HC Verma Questions and Solutions: Chapter 38: Electromagnetic Induction- 1 | HC Verma Solutions - JEE PDF Download

Let us consider a small element dx at a distance x from the centre of the rod rotating with angular velocity ω about its perpendicular bisector. The emf induced in the rod because of this small element is given by

The emf induced across the centre and end of the rod is given by

zero

Let us consider a small element dx at a distance x from the centre of the rod rotating with angular velocity ω about its perpendicular bisector. The emf induced in the small element of the rod because of its motion is given by 
The emf induced between the centre of the rod and one of its end is given by

The emf at both ends is the same. So, the potential difference between the two ends is zero.

When the switch is closed, the current will flow in downward direction in part AB of the circuit nearest to the closed loop.

Due to current in wire AB, a magnetic field will be produced in the loop. This magnetic field due to increasing current will be the cause of the induced current in the closed loop. According to Lenz's law, the induced current is such that it opposes the increase in the magnetic field that induces it. So, the induced current will be in clockwise direction opposing the increase in the magnetic field in upward direction.
Similarly, when the circuit is opened, the current will suddenly fall in the circuit, leading to decrease in the magnetic field in the loop. Again, according to Lenz's law, the induced current is such that it opposes the decrease in the magnetic field. So, the induced current will be in anti-clockwise direction, opposing the decrease in the magnetic field in upward direction.

According to Lenz's law, the induced current in the loop will be such that it opposes the increase in the magnetic field due to current flow in the circuit. Therefore, the direction of the induced current when the switch is closed is anti-clockwise.
Similarly, when the switch is open, there is a sudden fall in the current, leading to decrease in the magnetic field at the centre of the loop. According to Lenz's law, the induced current in the loop is such that it opposes the decrease in the magnetic field. Therefore, the direction of the induced current when the switch is open is clockwise.

As the magnet is moving under gravity, the flux linked with the copper tube will change because of the motion of the magnet. This will produce eddy currents in the body of the copper tube. According to Lenz's law, these induced currents oppose the fall of the magnet. So, the magnet will experience a retarding force. This force will continuously increase with increasing velocity of the magnet till it becomes equal to the force of gravity. After this, the net force on the magnet will become zero. Hence, the magnet will attain a constant speed.

For the circuit,
E = -L(di/dt)
The current will increase in the solenoid, flowing in clockwise direction in the circuit. Due to this increased current, the flux linked with the copper ring with increase with time, causing an induced current. This induced current will oppose the cause producing it. Hence, the current in the copper ring will be in anticlockwise direction. Now, because the directions of currents in the solenoid and ring are opposite, the ring will be repelled and hence will move away from the solenoid.

Statement A is true, as an emf can be induced by moving a conductor with some velocity v in a magnetic field B. It is given by
e = Bvl
Statement B is true, as an emf can be induced by changing the magnetic field that causes the change in flux ϕ through a conductor or a loop. It is given by
e = 

The induced emf across ends A and B is given by
E = Bvl
This induced emf will serve as a voltage source for the current to flow across resistor R, as shown in the figure. The direction of the current is given by Lenz's law and it is anticlockwise.
i = Bvl/R
If the wire is replaced by a semicircular wire, the induced current will remain the same, as it depends on the length of the wire and not on its shape (when B, v and R are kept constant).

Consider loops A and B placed coaxially as above. Let the direction of the current in loop A be clockwise when a battery is connected to it. According to the right-hand screw rule, the direction of the magnetic field due to this current will be towards the left. Now, the current through this loop will decrease with time due to an increase in resistance with temperature. So, the magnetic field due to this current will also decrease with time. This changing current will induce a current in loop B. Now, according to Lenz's law, the direction of the induced current in loop B will be such that it will oppose the decrease in the magnetic field due to loop A. Hence, current will be induced in a clockwise direction in loop B. Also, because the direction of the currents in the loops is the same, they will attract each other.

The magnetic field inside the solenoid is parallel to its axis. If the plane of the loop contains the axis of the solenoid, then the angle between the area vector of the circular loop and the magnetic field is zero. Thus, the flux through the circular loop is given by

Here,
B = Magnetic field due to the solenoid
A = Area of the circular loop
θ = Angle between the magnetic field and the area vector
Now, the induced emf is given by

We can see that the induced emf does not depend on the varying current through the solenoid and is zero for constant flux through the loop. Because there is no induced emf, no current is induced in the loop.

