Question 1

A metallic loop is placed in a nonuniform magnetic field. Will an emf be induced in the loop?

Accepted Answer

If the flux through the loop does not vary, there will be no induced emf. Because the magnetic field is non-uniform and does not change with time, there will be no change in the magnetic flux through the loop. Hence, no emf will be induced in the loop.

Question 2

An inductor is connected to a battery through a switch. Explain why the emf induced in the inductor is much larger when the switch is opened as compared to the emf induced when the switch is closed.

Accepted Answer

When we close the switch, the current takes some time to grow in the circuit. Due to this growth of current, the flux increases; hence, an emf is induced. On the other hand, when we open the switch, there is no path for the current to flow; hence, it suddenly drops to zero. This rate of decrease of current is much greater than the rate of growth of current when the switch is closed. So, when the switch is opened, the induced emf is more.

Question 3

The coil of a moving-coil galvanometer keeps on oscillating for a long time if it is deflected and released. If the ends of the coil are connected together, the oscillation stops at once. Explain.

Accepted Answer

When the ends of the coil are not connected, the coil acts as an inductor in which oscillations persist until the current decays slowly. When these ends are connected, the coil forms a close loop; hence, there is inductance across the ends and the coil does not behave like an inductor. Therefore, all oscillations stop at once.

Question 4

A short magnet is moved along the axis of a conducting loop. Show that the loop repels the magnet if the magnet is approaching the loop and attracts the magnet if it is going away from the loop.

Accepted Answer

Consider the above situation in which a magnet is moved towards a conducting circular loop. The north pole of the magnet faces the loop. As the magnet comes closer to the loop, the magnetic field increases; hence, flux through the loop increases. According to Lenz's law, the direction of induced current is such that it opposes the magnetic field that has induced it. Thus, the induced current produces a magnetic field in the direction opposite to the original field; hence, the loop repels the magnet.On the other hand, when the magnet is going away from the loop, the magnetic field decreases. Hence, flux through the loop decreases. According to Lenz's law, the induced current produces a magnetic field in the opposite direction of the original field; hence, the loop attracts the magnet.

Question 5

Two circular loops are placed coaxially but separated by a distance. A battery is suddenly connected to one of the loops establishing a current in it. Will there be a current induced in the other loop? If yes, when does the current start and when does it end? Do the loops attract each other or do they repel?

Accepted Answer

Consider loops A and B placed coaxially as shown above. Let the direction of the current in loop A be clockwise when the 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 left, as seen from the side of B. Due to a sudden flux through loop B, a current will be induced in it. It will only be induced for a moment when the current suddenly jumps from zero to a constant value. After it has attained a constant value, there will be no induced current. Now, according to Lenz's law, the direction of the induced current in loop B will be such that it will oppose the magnetic field due to loop A. Hence, a current will be induced in anti-clockwise direction in loop B. The induced current will flow in loop B as soon as the current grows in loop A and will end when the current through loop A becomes zero. Because the directions of the currents in the loops are opposite, they will repel each other.

Question 6

The battery discussed in the previous question is suddenly disconnected. Is a current induced in the other loop? If yes, when does it start and when does it end? Do the loops attract each other or repel?

Accepted Answer

When the battery is suddenly disconnected, a current is induced in loop B due to a sudden change in the flux through it. It is only induced for a moment when the current suddenly falls to zero. There is no induced current after it has fallen to zero. According to Lenz's law, the induced current is such that it increases the decreasing magnetic field. So, if the current in loop A is in clockwise direction, the induced current in loop B will also be in clockwise direction. Hence, the two loops will attract each other.

Question 7

If the magnetic field outside a copper box is suddenly changed, what happens to the magnetic field inside the box? Such low-resistivity metals are used to form enclosures which shield objects inside them against varying magnetic fields.

Accepted Answer

The varying magnetic field induces eddy currents on the walls of the copper box. There is a magnetic field due to the induced eddy currents, that is in opposite direction. As copper has good conductivity, thus the magnetic field due to the eddy currents will be strong.  The magnetic field induced due to eddy currents in the copper walls cancel the original magnetic field. Thus, magnetic field does not penetrate the enclosure made of copper. The magnetic field inside the box remains zero. This is how a copper box protects the inside material from varying magnetic fields.

Question 8

Metallic (nonferromagnetic) and nonmetallic particles in a solid waste may be separated as follows. The waste is allowed to slide down an incline over permanent magnets. The metallic particles slow down as compared to the nonmetallic ones and hence are separated. Discuss the role of eddy currents in the process.

Accepted Answer

When solid waste is allowed to slide over a permanent magnet, an emf is induced in metallic particles. This is because magnetic flux linked with the particles changes in this case. According to Lenz's law this induced emf opposes its cause i.e. downward motion along the inclined plane of the permanent magnet. On the other hand, non-metallic or insulating particles are free from such effects. As a result, the metallic particles slow down and hence get separated from the waste (or non-metallic particles).

