Let Fa be the force exerted by the charge of ‘+ O.1μC’ on the charge ‘+ 1 μC’ placed at centre O of the square.
Then using Coulomb's law for Electrostatic force,
If Fc is the force exerted by charge at C on charge at O, then
Both act in the same direction. The resultant of
Force exerted by the charge at B on the charge at O,
Force exerted by the charge at D on the charge at O,
Both act in the same direction.
Resultant of
The angle between F1 and F2 is clearly 90°.
So, the resultant F of F1 and F2 is given by
The question can be depicted as in figure above.
Given,
D.
oscillate between the plates without touching them.If a system contains two positive charges say, q1 and q2 then, the total charge of the system is the sum of these two charges.
i.e, q1+q2.
If a system contains n charges q1,q2,q3,.....qn, then the total charge of the system is q1+q2+q3+.....+qn.
Charge is a scalar quantity and they add up like real numbers and this property is known as Additivity of charges.
The fact that a charged body attracts light particles can be explained on the basis of induction. Consider, a negatively charged rod is brought near a small piece of metal foil as shown in the fig. below.
The free electrons present in the metal foil are repelled to the end D leaving a positive charge near C. Now, since the positive charge is near to the charged rod as compared to the negative charge, the resultant force in the foil is attractive. Thus, induction always precedes attraction of small particles.
When the vehicle is moving, due to friction in air, the body of the vehicle gets charged. If this accumulated charge becomes excessive sparking may occur and inflammable material will catch fire. Hence, mettalic chains dragging the ground is introduced so that the excessive charge gets leaked onto the ground and fire can be avoided.
Similarly, vehicles running on rubber(insulator) tyres also drag a chain along the ground so as to dissipate the static electricity that has been accumulated on tyres to the ground.
Consider two charges q1 and q2. Let the position vectors of the respective charges be r1 and r2.
Let, the force acting on q1 due to q2 be F12 and the force acting on q2 due to q1 be F21.
The vector leading from 1 to 2 is denoted by r21 :
r21= r2 - r1
ly, vector leading from 2 to 1 is denoted by r12:
r12=r1 - r2
and
r12=-r21.
The direction of a vector is specified by a unit vector along the vector.
Now, Coulomb's force law between two point charges q1 and q2 when force is acting on q2 due to q1 is expressed as:
And, force acting on q1 due to q2 can be expressed as:
Ratio of these forces are given by,
= .
The electric field lines of the point charge are as given below:
Electric field line or the line of force is the path along which a unit positive test charge would move when kept in an electrostatic field.
In other words, a field line is an imaginary line drawn such that its direction at any point is the same as the direction of the field at that point.
Line of field due to two equal positive charge is shown below:
Sponsor Area
Let q1 and q2 be the point charges. Then, force between the charges when kept at a distance 'd' apart is given by
a) Suppose that the force between two charges becomes 3F when kept at a distance 'x' apart
b) Suppose that the force between two charges becomes F/3 when kept at distance 'x' apart.
Electric dipole moment is defined as the product of either charge and the length of the electric dipole.
Electric field on equitorial line of an electric dipole:
Consider an electric dipole consisting of charges -q and +q seperated by a distance 2a. Let P be a point on equitorial line of the dipole at a distance 'r' from the centre of the dipole as shown in fig.
Let, 
To find the resultant electric field intensity due to the dipole at point P, we will represent by the two adjacent sides PL and PM of a parallelogram. Then, diagonal PN represents the resultant Electric field due to the dipole acting along Px'. The resultant electric field can also be found using the triangle law of addition of vectors.
In
Consider an electric dipole consisting of charges -q and +q and of length 2a placed in uniform electric field makng an angle with the direction of electric field.
Force acting on charge -q at A= (opposite to )
Force acting on charge +q at B= (along )
Thus, electric dipole is under the action of two equal and unlike parallel forces giving rise to a torque on the dipole.
The magnitude of the torque is given by
Torque experience maximum dipole when it placed perpendicular to the dirction of Electric field i.e,
and
Torque aligns the dipole along the direction of the electric field.
Two essential conditions for this to happen is:
i) Both charges cannot be of the same sign.
ii)The point where electric field intensity has to be 0 is closer to the smaller charge as compared to the charge larger in magitude.
An electric dipole is held in a uniform electric field. Show that no translation force acts on it.
Since the electric dipole is held in a uniform electric field no net force is acting on the dipole.
Force due to -q is -qE and force due to +q is qE. Hence the no net force is acting on it and the translational forces is 0.
The total number of electric field lines crossing an area placed normal to the electric field is termed as electric flux.
Electric flux is a scalar quantity, its SI unit is Nm2 C–1.
Electric flux does not depend on the size and shape of object. In this case, as the charge enclosed is same there will be no net change in the electric flux coming out of the surface.
In case the test charge is not vanishingly small it will produce it's own electric field. Therefore, the measured value of electric field at the observation point will be affected and will be different from the actual value of electric field at that point. Therefore we take the test charge to be negligibly small.
Consider a thin spherical shell of radius 'R' with centre O. Let, a charge +q be distributed uniformly on the surface of the shell.
According to gaussian theorem,
Surface area of the sphere = 4r2
Now, at a point on the surface area of the shell
Let, be the surface charge density on the shell,
then,
q = 4R2 .
Therefore,
which is the required electric field at the surface of the shell.
Since the weight of an oil drop is balanced in an electric field we will have
where, q is the charge on the oil drop.
Given, q1= +0.2 C
q2= +0.4 C
distance between the charges-d=0.1 m
a)Electric field at the mid point between these two charges:
Electric field due to q1-E1=
/C
Electric field due to q2-E2=
/C
Resultant Electric field at mid-point-E=
Since the net electric field is acting in opposite direction we have E= 1440
= /C
b) 
Let point P be on the line joining the charges such that it is 0.05m away from q2 and 0.15 m away from q1.
Electric field due to q1=
Electric field due to q2=/C
Since, Electric field is acting in the same direction
Resultant electric field intensity-E=
Let, the midpoint of the line joining the two charges be O.
a) Electric field at O due to charge at A-EA=
Electric field at O due to charge at B-EB=
Resultant intensity -E =EA+ EB
=
b) If a test charge of magnitude is placed at the centre, then force experienced is
Let the charge Q be kept along the line joining two charges such that it is at a distance x from q and at a distance (d-x) from 2q.
For the net force on q and 2q to be 0 the system of charges should be in equilibrium.
Force acting on Q due to q=
Force acting on Q du to 2q=
For, the system to be in equilibrium
Charge on dipole is
Distance between the charges = 1mm=
Dipole moment is given by-p = q(2a)
=
Given,
Electric field, E =
Torque ,
S1 and S2 are two hollow concentric spheres enclosing charges 2Q and 4Q respectively as shown in the figure.
(i) What is the ratio of electric flux through S1 and S2?
(ii) How will the electric flux through the sphere S1 change, if a medium of dielectric constant 6 is introduced in the space inside S1 in place of air?
(i)
(ii) If a medium of dielectric constant k is introduced then,
Using Gauss's theorem,
i.e,
Consider the equilibrium of sphere A.
Following forces act on the sphere A.
(i) Force F of repulsion on A due to B.
(ii) Weight mg acting vertically downwards.
(iii) Tension T in the string towards the point of suspension O.
Resolving the tension T into two rectangular components:
For equilibrium of A,
and
Dividing,
But
In all the problems, which involve distribution of charge, we choose an element of charge dq to find the element of the field dE produced at the given location. Then we sum all such dEs to find the total field E at that location.
We must note the symmetry of the situation. For each element dq located at positive x-axis. There is a similar dq located at the same negative value of x. The dEx produced by one dq is cancelled by the dEx in the opposite direction due to the other dq. Hence, all the dExcomponents add to zero.
So, we need to sum only the dEy components, a scalar sum since they all point in the same direction.
The element charge is dq = λ dx
The integral on the right hand side can be evaluated by substituting x = and dx = r sec2 .
[]
A rigid insulated wire frame in the form of a right-angled triangle ABC, is set in a vertical plane as shown in fig. Two beads of equal masses m each and carrying charges q1and q2 are connected by a cord of length l and can slide without friction on the wires.
Considering the case when the beads are stationary, determine
(i) the angle
(ii) the tension in the cord and
(iii) the normal reaction on the beads.
If the cord is now cut, what are the values of the charges for which the beads continue to remain stationary?
Let us consider an infinitesimal element of length dx at a distance x from the point P.
The charge on this element is dq = λdx,
where, is the linear charge density.
The magnitude of the electric field at P due to this element is
and its direction is to the right since X is positive.
The total electric field strength E is given by
Given,
Distance between the charges, r =
a) Since,
b)As,
For equilibrium of q, the forces on it exerted by P and Q must be co-linear, equal and opposite.
Force on q by P
Force on q by Q
towards Q
Hence charge q should be equidistant from P and Q.
For the system to be in equilibrium, the charges P and Q must also be in equilibrium.
Now,
Since Fpq and FPQ are oppositely directed along the same line, we have, for equilibrium,
or
Similarly for the equilibrium of Q, we would get
Thus in magnitude of either charge P or Q.
Stability: A slight displacement of q towards P increases the magnitude of FqP and decreases the magnitude of FqQ. Consequently, the displacement of q is increased. Thus the three charges are no longer in equilibrium. Hence the original equilibrium is unstable for displacement along the axis on which the charges are located. For a displacement of q along a direction normal to the line PQ, the resultant of the two forces of attraction Fqp and FqQ will bring the charge q back to its original position. Thus the equilibrium is stable for displacement in the vertical direction.
Two identical charged bodies have 12 μC and – 18 μC charge respectively. These bodies experience a force of 48 N at certain separation. The bodies are touched and placed at the same separation again. Find the new force between the bodies.
No, current will not be induced in the coil if the coil is rotated about it's axis.
Formula for flux is .
(i) If seen from the right hand side, current flows clockwise when South-pole moves towards the coil.
(ii) If seen from the left hand side, current flows clockwise when N-pole of the bar magnet moves away from the coil.
Will an induced current be always produced whenever there is change of magnetic flux linked with a coil?
Or
Does change in magnetic flux induce emf or current?
Given,
Length of loop, l = 8 cm = 8 x 10–2 m
Breadth of loop, b = 2 cm = 2 x 10–2 m
Strength of magnetic field, B = 0.3 T
Velocity of loop, v = 1 cm/sec = 10–2 m/sec
Let the field be perpendicular to the plane of the paper directed inwards.
(i)Then, the magnitude of induced emf is,
Time for which induced e.m.f. will last is equal to the time taken by the coil to move outside the field
(ii) The conductor is moving outside the field normal to the shorter side.
b = 2 x 10–2 m
∴ The magnitude of induced emf is
Time,
A horizontal straight wire 10 m long extending from east to west is falling with a speed of 5.0 ms–1 at right angles to the horizontal component of the earth's magnetic field 0.30 x 10–4 Wb m–2.
(a) What is the instantaneous value of the e.m.f. induced in the wire?
(b) What is the direction of the e.m.f?
(c) Which end of the wire is at higher electrical potential?
Here,
Initial current, I1 = 5 A
Final current, I2 = 0 A
t = 0.1 sec
Average emf, e= 200 V
Change in the value of current, dI/dt is,
Therefore,
which is the required self inductance of the circuit.
