Question

A copper ring is moved towards the north pole of a bar magnet. Then
A. The ring will not be affected.
B. The ring will tend to get warm.
C. An alternating current will flow in the ring.
D. The ring will be positively charged.

Answer 1

Recall Faraday’s law and Lenz’s law. Faraday's law states that the current gets induced in the loop when there is change in the magnetic flux. Recall Joule’s law of heating to answer this question.

Complete answer:
To answer this question, let’s revise Faraday’s law and Lenz’s law. According to the first law of electromagnetic induction given by Faraday, when the magnetic flux changes in the closed loop, an induced emf is produced in the loop. According to Lenz’s law, the direction of the induced current in the loop due to induced emf is such that the magnetic field produced by the induced current opposes the magnetic flux that causes the induced current in the loop.

When we move the copper ring towards the north pole of the bar magnet, the current gets induced in the copper ring in the anticlockwise direction. The induced current in the loop produces a magnetic dipole and the orientation of the magnetic dipole is such that it opposes the movement of the copper ring. Thus, we see that there is the induced current flow through the copper ring.

We know Joule’s law of heating. According to Joule’s law of heating, whenever there is flow of current through the conductor, the heat gets generated in the conductor. The copper is the conductor and it has some resistance. According to Joule’s law,
$H = {I^2}Rt$, where, I is the current, R is the resistance and t is the time.

Therefore, from the above equation, the heat gets generated in the copper ring and thus the copper ring will get warm. So, the correct answer is option B.

We have seen that the ring tends to get warm when it is moved towards the north pole of the bar magnet. Therefore, the option A is incorrect.

The induced current in the copper ring is not alternating current. The alternating current changes the direction but the induced current in the ring will not change the direction unless it is moved backward. Therefore, the option C is incorrect.

The ring does not get extra positive charges to get positively charged. Therefore, the option D is incorrect.

Hence, option B is the correct answer.

Note: The ring tends to get warm when it moves towards both the South Pole and North Pole. The direction of the induced current will also be the same in both the situations. To answer this question, both Faraday’s law and Lenz are equally important. Therefore, the students should remember both the laws.