Question 9

A pivoted aluminium bar falls much more slowly through a small region containing a magnetic field than a similar bar of an insulating material. Explain.

Accepted Answer

An aluminium bar falls slowly through a small region containing a magnetic field because of the induced eddy currents (or induced emf)  in it. According to Lenz's law this induced eddy current oppose its cause (its motion). Hence, it slows down while falling through a region containing a magnetic field. On the other hand, non-metallic or insulating materials are free from such effects.

Question 10

A metallic bob A oscillates through the space between the poles of an electromagnet (See the figure). The oscillations are more quickly damped when the circuit is on, as compared to the case when the circuit is off. Explain.

Accepted Answer

When the circuit is on, eddy currents are produced on the surface of the metallic bob. Due to these eddy currents, thermal energy is generated in it. This thermal energy comes at the cost of the kinetic energy of the bob; hence, oscillations are more quickly damped when the circuit is on compared to when the circuit is off.

Question 11

Two circular loops are placed with their centres separated by a fixed distance. How would you orient the loops to have (a) the largest mutual inductance (b) the smallest mutual inductance?

Accepted Answer

(a) For the largest mutual inductance, the two loops should be placed coaxially. In this case, flux through a loop due to another loop is the largest; hence, mutual inductance is the largest.(b) For the smallest mutual inductance, the two loops should be placed such that their axes are perpendicular to each other. In this case, flux through a loop due to another loop is the smallest (zero); hence, mutual inductance is the smallest.

Question 12

Consider the self-inductance per unit length of a solenoid at its centre and that near its ends. Which of the two is greater?

Accepted Answer

Self-inductance per unit length of solenoid is given as,lL​=μ0​n2A From the above equation we can see that, self inductance per unit length will depend on the permeability of free space (μ0​) , number of turns per unit length (n)  and area of the cross-section of the solenoid (A). All the above factors are constant at the centre and near any end of the solenoid therefore self inductance at both the points will be same.

Question 13

Consider the energy density in a solenoid at its centre and that near its ends. Which of the two is greater?

Accepted Answer

In a solenoid energy is stored in the form of magnetic field. If a constant current is flowing from a solenoid then magnetic field inside the solenoid is uniform. Therefore, energy per unit volume (or energy density) in the magnetic field inside the solenoid is constant.The energy per unit volume in the magnetic field is given as,u=2μ0​B​ , where B is uniform magnetic field inside the solenoid.Therefore, energy density all points inside a solenoid is same.

Question 14

A rod of length l rotates with a small but uniform angular velocity ω about its perpendicular bisector. A uniform magnetic field B exists parallel to the axis of rotation. The potential difference between the centre of the rod and an end is:

Accepted Answer

(b) 81​ωBl2 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 byde=Bvl=Bωxdx The emf induced across the centre and end of the rod is given by​∫de=∫0l/2​Bωxdx⇒E=Bω[2x2​]0l/2​⇒E=81​Bωl2​

Question 15

A rod of length l rotates with a uniform angular velocity ω about its perpendicular bisector. A uniform magnetic field B exists parallel to the axis of rotation. The potential difference between the two ends of the rod is:

Accepted Answer

(a) 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 de=Bωxdx The emf induced between the centre of the rod and one of its end is given by∫de=∫0l​Bωxdx ⇒e=Bω[2x2​]0l/2​ ⇒e=81​Bωl2 The emf at both ends is the same. So, the potential difference between the two ends is zero.

Question 16

Consider the situation shown in figure. If the switch is closed and after some time it is opened again, the closed loop will show.

Accepted Answer

(d) a clockwise current-pulse and then an anticlockwise current-pulseDue 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.

Question 17

Solve the previous question if the closed loop is completely enclosed in the circuit containing the switch.

Accepted Answer

(C) an anticlockwise current-pulse and then a clockwise current-pulse 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.

Question 18

A bar magnet is released from rest along the axis of a very long, vertical copper tube. After some time the magnet:

Accepted Answer

(b) will move with almost contant speedAs 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.

Question 19

Figure shows a horizontal solenoid connected to a battery and a switch. A copper ring is placed on a frictionless track, the axis of the ring being along the axis of the solenoid. As the switch is closed, the ring will:

Accepted Answer

(c) move away from the solenoidFor the circuit,E=−Ldtdi​ 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.

Question 20

Consider the following statements:-(A) An emf can be induced by moving a conductor in a magnetic field.(B) An emf can be induced by changing the magnetic field.

Accepted Answer

(a) Both A  and B  are true.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=−dtdϕ​

Chapter 16 : Electromagnetic Induction