Given, a pair of adjacent coils.
Mutual inductance, M = 1.5 H
Current in the coil, I = 20 A
Time ,t = 0.5 s
Using formula, Φ = MI we get,
= 1.5 20
= 30 H.
Given, a jet plane is travelling towards west.
Earth's field,
Angle of dip,
Length of the jet wing = 25 m
The vertical component (Bv) of earth's field is normal to both wings and the direction of motion.
Induced e.m.f produced
Using formula, we get
Induced emf,
Therefore,
Power loss =
Given, a square loop.
side of the loop = 12 cm
velocity with which the loop is moving = 8 cm s–1 = 8 x 10–2 m/s
Area of the loop, A = (12 x 10–2) = 144 x 10–4 m2
Gradient in magnetic field,
Induced e.m.f. due to change of magnetic field B with time t is,
...(I)
Induced e.m.f. due to change of magnetic field B. with distance (x) is,
Total e.m.f
Resistance of the loop, R= 4.5 mΩ
Therefore,
Induced current
Magnitude of flux of the flat coil,
Induced emf in the magnet,
This implies,
That is,
(b) Is there an excess charge built up at the ends of the rods when K is open? What if K is closed?
(c) With K open and the rod moving uniformly, there is no net force on the electrons in the rod PQ even though they do experience magnetic force due to the motion of the rod. Explain.
(d) What is the retarding force on the rod when K is closed?
(e) How much power is required (by an external agent) to keep the rod moving at the same speed (= 12 cm s–1) when K is closed? How much power is required when K is open?
(f) How much power is dissipated as heat in the closed circuit? What is the source of this power ?
(g) What is the induced emf in the moving rod if the magnetic field is parallel to the rails instead of being perpendicular?
(a) Here, speed with which the rod is moved, v = 12 cm s–1
The magnitude of the induced emf is given by ε = B v I
= 0.50 x 0.12 x 0.15 volt
= 9 x 10–3volt
= 9 mV
P is the positive end and Q is the negative end.
(b) Yes, an excess amount of charge is built up at the ends of the rod when key is open. When key is closed, the excess charge is maintained by the continuous flow of current.
(c) There is no net force as the magnetic force Fm = – e (v + B) is cancelled by the electric force [Fe = eE] which is set up due to the excess charge of opposite signs at the ends of the rod.
(d) Induced current,
Retarding force on the rod F = BIl
= 0.5 x 1 x 0.15
= 7.5 x 10 –2 N
(e) Power expended by an external agent against the above retarding force to keep the rod moving uniformly at 12 cm/s is given by,
P = F.v
P = 75 x 10–3 x 12 x 10–2 W
= 9.0 x 10–3 W
(f) Power dissipated as heat = I2 R
= 12 (9 x 10–3)
= 9 x 10–3 W
Source of this power is an external agent which keeps rod in motion, against magnetic retarding force.
(g) When the permanent magnet is rotated in a vertical position, the field becomes parallel to rails. The motion of rod will not cut across the lines of magnetic field and hence no e.m.f. is induced.
Given, an cored solenoid.
Length of the solenoid, l = 0.30 m
Area of cross-section of solenoid, A = 25 x 10-4m2
Number of turns of the coil, N = 500
Current carried by the coil, I = 2.5 A
Initial magnetic flux,
i.e.,
Now, on substituting the values,
Final magnetic flux,
Therefore, Change of flux is,
Corresponding time interval,
Hence,
Average e.m.f induced across the open switch
(a) Obtain an expression for the mutual inductance between a long straight wire and a square loop of side a as shown in Fig.
(b) Now assume that the straight wire carries a current of 50 A and the loop is moved to the right with a constant velocity, v = 10 m/s; calculate the induced emf in the loop at the instant when x = 0.2m. Take a = 0.1 m and assume that the loop has a large resistance.
Predict the direction of induced current in the situations described by the following figs.(a) to(f).
(a) South pole develops at end q and the induced current should flow clockwise. Therefore, induced current in the coil flows from qr to pq.
(b) Coil pq in this case would develop S-pole at q and coil XY would also develop S pole at X. Therefore, induced current in coil pq will be from q to p and induced current in the coil XY will be from Y to X.
(c) North pole is moving away from the coil hence, induced current in the right loop will be along XYZ.
(d) Induced current in the left loop will be along ZYX as seen from front.
(e) Induced current in the right coil is from X to Y.
(f) Since magnetic lines of force lie in the plane of the loop, no current is induced.
Use Lenz’s law to determine the direction of induced current in the situation described by Fig:
(a) A wire of irregular shape turning into a circular shape;
(b) A circular loop being deformed into a narrow straight wire.
(a) When a wire of irregular shape turns into a circular loop, area of the loop tends to increase. Therefore, magnetic flux linked with the loop increases. According to Lenz’s law, the direction of induced current must oppose the change in magnetic field, for which induced current should flow along adcba (anticlockwise).
(b) In this case, the magnetic flux tends to decrease. Therefore, induced current must support the magnetic field, for which induced current should flow along adcba (anticlockwise).
Given,
A large circular coil of radius R and a small circular coil of radius r.
Coefficient of mutual inducatnce,
Current flowing through the small coil, I = 0.5 A
Flux linked with the larger coil,
If the current through the small coil falls to zero then the induced current in the larger coil becomes zero.
How is the mutual inductance of a pair of coils affected when:
(i) separation between the coils is increased?
(ii) the number of turns of each coil is increased?
(iii) A thin iron sheet is placed between the two coils, other factors remaining the same? Explain your answer in each case.
(i) When seperation between the coils is increased, mutual inductance (M) decreases because the measure of flux linking to a coil due to the other one will decrease.
(ii) When the number of turns of each coil is increased, mutual inductance (M) increases because the overall flux density increases with increase in turns of the coil and, hence the mutual inductance increases.
(iii) When a thin iron sheet is placed betweent the two coils, M increases because iron is ferromagnetic in nature hence, it will increase the flux density.
Given, a coil of N turns having an area A, rotated with angular speed '' in a uniform magnetic field 'B' connected to a resistor 'R' .
The flux linking to the coil,
Therefore,
Induced EMF,
When, sint = 1, the induced emf value is maximum.
(i) The maximum EMF,
(ii) The power dissipated in the coil is given by,
(a) When bar magnet is moved towards left, the anticlockwise current will be in loop (1) and (2) {facing bar magnet} following Lenz’s law. Thus, the direction of induced current in loop (1) will be from A to B and that in (2) will be from D to C.
(b) When bar magnet is moved towards right, the clockwise current will be in loop (1) and (2) {facing bar magnet} following Lenz’s law. Thus, the direction of induced current in the loop (1) will be from B to A and C to D.
(i) What will be the directions of the induced currents in the loops when they are pulled away from the conductor with same velocity v?
(ii) Will the e.m.f. induced in the two loops be equal? Justify your answer.
Here,
Speed of the train,
Vertical component of earth's magnetic field, Bv = 0.2 x 10–4 T
Seperation between the rails, l = 1 m
The induced emf generated is given by
where, A is the area and B is the magnetic field.
If l is the distance between the rails and v is the speed of the train, then
Induced emf,
Hence, the millivoltmeter will read 1 mV.
Given,
Inductance of the coil, L = 0.25 H
Voltage of the battery, E = 18 V
Rate of growth of current, dI/dt = ?
Using formula,
Given, raise in current of the primary circuit, dl = 5 ampere
Mutual inductance, M = 1.5 henry
Time, dt = 1 milli sec = 10–3 sec.
Using formula,
Length of the metal rod PQ, l = 0.5 m
Magnetic field, B = 0.15 T
Resistance of the circuit, R = 3 Ω
Consider coil PQRS with its arm PQ movable as shown in the figure.
A magnetic field is applied normal to the surface of the coil. 
Area of the coil, ΔS = l x (x) and,
Flux, Φ = 8 Δ S = Blx
The rate of change of magnetic flux linked with the coil is given by
If e is the induced emf produced, then
Let R be the resistance of movable arm PQ of the rectangular conductor. Taking the resistance of other arms as negligibly small, the current in the loop is given by,
I = ...(I)
The arm PQ moves with the speed v.
The power required to move it is given by,
(From I)
(a) State Lenz’s law. Give one example to illustrate this law. “The Lenz’s law is a consequence of the principle of conservation of energy.” Justify this statement.
(b) Deduce an expression for the mutual inductance of two long coaxial solenoids but having different radii and different number of turns.
Whenever the magnetic flux linked with a closed circuit changes, an emf is set up across it which lasts only as long as the change in flux is taking place. This emf is called induced emf.
According to Faraday's law of electromagnetic induction, the magnitude of induced emf is equal to the rate of change of magnetic flux linked with the closed circuit (or coil).
Mathematically,
where,
N is the number of turns in the circuit and,
ΦB is the magnetic flux linked with each turn.
Suppose the conducting rod completes one revolution in time T.
Then,
Change in flux = B x Area swept
= B x l2
Therefore,
But,
The current in a coil of self-inductance L = 2H is increasing according to the law i = 2 sin t2.
Find the amount of energy spent during the period when the current changes from 0 to 2 ampere.
(a) Given, a toroidal solenoid.
Average radius, r = 0.15 m
Area of cross-section, A = 12 x 10–4 m2
Number of turns in the coil, n = 1200
Magnetic field is given by,
Total magnetic flux,
But,
How is mutual inductance of a pair of coils affected when
(i) separation between the coils is increased,
(ii) the number of turns of each coil is increased,
(iii) a thin iron sheet is placed between the two coils, other factors remaining the same. Explain your answer in each case.
(i) When separation between the coils is increased, magnetic flux linked with secondary coil decreases. Therefore, mutual inductance (M) of pair of coils decreases.
(ii) When number of turns of each coil is increased, mutual inductance increases, because
(iii) As 
therefore, mutual inductance will increase on placing a thin iron sheet between the two coils.
Given, a conductor and two conducting parallel rods.
Length of conductor, l = 0.5 m
resistance, r = 0.5 ohm
velocity, v = 2 m/s
Resistance of rod, R = 2.5 ohm
Magnetic field, B = 0.6 T
The conductor ef moves with a velocity v perpendicular to a uniform magnetic induction B and hence induces an emf E = Blv.
The resistance of the circuit = (R + r)
(i) Hence, the current in the circuit
(ii) The power spend in the system
A force of 0.06 N is required to maintain the motion of the conductor
(iii) The power generated
An electromagnet has stored 648 J of magnetic energy, when a current of 9 A exists in its coils.
What average emf is induced if the current is reduced to zero in 0.45 s?
Given, two different coils.
L1 = 8 mH; L2 = 2 mH
a) Now, using the formula,
Hence, ratio of induced volatge is 4:1 .
b) Since,
Therefore,
Ratio of current is 1:4 .
c) Energy is given by,
Thus ratio of energy is 1 :4 .
Given, a wire loop which is square in shape.
Side of the square loop = 5 cm
angle between B and A = 30o
Magnetic induction = 0.50 T
Area of the loop,
Magnetic flux,
where B cos θ is the component of magnetic induction perpendicular to the plane of the loop.
Therefore,
Average induced emf is
Thus, the induced emf is in an anticlockwise direction around the loop during the time when the downward magnetic flux is increasing.