Figure (a) shows the square loop moving in its plane with a uniform velocity v.
Figure (b) shows the equivalent circuit.
The induced emf across ends AB and CD is given by
E = Bvl
On applying KVL in the equivalent circuit, we get
E - E + iR = 0
⇒ i = 0
No current will be induced in the circuit due to zero potential difference between the closed ends.

It can be observed that the induced current is in anti-clockwise direction. So, the magnetic field induced in the copper ring is towards the observer.
According to Lenz's law, the current induced in a circuit due to a change in the magnetic flux is in such direction so as to oppose the change in flux.
Two cases are possible:-
(1) The magnetic flux is increasing in the direction from the observer to the circularcoil.
(2) The magnetic flux is decreasing in the direction from the coil to the observer.
So, from the above mentioned points, the following conclusions can be made:-
1. The south pole faces the ring and the magnet moves away from it.
2. The north pole faces the ring and the magnet moves towards it.

The strong magnetic pole will attract the magnet, so a force is needed to hold the sheet there if the metal is magnetic.
If we move the metal sheet (magnetic or nonmagnetic) away from the pole, eddy currents are induced in the sheet. Because of eddy currents, thermal energy is produced in it. This energy comes at the cost of the kinetic energy of the plate; thus, the plate slows down. So, a force is needed to move the sheet away from the pole with uniform velocity.

A conducting loop is placed in a uniform magnetic field with its plane perpendicular to the field. An emf is induced in the loop if it is rotated about its diameter.

Iron rod has high permeability. When it is inserted inside a solenoid the magnetic field inside the solenoid increases. As magnetic field increases inside the solenoid thus the magnetic flux also increases. The Self-inductance (L) of the coil is directly proportional to the permeability of the material inside the solenoid. As the permeability inside the coil increases. Therefore, the self-inductance will also increase.

Because the solenoids are identical, their self-inductance will be the same.
Resistance of a wire is given by
R = p(l/A)
Here,
l = Length of the wire
A = Area of cross section of the wire
ρ = Resistivity of the wire
Because ρ and l are the same for both wires, the thick wire will have greater area of cross section and hence less resistance than the thin wire.

The time constant for a solenoid is given by

Thus, time constants of the solenoids would be different if one solenoid is connected to one battery and the other is connected to another battery.
Also, because the self-inductance of the solenoids is the same and the same current flows through them, the magnetic field energy given by 1/2Li² will be the same.
Power dissipated as heat is given by
P = i²R
i is the same for both solenoids.
∴ P_thick < P_thin
Because the resistance of the coils are different, the rate of Joule heating will be different for the coils if the same current goes through them.

At time t = 0, the current in the L-R circuit is zero. The magnetic field energy is given by U = (1/2)Li², as the current is zero the magnetic field energy will also be zero. Thus, the power delivered by the battery will also be zero. As, the LR circuit is connected to the battery at t = 0, at this time the current is on the verge to start growing in the circuit. So, there will be an induced emf in the inductor at the same time to oppose this growing current.

Due to electromagnetic induction, emf e is induced across the ends of the rod. This induced emf is given by e = Bvl
The direction of this induced emf is from A to B, that is, A is at the higher potential and B is at the lower potential. This is because the magnetic field exerts a force equal to qvB  on each free electron where q is -1.6 × 10^-16C. The force is towards AB by Fleming's left-hand rule; hence, negatively charged electrons move towards the end B and get accumulated near it. So, a negative charge appears at B and a positive charge appears at A.

The potential difference across the two ends is given by e = Bvl
It is non-zero only
(i)  if the rod is moving in the direction perpendicular to the magnetic field 
(ii) if the velocity of the rod is in the direction perpendicular to the length of the rod 
(iii) if the magnetic field is perpendicular to the length of the rod 
Thus, none of the above conditions is satisfied in the alternatives given.

The time constant of the RC circuit is given by

On taking the reciprocal of the above relation, we get

f₁ will have the dimensions of the frequency.
The time constant of the LR circuit is given by

On taking the reciprocal of the above relation, we get

f₂ will have the dimensions of the frequency.
On multiplying eq. (1) and (2), we get

Thus,  will have the dimensions of the frequency.

The charge on the capacitor at time ''t'' after connecting it with a battery is given by,

Just after t = 0, the charge on the capacitor will be

For a long after time, t → ∞
Thus, the charge on the capacitor will be

The current in the inductor at time ''t'' after closing the switch is given by

Just before the time t₀, current through the inductor is given by

It is given that the time t₀ is very long.
∴ t₀ → ∞

When the switch is opened, the current through the inductor after a long time will become zero.