Two long parallel horizontal rails, distance d apart and each having a resistance A per unit length, are joined at one end by a resistance R. A perfectly conducting rod MN of mass m is free to slide along the rails without friction (see Fig.). There is a uniform magnetic field of induction B normal to the plane of the paper and directed into the paper. A variable force F is applied to the rod MN such that as the rod moves, constant current flows through R.
(i) Find the velocity of the rod and the applied force F as function of the distance x of the rod from R.
(ii) What fraction of the work done per second by F is converted into heat?
A line charge λ per unit length is lodged uniformly onto the rim of a wheel of mass M and radius R. The wheel has light non-conducting spokes and is free to rotate without friction about its axis (Fig a). A uniform magnetic field extends over a circular region within the rim. It is given by,
B = – B0 k (r ≤ a; a < R)
= 0 (otherwise)
What is the angular velocity of the wheel after the field is suddenly switched off?
Line charge per unit length is given by,
where,
‘r’ is the distance of the point from the wheel,
M is the mass of the wheel,
R is the radius of the wheel, and
is the magnetic field.
At distance r, the magnetic force is balanced by the centripetal force.
i.e.
where, v is the linear velocity of the wheel.
Therefore,
Angular velocity,
For we get
,
is the required angular velocity of the wheel after the field is suddenly switched off.
Obtain the formula for the electric field due to a long thin wire of uniform linear charge density λ without using Gauss’s law. [Hint: Use Coulomb’s law directly and evaluate the necessary integral.]
Consider a long thin wire of uniform linear charge density, .
To find: Formula for electric field due to this wire at any point P at a perpendicular distance PC = r from the wire.
Consider a small element of length dx of the wire with centre O, such that OC = x.
Charge on the element, q =
So, electric intensity at P due to the element is given by,
Now, can be resolved into two rectangular components, that is in a perpendicular direction and in a parallel direction.
The parallel component will be cancelled by the parallel component of the field due to charge on a similar element dx of wire on the other half.
The radial components get added.
Therefore,
Effective component of electric intensity due to the charge element,
...(1)
From
From equation (1), we have
Since the wire has infinite length, it’s ends A and B are infinite distances apart.
Therefore, varies from
So, Electric Intensity at P due to the whole wire is given by,
, is the required electric field intensity.
Two charges of magnitudes – 2Q and + Q are located at points (a, 0) and (4a, 0) respectively. What is the electric flux due to these charges through a sphere of radius ‘3a’ with its centre at the origin?
Using gauss law, we have
Electric flux is given by, 
A rectangular conductor LMNO is placed in a uniform magnetic field of 0.5 T. The field is directed perpendicular to the plane of the conductor. When the arm MN of length of 20 cm is moved towards left with a velocity of 10 ms–1, calculate the emf induced in the arm. Given the resistance of the arm to be 5 (assuming that other arms are of negligible resistance) find the value of the current in the arm.
Emf induced in a moving rod is given by,
E = -Blv
Emf induced in a moving rod is given by,
E = + Blv
= 0.5 x 0.2 x 10
= 1 v
Therefore, current in the rod is given by,
I = 
Using Gauss’ laws deduce the expression for the electric field due to a uniformly charged spherical conducting shell of radius R at a point (i) outside and (ii) inside the shell.
Plot a graph showing variation of electric field as a function of r > R and r < R. (r being the distance from the centre of the shell).
i) Consider a uniformly charged thin spherical shell of radius R carrying charge Q. To find the electric field outside the shell, we consider a spherical Gaussian surface of radius (r >R), concentric with given shell. If E is electric field outside the shell, then by symmetry electric field strength has same magnitude E0 on the Gaussian surface and is directed radially outward.
So, electric flux through Gaussian surface is given by, 
Therefore, 
Charge enclosed by the Gaussian surface is Q.
Therefore, using gauss’s theorem, we have
Thus, electric field outside a charged thin spherical shell is the same as if the whole charge Q is concentrated at the centre.
ii) Electric field inside the shell:
The charge resides on the surface of a conductor. Thus, a hollow charged conductor is equivalent to a charged spherical shell. Let’s consider a spherical Gaussian surface of radius (r < R). If E is the electric field inside the shell, then by symmetry electric field strength has the same magnitude Ei on the Gaussian surface and is directed radially outward.
Electric flux through the Gaussian surface is given by, 
= 
Now, Gaussian surface is inside the given charged shell, so charge enclosed by Gaussian surface is zero.
Therefore, using Gauss’s theorem, we have
Thus, electric field at each point inside a charged thin spherical shell is zero.
The graph above shows the variation of electric field as a function of R.
Given, cube encloses an electric dipole.
So, the total charge enclosed by the cube is zero.
That is, Q= 0.
Now, using the formula for electric flux as per the Gauss’s law of electrostatics, we have
(a) Define mutual inductance and write its S.I. units.
(b) Derive an expression for the mutual inductance of two long co-axial solenoids of same length wound one over the other.
(c) In an experiment, two coils C1 and C2 are placed close to each other. Find out the expression for the emf induced in the coil C1 due to a change in the current through the coil C2.
a) Mutual inductance of two coils is equal to the e.m.f induced in one coil when 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 S1 completely.
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
= Flux linked with S2 due to current flowing in S1
is the coefficient of mutual inductance of two solenoids.
When current is passed through S1, emf is induced in S2 .
Magnetic field inside solenoid S1 is given by, 
Magnetic flux linked with each turn of the solenoid = 
Total magnetic flux liked with S2 is given by,
Similarly, mutual inductance between two solenoids, when current is passed through S2 and emf induced in solenoid S1 is given by, 
Hence, coefficient of mutual induction between the two long solenoids is given by, 
c) It is found that, 
where, I is the strength of current in coil 2, and
is the total amount of magnetic flux linked with coil 1.
Emf induced in the neighboring coil C1 is,
(a) An electric dipole of dipole moment 
consists of point charges +q and –q separated by a distance 2a apart. Deduce the expression for the electric field 
due to the dipole at a distance x from the center of the dipole on its axial line in terms of the dipole moment 
. Hence show that in the limit x >> a, 
2 
/ (4p 
0 x3).
(b) Given the electric field in the region 
= 2x i , find the net electric flux through the cube and the charge enclosed by it.
Electric field on axial line of an electric dipole is given by, 
Suppose, P is a point at distance r from the center of the dipole on the side of charge –q.
Electric field at P due to –q is given by, 
is the unit vector along the dipole axis.
Electric field at P due to +q is given by,
Therefore, total electric field at point P is,
b) Since, the electric field is parallel to the faces parallel to xy and xz planes, the electric flux through them is zero.
Electric flux through the left face, 
Electric flux through the right face, 
Net flux is given by, 
"For any charge configuration, equipotential surface through a point is normal to the electric field." Justify.
Work done (W) in moving a test charge along an equipotential surface is zero.
Work done is given by, 
F is the electric force and s is the magnitude of displacement.
For non-zero displacement, this is possible only when cos 
is equal to 0.
= 90°
Thus, the force acting on the point charge is perpendicular to the equipotential surface.
Electric field lines give us the direction of electric force on a charge.
Thus, for any charge configuration, equipotential surface through a point is normal to the electric field.
Two spherical bobs, one metallic and the other of glass, of the same size are allowed to fall freely from the same height above the ground. Which of the two would reach earlier and why?
The glass bob would reach the ground earlier. Glass bob which is non-conducting in nature will only experience Earth’s gravitational pull unlike the metallic bob which is conducting.
Since the metallic bob is conducting in nature, eddy current is induced as it falls through the magnetic field of the Earth. As per Lenz’s law, current is induced in a direction opposite to the motion of the metallic bob. Hence, there is a delay.
Given a uniform electric field E =5 ×103 
N/C, find the flux of this field through a square of 10 cm on a side whose plane is parallel to the y-z plane. What would be the flux through the same square if the plane makes a 30° angle with the x-axis?
Given, Electric flux, 
We have, Electric flux, ϕ = E.A = 
= 50 Weber
When the plane makes a 30° angle with the x-axis, the area vector makes 60° with the x-axis
ϕ = E. A
⇒ ϕ=EA cos θ
⇒ ϕ= (5×103)(10−2) cos 60°
⇒ ϕ= 
⇒ ϕ = 25 Weber
(a) Describe a simple experiment (or activity) to show that the polarity of emf induced in a coil is always such that it tends to produce a current which opposes the change of magnetic flux that produces it.
(b) The current flowing through an inductor of self-inductance L is continuously increasing. Plot a graph showing the variation of
(i) Magnetic flux versus the current
(ii) Induced emf versus dI/dt
(iii) Magnetic potential energy stored versus the current.
(a) The statement ‘polarity of the induced emf is such that it opposes a change in magnetic flux’ is given by Lenz law.
The given activity demonstrates the above statement. The amount of magnetic flux linked with the coil increases, when the north pole of a bar magnet is brought near the coil. Current in the coil is induced in a so as to opposes the increase in magnetic flux. This is possible only when the current induced in the coil is in anti-clockwise direction, with respect to an observer. The magnetic moment associated with this induced emf has north polarity, towards the north pole of the approaching bar magnet.
Similarly, magnetic flux linked with the coil decreases when the north pole of the bar magnet is moved away from the coil. Inorder to oppose this decrease in magnetic flux, current is induced in the coil in clockwise direction so that its south pole faces the receding north pole of the bar magnet. This would result in an attractive force which opposes the motion of the magnet and the corresponding decrease in magnetic flux.
(b) (i) Since ϕ = L I ;
where,
I = Strength of current through the coil at any time,
ϕ = Amount of magnetic flux linked with all turns of the coil at that time, and
L = Constant of proportionality called coefficient of self-induction.
(ii) Induced emf, e=− = 
i.e., e = − 
(iii) Since magnetic potential energy is given by , 
(a) Draw a schematic sketch of an ac generator describing its basic elements. State briefly its working principle. Show a plot of variation of
(i) Magnetic flux and
(ii) Alternating emf versus time generated by a loop of wire rotating in a magnetic field.
(b) Why is choke coil needed in the use of fluorescent tubes with ac mains?(a) The working of AC generator is based on the principle of electromagnetic induction.
Construction:
AC generator mainly consists of:
Armature − The rectangular coil ABCD
Filed Magnets − Two pole pieces of a strong electromagnet
Slip Rings − The ends of the coil ABCD are connected to two hollow metallic rings R1 and R2.
Brushes − B1 and B2 are two flexible metal plates or carbon rods. They are fixed and are kept in tight contact with R1 and R2, respectively.
Working –
The angle θ between field and the normal to the coil changes continuously as the armature coil is rotated in a magnetic field. Therefore, magnetic flux linked with the coil changes and an emf is induced in the coil. According to Fleming’s right hand rule, current is induced from A to B in AB and from C to D in CD. In the external circuit, current flows from B2 to B1.
Magnetic flux linked with the coil is given by,
... (1)
Graph between magnetic flux and time, according to equation (i), is shown below:
As the coil rotates, angle θ changes. Therefore, magnetic flux Φ linked with the coil changes and an emf is induced in the coil. At this instant t, if e is the emf induced in the coil, then
The graph between alternating emf vs. time is shown below:
(b) A choke coil enables us to control the current in an ac circuit. If resistance R is used, then a lot of energy will be wasted in the form of heat.
Why should electrostatic field be zero inside a conductor?
In a conductor, charge resides on the surface. So, charge inside the conductor is zero.
Then, according to the Gauss Theorem, electrostatic field is zero. 
Predict the directions of induced currents in metal rings 1 and 2 lying in the same plane where current I in the wire is increasing steadily. 
The direction of induced current is predicted using Lenz’s law. So, current is induced in a direction so as to oppose the increasing magnetic flux.
Therefore, direction of current is clockwise in ring 1 and anticlockwise in ring 2.
Draw a plot showing the variation of (i) electric field (E) and (ii) electric potential (V) with distance r due to a point charge Q.
For a point charge Q,
Electric potential is inversely proportional to r and Electric field is inversely proportional to r2.
|
Electric Potential |
Electric field |
 |
 |
The graph below shows us the variation of E and V with distance ‘r’.
A metallic rod of ‘L’ length is rotated with angular frequency of ‘ 
’ with one end hinged at the centre and the other end at the circumference of a circular metallic ring of radius L, about an axis passing through the centre and perpendicular to the plane of the ring. A constant and uniform magnetic field B parallel to the axis is present everywhere. Deduce the expression for the emf between the centre and the metallic ring.
Emf is induced is given by, 
where, 
is the rate of change of area of the loop formed by the sector OPQ.
At any instant of time t, let be the angle between the rod and the radius of the circle at P.
Area of the sector OPQ = 
;R is the radius of the circle.
Therefore, emf induced is, 
The figure shows a series LCR circuit with L = 5.0 H, C = 80 mF, R = 40 W connected to a variable frequency 240V source. Calculate. 
(i) The angular frequency of the source which drives the circuit at resonance.
(ii) The current at the resonating frequency.
(iii) The rms potential drop across the capacitor at resonance.
Given a series LCR circuit,
i) At resonance, angular frequency is given by,
ii) Current at resonating frequency is given by,
iii) RMS potential drop across the capacitor at resonance is given by, 
(a) Define electric flux. Write its S.I. units.
(b) Using Gauss’s law, prove that the electric field at a point due to a uniformly charged infinite plane sheet is independent of the distance from it.
(c) How is the field directed if (i) the sheet is positively charged, (ii) negatively charged?a) Electric flux is defined as the total number of electric field lines passing through an area normal to them.
SI unit is Nm2/C.
b) Gauss’s theorem states that the total electric flux through a closed surface is equal to times the net charge enclosed by the surface. 
Consider a uniformly charged infinite plane sheet of charge density 
.
Consider a Gaussian surface inside as shown in the figure which is in the form of a cylinder.
On applying Gauss’s law, we have
Thus, electric field strength due to an infinite flat sheet of charge is independent of the distance of the point and is directed normally away from the charge.
If the surface charge density s is negative the electric field is directed towards the surface charge.
c) i) Away from the charged sheet.
ii) towards the plane sheet.
Two small identical electrical dipoles AB and CD, each of dipole moment 'p' are kept at an angle of 120o as shown in the figure. What is the resultant dipole moment of this combination? If this system is subjected to electric field (
) directed along + X direction, what will be the magnitude and direction of the torque acting on this? 
Given, AB and CD are dipoles kept at an angle of 120o to each other.
Resultant magnetic dipole moment is given by, 
Resultant magnetic dipole makes an angle 60o with Y-axis or 30o with x-axis.
Now, torque is given by, 
Direction of torque is along negative Z-direction.
Using Gauss's law obtains the expression for the electric field due to a uniformly charged thin spherical shell of radius R at a point outside the shell. Draw a graph showing the variation of electric field with r, for r > R and r < R.
Electric field intensity at a point outside a uniformly charged thin spherical shell:
Consider a uniformly charged thin spherical shell of radius R carrying charge Q. Let us assume a spherical Gaussian surface of radius r (>R), concentric with the given shell inorder to find the electric field outside the shell.
If is the electric field outside the shell, then by symmetry electric field strength has same magnitude Eo on the Gaussian surface and is directed radially outward. Also, the direction of normal at each point is radially outward. So, angle between 
Therefore, electric flux through Gaussian surface, 
Now, charge enclosed by the Gaussian surface is Q [Gaussian surface is outside the given charged shell].
Therefore, using Gauss theorem, 
This is the required electric field outside a given thin charged shell.
If 
is the surface charge density of the spherical shell, then
Electric field inside the shell:
If is the electric field inside the shell, then by symmetry electric field strength has the same magnitude Ei on the Gaussian surface and is directed radially outward. Also the directions of normal at each point are radially outward. So angle between Ei and is zero at each point.
Thus, electric flux through Gaussian surface is 0 because of the absence of the charge.
The below graph shows the variation of electric field with r, for r >R and r < R.
(i) With the help of a labelled diagram, describe briefly the underlying principle and working of a step up transformer.
(ii) Write any two sources of energy loss in a transformer.
(iii) A step up transformer converts a low input voltage into a high output voltage.
Does it violate law of conservation of energy? Explain.
i) Underlying principle of a step-up transformer:
A transformer which increases the ac voltage is known as a step up transformer.
Working of step-up transformer is based on the principle of mutual inductance and it converts the alternating low voltage to alternating high voltage. The number of turns in the secondary coil is greater than the number of turns in the primary coil.
i) Underlying principle of a step-up transformer: A transformer which increases the ac voltage is known as a step up transformer.
Working of step-up transformer is based on the principle of mutual inductance and it converts the alternating low voltage to alternating high voltage. The number of turns in the secondary coil is greater than the number of turns in the primary coil.
Working: When an A.C source is connected to the ends of the primary coil, the current changes continuously in the primary coil. Hence, the magnetic flux which is linked with the secondary coil changes continuously. So, the emf which is developed across the secondary coil is same as that in the primary coil. The emf is induced in the coil as per Faraday’s law.
Assumption: We assume that there is no leakage of flux so that, the flux linked with each turn of primary coil is same as flux linked with secondary coil.
ii) Two sources of energy loss in the transformer:
Joule Heating – Energy is lost in resistance of primary and secondary windings in the form of heat.
H = I2 Rt
Flux leakage - Energy is lost due to coupling of primary and secondary coils not being perfect, i.e., whole of magnetic flux generated in primary coil is not linked with the secondary coil.
iii) Conservation of law of energy is not violated in step-up transformer. When output voltage increases, the output current automatically decreases. Thus, there is no loss of energy.
State Lenz’s Law. A metallic rod held horizontally along east-west direction, is allowed to fall under gravity. Will there be an emf induced at its ends? Justify your answer.
The polarity of the induced emf at the open ends of a closed loop is such that it tends to produce a current which opposes the change in magnetic flux that produced it.
Emf will be induced at the ends. Magnetic flux changes in a metallic rod falling freely under gravity and hence, emf is induced.
A convex lens of focal length 25 cm is placed coaxially in contact with a concave lens of focal length 20 cm. Determine the power of the combination. Will the system be converging or diverging in nature?
Given, a convex lens placed coaxially in contact with a concave lens.
Power of convex lens, P1 = 
Power of concave lens, P2 = 
So, Power of combination, P = P1 + P2
= 
Since, power has negative magnitude; system of lenses is diverging in nature.
An ammeter of resistance 0.80 can measure current upto 1.0A.
(i) What must be the value of shunt resistance to enable the ammeter to measure current upto 5.0A?
(ii) What is the combined resistance of the ammeter and the shunt?
Given, Resistance, R = 0.80
Maximum current, Imax = 1 A
So, Voltage across ammeter, V = IR = 1.0 0.80 = 0.8 V
i) If a shunt is connected in parallel,
Current I flowing through S is (I – I1)
For parallel combination of resistors,
I1. RA = ((I – I1) S
This is the required value of shunt resistance.
ii) Combined resistance of the ammeter and the shunt, 
A metallic rod of length ‘l’ is rotated with a frequency v with one end hinged at the centre and the other end at the circumference of a circular metallic ring of radius r, about an axis passing through the centre and perpendicular to the plane of the ring. A constant uniform magnetic field B parallel to the axis is present everywhere. Using Lorentz force, explain how emf is induced between the centre and the metallic ring and hence obtain the expression for it.
As the rod is rotated, free electrons in the rod move towards the outer end due to Lorentz force and get distributed over the ring. Thus, the resulting separation of charges produces an emf across the ends of the rod. At a certain value of emf, there is no more flow of electrons and a steady state is reached.
Expression for Induced emf in a Rotating Rod
Consider a metallic rod OA of length l, which is rotating with angular velocity in a uniform magnetic field B, the plane of rotation being perpendicular to the magnetic field. A rod may be supposed to be formed of a large number of small elements. Consider a small element of length dx at a distance x from centre.
If v is the linear velocity of this element, then area swept by the element per second = v.dx
Emf induced across the rod , 
But, v = 
Therefore,
Emf induced across the rod is given by, 
Power dissipated, when circuit is closed is, P = 
.
While travelling back to his residence in the car, Dr. Pathak was caught up in a thunderstorm. It became very dark. He stopped driving the car and waited for thunderstorm to stop. Suddenly he noticed a child walking alone on the road. He asked the boy at his residence. The boy insisted that Dr. Pathak should meet his parents. The parents expressed their gratitude to Dr. Pathak for his concern for safety of the child.
Answer the following questions based on the above information:
(a) Why is it safer to sit inside a car during a thunderstorm?
(b) Which two values are displayed by Dr. Pathak in his actions?
(c) Which values are reflected in parents’ response to Dr. Pathak?
(d) Give an example of a similar action on your part in the past from everyday life.
a) On the basis of electrostatic screening, no electric field exists inside the charged conducting body. During lightening a shower of the charged particles falls on the earth. So it would be safer to sit inside the car.
(b) Dr. Pathak knows the result of lightening during thunderstorm; so he displayed two actions;
(i) Shows love, kindness and sympathy to the child.
(ii) Keeping in view the safety of the child, he allow the boy to sit in the car till the thunderstorm stopped.
(c) Parent meets Dr. Pathak; and express their gratitude and heart felt thank for providing the safety to the child from lightning and thunderstorm.
(d) Many of us have read in the newspaper that the person either working in the field or in open space have lost their life during thunderstorm. So the persons belonging to villages must be given advices that they should remain inside the houses during thunderstorm.
The field lines of a negative point charge are as shown in the figure. Does the kinetic energy of a small negative charge increase or decrease in going from B to A? 
The electric field present due to the given point charge will be directed towards the centre. A negative charge always experiences a force in the direction opposite to that of the external electric field present; that is away from the centre. This will cause its motion to retard while moving from B to A. Hence, its kinetic energy will decrease in going from B to A.
State Lenz's law. Illustrate, by giving an example, how this law helps in predicting the direction of the current in a loop in the presence of a changing magnetic flux.
In a given coil of self-inductance of 5 mH, current changes from 4 A to 1 A in 30 ms. Calculate the emf induced in the coil.
OR
In what way is Gauss's law in magnetism different from that used in electrostatics? Explain briefly.
The Earth's magnetic field at the Equator is approximately 0.4 G. Estimate the Earth's magnetic dipole moment. Given: Radius of the Earth = 6400 km.
Lenz law states that,polarity of the induced emf is such that it opposes a change in magnetic flux.
The given activity demonstrates the above statement. The amount of magnetic flux linked with the coil increases, when the north pole of a bar magnet is brought near the coil. Current in the coil is induced in a so as to opposes the increase in magnetic flux. This is possible only when the current induced in the coil is in anti-clockwise direction, with respect to an observer. The magnetic moment associated with this induced emf has north polarity, towards the north pole of the approaching bar magnet.
Similarly, magnetic flux linked with the coil decreases when the north pole of the bar magnet is moved away from the coil. Inorder to oppose this decrease in magnetic flux, current is induced in the coil in clockwise direction so that its south pole faces the receding north pole of the bar magnet. This would result in an attractive force which opposes the motion of the magnet and the corresponding decrease in magnetic flux.
Given,
Self-inductance, L = 5 mH = 510-3 H
Change in current, dI = (4-1) = 3 A
Change in time, dt = 30 ms = 3010-3 s
So, emf induced in the coil is given by, 
OR
i) Gauss’s law for magnetism states that magnetic flux through any closed surface is 0.
That is,
Gauss law for electrostatics states that electric flux through a closed surface is give n by,
Therefore, electric flux is zero when the surface encloses an electric dipole.
Magnetic flux is zero implies that, isolated magnetic poles do not exist.
ii) Given,
Magnetic field of Earth = 0.4 G = 0.4 10-4 G
Equatorial magnetic field of earth is given by, 
where, d = 6400km = 6.4 x 106 m
Therefore, Earth’s magnetic dipole moment is given by, 
The electric current flowing in a wire in the direction from B to A is decreasing. Find out the direction of the induced current in the metallic loop kept above the wire as shown. 
The direction of induced current is so as to oppose the change in magnetic flux through the wire.
Define the term ‘mutual inductance’ between the two coils. Obtain the expression for mutual inductance of a pair of long coaxial solenoids each of length l and radii r1 and r2 (r2 >> r1). Total number of turns in the two solenoids is N1 and N2 respectively.
The ratio of magnetic flux passing through one coil to the current passing through the other is known as mutual inductance between the two coils.
Consider current I2 flowing through the outer coil.
Magnetic field due to current I2 is given by, 
Therefore, resultant magnetic flux linked with the inner coil is, 
(a) Deduce the expression for the torque acting on a dipole of dipole moment p in the presence of a uniform electric field E⃗.
(b) Consider two hollow concentric spheres, S1 and S2, enclosing charges 2Q and 4Q respectively as shown in the figure.
(i) Find out the ratio of the electric flux through them.
(ii) How will the electric flux through the sphere S1 change if a medium of dielectric constant 'εr' is introduced in the space inside S1 in place of air ?
Deduce the necessary expression.
a) Dipole in a uniform electric field:
Consider an electric dipole consisting of charges −q and +q and of length 2a placed in a uniform electric field making an angle θ with electric field.
Forces acting on the two charges of the dipole, are +qE and -qE.
That is, the net force on the dipole is equal and opposite.
So, Force = 0
Two forces are equivalent to torque having magnitude given by,
Therefore, torque acting on the dipole is given by, 
b) i) Charge enclosed by sphere S1 = 2Q
Charge enclosed by sphere S2 = 2Q + 4Q = 6Q
Now, using Gauss law, electric flux enclosed by sphere S1 and S2 is given by, 
ii) If a medium of dielectric constant 'εr' is introduced in the space inside S1 in place of air, electric flux becomes
That is, electric flux decreases.
A bar magnet is moved in the direction indicated by the arrow between two coils PQ and CD.
Predict the directions of induced current in each coil.
In the figure, N pole is receding away coil (1), so in coil (1), the nearer faces will act as S-pole and in coil (2) the nearer face will also act as S-pole to oppose the approach of magnet towards coil (2), so current in coils will flow clockwise as seen from the side of magnet. The direction of current will be from P to Q in coil (1) and from C to D in coil (2).
An electric dipole is held in a uniform electric field.
(i) Show that the net force acting on it is zero.
(ii) The dipole is aligned parallel to the field. Find the work done in rotating it through the angle of 180°.
i) Dipole moment of dipole is given by,
i) 
Force on –q at A = -qE
Force on +q at B = + qE
So,
Net force = qE – qE = 0
ii)
Work done on dipole is given by, 
Define electric dipole moment. Write its S.I. unit.
Electric dipole moment is defined as the product of charge and the distance between the charges, and is directed from negative to positive charge.
The SI unit of electric dipole moment is coulomb metre (Cm).
A thin straight infinitely long conducting wire having charge density 
is enclosed by a cylindrical surface of radius r and length l, its axis coinciding with the length of the wire. Find the expression for the electric flux through the surface of the cylinder.
Given, charge density = 
Radius and length of the cylindrical surface are r and l respectively.
Charge enclosed by the cylindrical surface = 
l
Now, using Gauss theorem,
Electric flux, 
Plot a graph showing the variation of coulomb force (F) versus 
, where r is the distance between the two charges of each pair of charges: (1 
C, 2 
C) and (2 
C – 3 
C). Interpret the graphs obtained.
Force between charged particles is given by, 
The graph between F and 
is a straight line of slope passing through the origin.
Magnitude of slope is more for attraction, therefore attractive force is greater than the repulsive force.
What are eddy currents? Write any two applications of eddy currents.
Eddy currents are the currents induced in a metallic plate when it is kept in a time varying magnetic field. Magnetic flux linked with the plate changes and so the induced current is set up. Eddy currents are sometimes so strong, that metallic plate becomes red hot.
Applications:
State the working of a.c. generator with the help of a labelled diagram.
The coil of an a.c. generator having N turns, each of area A, is rotated with a constant angular velocity. Deduce the expression for the alternating e.m.f. generated in the coil.
What is the source of energy generation in this device?
OR
(a) Show that in an a.c. circuit containing a pure inductor, the voltage is ahead of current by 
/2 in phase.
(b) A horizontal straight wire of length L extending from east to west is falling with speed v at right angles to the horizontal component of Earth’s magnetic field B.
(i) Write the expression for the instantaneous value of the e.m.f. induced in the wire.
(ii) What is the direction of the e.m.f.?
(iii) Which end of the wire is at the higher potential?
Working of ac generator:
When the armature coil is rotated in the strong magnetic field, the magnetic flux linked with the coil changes and the current is induced in the coil, its direction being given by Fleming’s right hand rule. Considering the armature to be in vertical position and as it rotates in anticlockwise direction, the wire ab moves upward and cd downward, so that the direction of induced current is shown in fig. In the external circuit, the current flows along B1RL B2.
The direction of current remains unchanged during the first half turn of armature. During the second half revolution, the wire ab moves downward and cd upward, so the direction of current is reversed and in external circuit it flows along B2RL B1. Thus the direction of induced emf and current changes in the external circuit after each half revolution.
Let, N be the number of turns in the coil,
f is the frequency of rotation,
A is the area of coil,
B is the magnetic induction.
Then, the induced emf is given by, 
We can see that the emf and current produced is alternating.
Current produced by an ac generator cannot be measured by moving coil ammeter; because the average value of ac over full cycle is zero.
The source of energy generation is the mechanical energy of rotation of armature coil.
OR
Consider a coil of self-inductance L and negligible ohmic resistance. An alternating potential difference is applied across the ends. The magnitude and direction of AC is changing periodically, changing the magnetic flux. So, an emf is induced and produced in the coil. The instantaneous value of alternating voltage applied is given by,
V = Vo sin w t
Instantaneous induced emf is, 
According to Kirchhoff’s second law in closed circuit, we get
Integrating with respect to time ‘t’,
which is the required expression for current.
;
where, 
is the peak value of alternating current.
From the above equations we can say that the current lags behind the applied voltage by an angle .
b)
i) Induced emf, = BHvL ; where BH is horizontal component of earth’s magnetic field directed from South to North.
ii) West to East
iii) East end of the wire is at higher potential
According to Gauss's law, flux through a closed surface is given by,
Consider a ring of radius 'a' which carries uniformly distributed positive total charge Q. 
To find: electric field due to a ring at a point P lying at a distance x from its centre along the central axis perpendicular to the plane of the ring.
i) Define mutual Inductance.
ii) A pair of adjacent coils has a mutual inductance of 1.5 H. If the current in one coil changes 20 A in 0.5 s, what is the change of flux linkage with the other coil?
i) Mutual Induction: It is the phenomenon in which a change of current in one coil induces an emf in another coil places near it. The coil in which the current changes is called the primary coil and the coil in which the emf is induced is called the secondary coil.
ii)
EMF induced is, e = - M . 
So, the flux linked with the other coil is given by, 
, where 
is the electric flux produced during charging of the capacitor plates.
A moving charge produces both electric and magnetic fields, and an oscillating charge produces oscillating magnetic and electric fields. These oscillating electric and magnetic fields with respect to space and time produce electromagnetic waves.
The propagation of electromagnetic waves can be shown as:
OR
Maxwell's generalization of ampere's circuital law given by,
Consider that a parallel capacitor C is charging in a circuit.
The magnitude of electric field between the two plates will be, 
, is perpendicular to the surface of the plate.
Electric flux through the surface will be, 
Ram is a student of class X in a village school. His uncle gifted him a bicycle with a dynamo fitted in it. He was very excited to get it. While cycling during night, he could light the bulb and see the objects on the road. He, however, did not know how this device works. He asked this question to his teacher. The teacher considered it an opportunity to explain the working to the whole class.
Answer the following questions:
(a) State the principle and working of a dynamo.
(b) Write two values each displayed by Ram and his school teacher.
Dynamo:
a) Principle: Whenever a coil is rotated in a magnetic field, an emf is induce in it due to change in magnetic flux linked with the coil.
Working:
As the coil in the dynamo rotates, its inclination (
) with respect to the field changes. Therefore, a varying emf is obtained which is given by,
b) Values displayed by:
Ram: curiosity, scientific aptitude, keenness to learn.
Teacher: Depth of knowledge, motivational approach, generous, dedicated.
(a) Explain the meaning of the term mutual inductance. Consider two concentric circular coils, one of radius r1 and the other of radius r2 (r1 < r2) placed coaxially with centres coinciding with each other. Obtain the expression for the mutual inductance of the arrangement.
(b) A rectangular coil of area A, having number of turns N is rotated at ‘ f ’ revolutions per second in a uniform magnetic field B, the field being perpendicular to the coil. Prove that the maximum emf induced in the coil is 2
f NBA.
is the flux linked with S2 due to current flowing through S1
(i) Use Gauss’s law to find the electric field due to a uniformly charged infinite plane sheet. What is the direction of field for positive and negative charge densities?
(ii) Find the ratio of the potential differences that must be applied across the parallel and series combination of two capacitors C1 and C2 with their capacitances in the ratio 1 : 2 so that the energy stored in the two cases becomes the same.
on both sides of the sheet. Let a point be at a distance a from the sheet at which the electric field is required.
Area of the cross section of the gaussian cylinder.
(i) If two similar large plates, each of area A having surface charge densities 1s and 2s are separated by a distance d in air, find the expressions for
(a) field at points between the two plates and on outer side of the plates. Specify the direction of the field in each case.
(b) the potential difference between the plates.
(c) the capacitance of the capacitor so formed.
(ii) Two metallic spheres of radii R and 2R are charged so that both of these have same surface charge density s. If they are connected to each other with a conducting wire, in which direction will the charge flow and why ?
Tips: -
Drive the expression for electric field at a point on the equatorial line of an electric dipole.
The magnitudes of the electric field due to the two charges +q and -q are given by,
The directions of E+q and E-q are as shown in the figure. The components normal to the dipole axis cancel away. The components along the dipole axis add up.
Therefore, Total electric field.
E = - (E+q + E-q) cosθ p [Negative sign shows that field is opposite to p]
At large distances (r>>a),this reduces to
Depict the orientation of the dipole in (i) stable, (ii) unstable equilibrium in a uniform electric field.
Depicting the orientation of the dipole in (i) stable equilibrium in a uniform electric field.
Net force is zero in this case as qE-qE = 0
Net torque =pEsinθ as θ = 0
(ii) Unstable equilibrium in a uniform electric field.
Net force is negative in this case as-qE-qE = -2qE
Net torque = pEsinθ asθ = 180
Four point charges Q, q, Q and q are placed at the corners of a square of side ‘a’ as shown in the figure
Find the
(a) resultant electric force on a charge Q,
(b) the potential energy of this system.
F1 and F2 are perpendicular to each other so their resultant will be
F3 and resultant of F1 and F2 will be in the same direction
Net force
F = F' + F3
(b) The potential energy of the given system,
Three point charges q, – 4q and 2q are placed at the vertices of an equilateral triangle ABC of side ‘l’ as shown in the figure. Obtain the expression for the magnitude of the resultant electric force acting on the charge q.
(b) Find out the amount of the work done to separate the charges at infinite distance.
(b) The amount of work done to separate the charges at infinity will be equal to potential energy.
Define electric flux. Is it a scalar or a vector quantity?
A point charge q is at a distance of d/2 directly above the centre of a square of side d, as shown in the figure. Use Gauss’ law to obtain the expression for the electric flux through the square.
b) If the point charge is now moved to a distance 'd' from the centre of the square and the side of the square is doubled, explain how the electric flux will be affected.
The total number of electric field lines crossing an area placed normal to the electric field is termed as electric flux.
It is denoted by Φ.
Electric flux is a scalar quantity, its SI unit is Nm2C-1
Electric flux through square is
b) Flux will not be changed,
i.e.,
a) Use Gauss’ law to derive the expression for the electric field due to a straight uniformly charged infinite line of charge density λ C/m.
(b) Draw a graph to show the variation of E with perpendicular distance r from the line of charge.
(c) Find the work done in bringing a charge q from perpendicular distance r1 to r2 (r2>r1).
Gauss’s law in electrostatics: It states that total electric flux over the closed surface S is times the total charge (q) contained inside S.
Electric field due to an infinitely long straight wire.
Electric field due to an infinitely long straight wire.
Let us consider an infinitely long line charge having linear charge density λ. Assume a cylindrical Gaussian surface of radius r and length 1 coaxial with the line charge.
By symmetry, the electric field E has the same magnitude at each point of the curved surface S1 and is directed radially outward. So angle at surfaces between is zero, and angle of
Total flux through the cylindrical surface,
Since λ is the charge per unit length and l is the length of the wire.
Thus, the charge enclosed
q = λl
According to Gaussian law,
b) 
c)
The region between two concentric spheres of radii ‘a’ and ‘b’, respectively (see figure), has volume charge density ρ = A/r , where A is a constant and r is the distance from the centre. At the centre of the spheres is a point charge Q. The value of A such that the electric field in the region between the spheres will be constant is:
A.
A Gaussian surface at distance r from centre.
At the centre of the sphere is a point charge Q. The value of A such that the electric field in the region between the spheres will be constant is
As, Q = 2πAa2
i.e A = Q/2πa2
A long cylindrical shell carries positive surface charge in the upper half and negative surface charge in the lower half. The electric field lines around the cylinder will look like figure given in: (figures are schematic and not drawn to scale)
D.
Field lines should originate from a positive charge and terminate negative charge.
Two charges, each equal to q, are kept at x = −a and x = a on the x-axis. A particle of mass m and charge qo =-q/2 is placed at the origin. If charge qo is given a small displacement (y<< a) along the y-axis, the net force acting on the particle is proportional to
y
-y
1/y
-1/y
A.
y
A coil is suspended in a uniform magnetic field, with the plane of the coil parallel to the magnetic lines of force. When a current is passed through the coil it starts oscillating; it is very difficult to stop. But if an aluminium plate is placed near to the coil, it stops. This is due to
development of air current when the plate is placed.
induction of electrical charge on the plate
shielding of magnetic lines of force as aluminium is a paramagnetic material.
electromagnetic induction in the aluminium plate giving rise to electromagnetic damping.
D.
electromagnetic induction in the aluminium plate giving rise to electromagnetic damping.
The oscillating coil produces time variable magnetic field. It causes eddy current in the aluminium plate which causes anti–torque on the coil, due to which it stops.
A boat is moving due east in a region where the earth's magnetic field is 5.0 × 10-5 NA-1m-1 due north and horizontal. The boat carries a vertical aerial 2m long. If the speed of the boat is 1.50 ms-1, the magnitude of the induced emf in the wire of aerial is
0.75 mV
0.50 mV
0.15 mV
1 mV
C.
0.15 mV
Eind = B × v × l
= 5.0 × 10-5 × 1.50 × 2
= 10.0 × 10-5 × 1.5
= 15 × 10-5 vot.
= 0.15 mv
Let there be a spherically symmetric charge distribution with charge density varying as 
upto r = R, and ρ(r) = 0 for r > R, where r is the distance from the origin. The electric field at a distance r ( r < R) from the origin is given by
B.
Apply shell theorem, the total charge upto distance r can be calculated as follows
Charges are placed on the vertices of a square as shown. Let E be the electric field and V the potential at the centre. If the charges on A and B are interchanged with those on D and C respectively, then
remains unchanged, V changes
Both 
and V change
and V remain unchanged
changes, V remains unchanged
D.
changes, V remains unchanged
From a uniform circular disc of radius R and mass 9M, a small disc of radius R/3 is removed as shown in the figure. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through centre of disc is:
4MR2
10 MR2
B.
4MR2
let σ be the mass per unit area.
The total mass of the disc = σ x πR2 = 9M
The mass of the circular disc out
Let us consider the above system as a complete disc of mass 9M and a negative mass M superimposed on it. Moment of Inertia (I1) of the complete disc (9MR2)/2 about an axis passing through O and perpendicular to the plane of the disc.
M.I. of the cut-out portion about an axis passing through O and perpendicular to the plan of disc
Therefore, M.I (I2) of the cut out portion about an axis passing through O and perpendicular to the plane of disc
Using perpendicular axis theorem
Therefore, the total M.I. of the system about an axis passing through O and perpendicular to the plane of the disc is
I = I1 + I2
A long solenoid has 1000 turns. when a current of 4 A flows through it, the magnetic flux linked with each turn of the solenoid is 4x10-3 Wb. The self inductance of the solenoid is,
3H
2H
1H
4H
C.
1H
Given,
Number of turns in the solenoid, N= 1000
Current, I = 4 A
Magnetic flux, 
= 4 x 10-3 Wb
Self-inductance of solenoid is given by,
... (i)
Putting the value of equation in (i), we get
L = 
= 1 H
An electron moving in a circular orbit of radius r makes n rotations per second. The magnetic field produced at the centre has magnitude.
zero
D.
As I = q/t
So, for an electron revolving in a circular orbit of radius r
q = e and t =T
What is the flux through a cube of side a if a point charge of q is a one of its corner?
B.
Charge enclosed = q/8
therefore,
Two thin dielectric slabs of dielectric constants K1 and K2 (K1 < K2) are inserted between plates of a parallel plate capacitor, as shown in the figure. The variation of electric field E between the plates with distance d as measured from plate P is correctly shown by:
B.
The electric field inside the dielectrics will be less than the electric field outside the dielectrics. The electric field inside the dielectric could not be zero.
As K2 > K1 the drop in electric field for K2 dielectric must be more than K1.
A conducting sphere of radius R is given a charge Q. The electric potential and the electric field at the centre of the sphere respectively are:
zero and 
Both are zero
B.
In a region, the potential is represented by V (x,y,z) = 6x - 8 xy - 8y + 6yz, where V is in volts and x, y, z are in metres. The electric force experienced by a charge of 2 coulomb situated at point (1,1,1) is,
N
30 N
24 N
N
D.
N
We know,
F = qE ... (i)
E = 
... (ii)
The values of Ex, Ey and Ez at (1,1,1) are
Ex = 6 - 8x(1) = -2
Ey = -8(1) - 8 + 6(1) = -10
Ez = 6 x 1 = 6
Therefore,
A thin semicircular conducting ring (PQR) of radius r is falling with its plane vertical in a horizontal magnetic field B, as shown in figure. The potential difference developed across the ring when its speed is v, is
zero
is at higher potential
and R is at higher potential
is at higher potential
D.
is at higher potential
For motional emf,e = Bv x (2r)
= 2rBv
R will be at higher potential.
The electrons of the wire will move towards end P due to due to electric force and at end R the excess positive charge will be left.
The current (I) in the inductance is varying with time according to the plot shown in a figure.
D.
For inductor as we know induced voltage
For t= 0 to t= T/2
A coil of self -inductance L is connected in series with a bulb b and an AC source. Brightness of the bulb decreases when
frequency of the AC source is decreases
number of turns in the coil is reduced
a capacitance of reactance Xc=XL is included in the same circuit
an iron is inserted in the coil
D.
an iron is inserted in the coil
A current loop in a magnetic field
experiences a torque whether the field is uniform or non-uniform in all orientations
can be in equilibrium in one orientations
can be an equilibrium in two orientations, both the equilibrium states are unstable
can be in equilibrium in two orientations, one stable while the other is unstable
D.
can be in equilibrium in two orientations, one stable while the other is unstable
For parallel M is stable and for antiparallel is unstable.
If potential (in volts) in a region is expressed as V (x, y, z) = 6 xy-y +2 yz, the electric field (in N/C) at point (1,1,0) is
B.
Given, potential in a region, V = 6xy - y + 2yz
Electric field in a region
A beam of cathode rays is subjected to crossed Electric (E) and magnetic field (B). The fields are adjusted such that the beam is not deflected. The specific charge of the cathode rays is given by
(Where V is the potential difference between cathode and anode)
B2/2VE2
2VB2/E2
2VE2/B2
2E2/2VB2
D.
2E2/2VB2
As the electron beam is not deflected,
then
Fm = Fe
Or BeV = Ee
Or v= E/B .... (i)
As the electron moves from the cathode to anode, its potential energy at the cathode appears as its kinetic energy at the anode. If V is the potential difference between the anode and cathode, then potential energy of the electron at cathode = eV. Also, the kinetic energy of the electron at anode = mv2 /2. According to law of conservation of energy
Which of the following statement is false or the properties of electromagnetic waves?
Both electric and magnetic field vectors attain the maxima and minima at the same place and same time
The energy in electromagnetic waves is divided equally between electric and magnetic vectors
Both electric and magnetic field vectors are parallel to each other and perpendicular to the direction of propagation of wave
These waves do not require any material medium for propagation
C.
Both electric and magnetic field vectors are parallel to each other and perpendicular to the direction of propagation of wave
The time varying electric and magnetic fields are mutually perpendicular to each other and also perpendicular to the direction of propagation of the waves.
The electric field at a distance 3R/2 from the centre of a charged conducting spherical shell of radius R is E. The electric field at a distance R/2 from the centre of the sphere is
zero
E
E/2
E/3
A.
zero
The radius of a spherical shell is R. Because inside the conducting charge sphere, an electric field is zero. So, the electric field at a distance R/2 from the centre of the sphere will be zero.
The electric field of an electromagnetic wave in free space is given by 
, where t and x are in seconds and meters respectively. It can be inferred that
1) the wavelength λ is 188.4 m.
2)The wave number k is 0.33 rad/m.
3)The wave amplitude is 10 V /m.
4)The wave is propagating along +x direction
(3) and (4)
(1) and (2)
(2) and (3)
(1) and (3)
D.
(1) and (3)
The electric field of electromagnetic wave
A conducting circular loop is placed in uniform magnetic field 0.04 T with its plane perpendicular to the magnetic field. The radius of the loop starts shrinking at 2 mms-1. The induced emf in the loop when the radius is 2 cm is
3.2 π μ V
4.8 π μ V
0.8 π μ V
1.6 π μ V
A.
3.2 π μ V
According to faraday's second law of electromagnetic induction the induced emf is given by rate of change of magnetic flux linked with the circuit.
Here, B = 0.04 T and -dr /dt = 2 mms-1
Induced emf, e = dΦ / dt = -BdA/dt = -Bd(πr2)/dt
=- Bπ 2r dr / dt
now, if r = 2 cm
e = -0.04 x π x 2 x 2 x 10-2 x2 x 10-3
3.2 π μ V
The electric field part of an electromagnetic wave in a medium is represented by
Ex = 0;
moving along y direction with frequency 2π x 106 Hz and wavelength 200 m.
moving along x direction with frequency 106 Hz and wavelength 200 m.
moving along x direction with frequency 106 Hz and wavelength 200 m
moving along x- direction with frequency 106 Hz and wavelength 200 m
C.
moving along x direction with frequency 106 Hz and wavelength 200 m
Comparing the given equation
Hence, the wave is moving along positive x -direction with frequency 106 Hz and wavelength 200 m.
See the electrical circuit shown in this figure. Which of the following equations is a correct equation for it?
A.
Apply Kirchhoff;s second law also called loop rule.
The algebraic sum of the changes in potential in complete transversal fo a mesh (closed loop) is zero ie ,
Here, 
The electric potential at a point (x, y, Z) is given by
V = - x2y -xz3 + 4
The electric field E at that point is
A.
The electric field at a point is equal to the negative gradient of the electrostatic potential at that point.
Potential gradient relates to electric field according to the following relation E = -dV/dr
In the phenomenon of electric discharge through gases at low pressure, the coloured glow in the tube appears as a result of
excitation of electrons in the atoms
a collision between the atoms of the gas
a collision between the charged particle emitted from the cathode and the atoms of the gas
a collision between different electrons of the atoms of the gas
C.
a collision between the charged particle emitted from the cathode and the atoms of the gas
The discharge of electricity through rarefied gases is an interesting phenomenon which can be systematically studied with the help of a discharge tube. In discharge tube collisions between the charged particles emitted from the cathode and the atoms of the gas results to the coloured glow in the tube.
A cell can be balanced against 110 cm and 100 cm of potentiometer wire, respectively with and without being short-circuited through a resistance of 10 Ω. Its internal resistance is
1.0 Ω
0.5 Ω
2.0 Ω
zero
A.
1.0 Ω
This is a problem is a base on the application of potentiometer in which we find the internal resistance of a cell.
In potentiometer experiment in which we find internal resistance of a cell, let E be the emf of the cell and V the terminal potential differences, then
E/V = l1/l2
where l1 and l2 are the length of potentiometer wire with and without short-circuited through a resistance
A circular disc of radius 0.2 m is placed in a uniform magnetic field of induction in
such a way that its axis makes an angle of 60o with.
The magnetic flux linked with the disc is
0.02 Wb
0.06 Wb
0.05 Wb
0.01 Wb
A.
0.02 Wb
The magnetic flux Φ passing through a plane surface of area A placed in a uniform magnetic field B is given by
Φ = BA Cos θ
where θ is the angle between the direction of B and the normal to the plane.
Here,
What is the value of inductance L for which the current is a maximum in a series LCR circuit with C =10μ F and ω = 1000 s-1
100 mH
1 mH
cannot bec calculated unless R is known
10 mH
A.
100 mH
In resonance condition, maximum current flows in the circuit.
Current in LCR series circuit.
where V is rms value of current, R is resistance XL is inductive reactancea and Xc is capacititve reactance.
For current to be maximum denominator should be minimum can be done, if
XL = Xc
This happens in resonance state of the circuit ie,
Under the influence of a uniform magnetic field. a charged particle is moving in a circle of radius R with constant speed v. The time period of the motion.
depends on v and not on R
depends on both R and v
is independent of both R and v
depends on R and not on v
C.
is independent of both R and v
To move on circular path in a magnetic field, a centripetal force is a provided by the magnetic force.
when magnetic field is perpendicular to motion of charge particle, then
i,e, 
It is independent of bot R and v.
A transforner is used to light a 100 W and 110 V lamp from a 220 V mains. If the main current is 0.5 A, the efficiency of the transformer is approximately:
30%
50 %
90 %
10 %
C.
90 %
The efficiencvy of transformer
Energy obtained from the secondary coil
__________________________________
Energy given to the primary coil
n = output
_______
Input
Given VsIs = 100 W, Vp = 220 V, IP = 0.5 A
Hence, n = 100 / 220 x 0.5 = 90 %
When a charged particle moving with velocity 
is subjected to a magnetic field of induction 
, the force on it is non-zero. This implies that
angle between 
is necessarily 
angle between 
can have any value other than 
angle between 
can have any value othe than zero and 
angle between 
is either zero or 
C.
angle between can have any value othe than zero and
When a charged particle q is moving in a uniform magnetic field 
with velocity 
such that angle between 
be 
, then due to interaction between the magnetic field produced due to moving charge and magnetic field applied, the charge q experiences a force which is given by
Since, force on charged particle is non-zero, so angle between 
can have any value other than zero and 
Two coils of self-inductances 2 mH and 8 mH are placed so close together that the effective flux in one coil is completely linked with the other. The mutual inductance between these coils is
10 mH
6 mH
4 mH
16 mH
C.
4 mH
When the total flux associated with one coil links with the other i.e, a case of maximum flux linkage, then
Similarly, 
If all the flux of coil 2 links coil 1 and vice-versa then
Since, 
hence we have
In a discharge tube ionization of enclosed gas is produced due to collisions between
positive ions and neutral atoms/molecules
negative electrons and neutral atoms/molecules
photons and neutral atoms/molecules
neutral gas atoms/molecules
B.
negative electrons and neutral atoms/molecules
In a discharge tube, after being accelerated though a high potential difference the ions in the gas strike the cathode with huge kinetic energy. This collision liberates electrons from the cathode. These free electrons can further liberate ions from gas molecules through collisions. The positive ions are attracted towards the cathode and negatively electrons move towards anode. Thus, ionization of gas results.
An electron falls from rest through a vertical distance h in a uniform and vertically upward directed electric field E. The direction of electric field is now reversed, keeping its magnitude the same. A proton is allowed to fall from rest in it through the same vertical distance h. The time of fall of the electron, in comparison to the time of fall of the proton is
Smaller
5 times greater
Equal
10 times greater
A.
Smaller
As we know, F = qE = ma
i.e., time as 'q' is the same for electron and proton. Since the electron has a smaller mass so it will take a smaller time.
A certain charge Q is divided into two parts q and Q-q. How the charge Q and q must be related so that when q and (Q-q) is placed at a certain distance apart experience maximum electrostatic repulsion?
Q = 2q
q = 3q
Q = 4q
Q= 4q + c
A.
Q = 2q
The electrostatic force of repulsion between the charge q and (Q-q) at separation r is given by
If F is maximum then
i.e.
The working of dynamic is based on the principle of
Electromagnetic induction
chemical effect of current
magnetic effect of current
heating effect of current
A.
Electromagnetic induction
The dynamo operates on the principle of the production of dynamically induced emf. Hence whenever flux is cut by the conductor, emf is produced in it according to the law of electromagnetic induction.
When a voltmeter connected across the terminals of cell, measures 5 V and an ammeter connected measures 10 A. A resistance of 2 Ω is connected across the terminal of the cell. The current flowing through this resistance is
7.5amp
5.0amp
2.5amp
2.0amp
D.
2.0amp
Internal resistance of the cell is
In a neon discharge tube 2.9 x 10 Ne18 ions move to be the right per second while 1.2 x 108 electron move to the left per second electric charge is 1.6 x 10-19C. The current in discharge tube is
0.66A towards left
0.66 A towards left
1A towards right
Zero
B.
0.66 A towards left
Current due to both types of ions are in the same direction towards the right so
i=i1+i2
X-rays are
electromagnetic radiation
stream of electron
stream of proton
stream of uncharged particle
A.
electromagnetic radiation
X-rays consist of oscillating electric and magnetic field at right angles of each other and in the direction of propagation of the x-rays. Hence, X-rays are electromagnetic radiations.
The plates of a parallel plate capacitor of capacity of 50 µF are charged by a battery to a potential of 100 volt. The battery remains connected the plates are separated from each other so that the distance between them is doubled. Then, the energy spent by battery in doing so, will be
42.5×10-2J
25×10-2J
D.
25×10-2J
When the separation between the plates is doubled the capacitance becomes one half.
It means C'=50/2=25μF
Now energy spent by the battery =qV=(C'V)V
=C'V2=25×10-6×(100)2
=25×10-2J
A charged particle is suspended equillibrium in a uniform vertical electric field of intensity 2000V/m. If mass of the particle is 9.6×10-16 kg, the charge on it and excess number of electrons on the particle respectively are (g=10m/s2)
4.8×10-19 C,3
5.8×10-19 C,4
3.8×10-19C, 2
2.8×10-19 C, 1
A.
4.8×10-19 C,3
Charge of the particle is given by
A transmitting station transmits radiowaves of wavelength 360 m. Calculate the inductance of coil required with a condenser of capacity 1.20 μF in the resonant circuit to receive them
3.07×10-8 H
2.07×10-8 H
4.07×10-8 H
6.07×10-8 H
A.
3.07×10-8 H
The frequency of radiowaves (speed c=3×108 m/s) transmitted
The peak value of alternating current is ampere. The mean square value of current will be
5 A
2.5 A
A
None of the above
A.
5 A
Peak value of alternating current is known as "equivalent" or DC "equivalent" value of AC voltage or current. For sine wave, the RMS value is approximately 0.707 of it's peak value. The crest factor of an AC waveform is the ratio of its peak (crest) to its RMS value.
If the current 30A flowing in the primary coil is made zero in 0.1 sec, the emf induced in the secondary coil is 1.5 volt. The mutual inductance between the coils is
0.05 H
1.05 H
0.1 H
0.2 H
A.
0.05 H
When the emf is induced in a single isolate coil due to change of flux through the coil by means of varying the current through the same coil, this phenomenon is called self-induction & mutual induction as the current flowing in one coil that induces in an adjacent coil.
here :- e= 1.5 volt ,Δi =0-3=-3A, Δt=0.1 sec
Now the emf induced in secondary coil
Two equal resistances R are joined with voltage source V in (i)series (ii)parallel, the ratio of electrical power consumed in two cases will be
1:4
4:1
2:1
1:2
A.
1:4
(a)The same current flows through each part of a series circuit. The total resistance a series circuit is equal to the sum of individual resistances. Voltage applied to a series circuit is equal to the sum of the individual voltage drop. Two resistors are said to be if only one of the endpoints is closed.
In series:- total resistance= R+R=2R
(b)Two or more resistors are said to be in parallel if one end of all the resistors is joined together and similarly the other ends joined together.
In parallel:- potential difference across each resistance will be V
So, power consumed in each resistance is
A proton enters a magnetic field of intensity 1.5 Wb/m2 with a velocity 2×107 m/s in a direction at angle 30o with the field. The force on the proton will be (charge on proton is 1.6×10-19 C)
2.4×10-12 N
4.8×10-12 N
1.2×10-12 N
7.2×10-12 N
A.
2.4×10-12 N
Here :- q=1.6×10-19 C, B=1.5Wb/m2
v=2×107 m/s, θ=30o or Sin300 =
Force on proton is given by
F= q.v.BSinθ
=1.6 × 10-19 ×2 × 107 × 1.5 ×
F= 2.4×10-12 N
20 ampere current is flowing in a long straight wire. The intensity of magnetic field at a distance 10 cm from the wire will be
2×10-5 Wb/m2
9×10-5 Wb/m2
4×10-5 Wb/m2
6×10-5 Wb/m2
C.
4×10-5 Wb/m2
Here :- i=20A, r=10cm=0.1m
Intensiy of magnetic field produced due to straight current carrying wire is
The resistance of a galvanometer is 50 Ω. When 0.01A current flows in it, full scale deflection is obtained in galvanometer, the resistance of shunt connected to convert galvanometer into an ammeter of range 5A, will be
0.1 Ω
0.2 Ω
0.3 Ω
0.4 Ω
A.
0.1 Ω
Galvanometer is a very sensitive instrument therefore it cannot measure heavy currents. In order to convert a Galvanometer into an Ammeter, a very low shunt resistance is connected in parallel to Galvanometer. Value of shunt is so adjusted that most of the current passes through the shunt.
Here :- Ig =0.01A, G=50Ω
If resistance of galvanometer is Rg and it gives full-scale deflection when current Ig is passed through it. Then,
V=IgRg
V=0.01A × 50Ω
Let a shunt of resistance (Rs) is conected in parallel to galvanometer. If total current through the circuit is I
I =5A= Is +Ig
V=IsRs =(I-Ig)Rs
(I-Ig)Rs = IgRg
(5-0.01)Rs= 0.01×50
Rs =0.01Ω
When a resistance of α Ω is connected at the ends of a battery, its potential difference decreases from 40 V to 30V. The internal resistance of the battery is
3Ω
6Ω
1.5Ω
4Ω
A.
3Ω
Given:- E=40 V, V=30V , r=α Ω
Internal resistance
The equivalent resistance and potential difference between A and B for the circuit
respectively are
4Ω, 8V
8Ω, 4V
2Ω, 2V
16Ω, 2V
A.
4Ω, 8V
The resistance in parallel given by
A wire of resistance R is divided in equal parts, then these parts are joined in parallel, the equivalent resistance of the combination will be
nR
D.
A wire is divided into equal parts.
Let us consider the number of parts= n
Resistance of one part
Now equivalent resistance of n parts each of resistance in parallel is given by
The capacity of an air condenser is 2.0 µF. If a medium is placed between its plates, the capacity becomes 12 μF. The
dielectric constant of the medium will be
5
4
3
6
D.
6
When dielectric is placed in an electric field, electric charges do not flow through the material as they do in an electrical conductor but only slightly shift from their average equilibrium positions causing dielectric polarization
Dielectric constant of medium
Taking the earth to be a spherical conductor of diameter 12.8×103 km. Its capacity will be
711 μF
611 μF
811 μF
511 μF
A.
711 μF
Here :- Diameter of earth =12.8×103 km
A conducting sphere of radius R = 20 cm is given by a charge Q = 16μC. What is E at centre?
3.6×106 N/C
1.8×106 N/C
Zero
0.9×106 N/C
C.
Zero
The value of E at the centre of the sphere is zero. Since the centre of the sphere is placed within the Gaussian surface. The use of Gauss law to examine the electric field of a charged sphere shows the electric field enviornment outside the sphere is identical to that of point charge.
Therefore the potential is the same as that of a point charge
hence E=0
If an insulated non-conducting sphere of radius R has charge density ρ. The electric field at a distance r from the centre of sphere (r > R) will be
C.
Gauss law allows us to determine the electric field at a point when there is sufficient symmetry.
Charge which is enclosed in the surface is given by
Two spherical conductors B and C having equal radii and carrying equal charges in them repel each other with a force F when kept apart at some distance. A third spherical conductor having same radius as that of B but uncharged, is brought in contact with B, then brought in contact with C and finally removed away from both. The new force of repulsion between B and C is
D.
Let the spherical conductors Band C have same charge as q. The electric force between them is
r - being the distance between them.
When third uncharged conductor A is brought in contact with B, then charge on each conductor.
When this conductor A is now brought in contact with C, then charge on each conductor
Hence, electric force acting between B and C is
A charge q is located at the centre of a cube. The electric flux through any face is
A.
According to Gauss's theorem, electric flux through the cube (closed surface)
q = total charged enclosed by surface S
The law implies that the total electric flux through a closed surface is zero if no change is enclosed by the surface.
Since, cube has six surfaces and all the faces are symmetrical, therefore electric flux through any face
Three equal charges, each having a magnitude of 2.0 x 10-6 C, are placed at the three corners of a right angled triangle of sides 3 cm, 4 cm and 5 cm. The force (in magnitude) on the charge at the right angled corner is
50 N
26 N
29 N
45.9 N
D.
45.9 N
Consider the diagram
q
qA = qB = qC = 2 × 10-6 C
The sides of right angle triangle
AB = 3 cm
BC = 4 cm
AC = 5 cm
The force on A due to B is
FA =
= 9 × 109 ×
FA = 40 N ( along BA)
The force on B due to C is
FC =
=
FC = 22.5 N ( along BC )
The resultant force on the charge at B,
F =
=
F = 45.9 N
Two charges of + 10 μC and +20 μC are separated by a distance 2 cm. The net potential (electric) due to the pair at the middle point of the line joining the two changes, is
27 MV
18 MV
20 MV
23 MV
A.
27 MV
Using the equation
V =
Q - charge
ε0 - permittivity of free surface
The potential due to + 10 µC is
V1 =
V1 = 9MV
The potential due to + 20 µC is
V2 =
V2 = 18 MV
The net potential at the given point is
9 MV + 18 MV = 27 MV
The charges on two spheres are + 7 µC and -5C respectively. They experience a force F. If each of them is given an additional charge of -2 µC, then the new force of attraction will
F
2F
A.
F
Given:-
q1 = + 7gµC = +7 X 10-6 C
q2 = -5 μC = -5 × 10-6 C
New force F'= ?
We know that,
F =
F =
= N
F' =
= N
⇒ F' = F
A square loop ABCD, carrying a current I2, is placed near and coplanar with a long straight conductor XY carrying a current I1, as shown in figure. The net force on the loop will be
D.
Force on arm AB due to current in conductor XY is
F1 =
F1 =
This force acting towards the wire in the plane of loop.
Force on arm CD due to current in conductor XY is
F2 =
F2 =
F2 is away from the wire the plane loop.
Ner force on the loop is
F = F1 F2
=
=
F =
Three identical charges are placed at the vertices of an equilateral triangle. The force experienced by each charge, ( if k = 1/4 ) is
C.
The magnitude of the force is given by
F =
Force on charge q at A:
( i ) Due to B
f =
Along BA
f =
(ii) Due to C
f =
Along CA
f =
∴ Resultant force = F
F2 = f2 + f2 + 2f × f × cos 60o
F2 = 2f2 +
F2 = 3 f2
⇒ F = f
⇒ Resultant force =
The angle of dip at a certain place where the horizontal and vertical components of the earth's magnetic field are equal is
30o
75o
60o
45o
D.
45o
Given:- Horizantol and vertical components of the earth's magnetic field are equal.
BH = Bv
tanδ =
=
tanδ = 1
∴ δ = 45o ( tan45o = 1 )
Assertion: Lines of force are perpendicular to conductor surface.
Reason: Generally electric field is perpendicular to equipotential surface.
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.
The electric field lines are perpendicular to its surface, ending or beginning on charges on the surface.
Above figure shows electric field E is applied to a conductor, free charges inside the conductor move the field is perpendicular to the surface.
(a) The electric field is a vector quantity, with both parallel and perpendicular components. The parallel components (Ell) exerts a force (Fll) on the free charge q, which moves the charge until Fll = 0
(b) The resulting field is perpendicular to the surface. The free charge has been brought to the conductor's surface, leaving electrostatics forces in equilibrium.
Equipotential lines are always perpendicular to the electric field. In three dimensions, the lines form euipotential surfaces
The force per unit length between two straight conductors separated by a distance of 0.02 m is 2 x 10-3 N m-1. The current in one conductor is 10 A and that in the other conductor is
20 A
5 A
2 A
10 A
A.
20 A
Force per unit length between two parallel current carrying wire is given by
F =
Hence = 10-7 N m A-2
r = 0.02 m
D = 2 × 10-3 N m-1
I1 = 10 A
I2 = ?
so, 2 × 10-3 = 2 × 10-7 ×
0.02 × 104 = 10 I1
∴ I1 = 20 A
The angle of dip at the poles and the equator respectively are
30o, 60o
0o, 90o
45o, 90o
90o, 0o
D.
90o, 0o
Since the angle of dip at a place is defined as the angle δ, which is the direction of the total intensity of earth's magnetic field B makes with Z horizontal line in magnetic meridian,
At poles
B = Bv
and B = B sinδ
∴ sinδ = 1
⇒ δ = 90o
At equator
B = BH
and BH = B cosδ
∴ cosδ = 1
⇒ δ = 0o
Angle of dip at the two poles is 90o. On the magnetic equator, the angle of dip is 0o as the needle would rest horizontally at the magnetic equator. At other places the value of dip angle lies between 0o to 90o.
Assertion: The conductivity of an electrolyte is very low as compared to a metal at room temperature.
Reason: The number density of free ions in electrolyte is much smaller as compared to the number density of free electrons in metals. Further, ions drift much more slowly, being heavier.
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.
A current can be carried through an electrolyte is via the movement of ions: which will be slow, and the concentration of ions is low compared to the electrons that are present in bulk metal.
Sponsor Area
Sponsor Area
, with a velocity 
perpendicular to both 
, and comes out without any change in magnitude or direction of 
.Then
