Question

A bar magnet is moved along the axis of a copper ring placed far away from the magnet. Looking from the side of the magnet, an anticlockwise current is found to be induced in the ring. Which of the following may be true ?

• A. The south pole faces the ring and the magnet moves towards it.
• B. The south pole faces the ring and the magnet moves away from it.
• C. The north pole faces the ring and the magnet moves towards it.
• D. The north pole faces the ring and the magnet moves away from it.

Answer 1

Answer: (B), (C)

The north pole faces the ring and the magnet moves towards it.



If the north pole of the bar magnet is moved towards the cooper ring, the magnetic flux through the ring increases, as shown in the figure.

The change in the magnetic flux will induce emf in the ring and which leads to induce the induced current in it.

According to the Lenz's law the direction of the induced current must be such that, the magnetic field $\left(B_{induced}\right)$ produced by it must oppose the magnetic field of the bar magnet, $B_{bar}$. It is possible only when induced current flows in anti-clockwise, as shown in the figure.

So, option $\left(B\right)$ is correct.



Similarly, we can explain if south pole is placed in front of the rind and move it away of the ring as shown in the figure. The magnetic flux will decrease in the rightwards direction.

According to the Lenz's law, the induced current in the ring such that the induced magnetic field produced by it must oppose the change due to the magnetic flux produced by the bar magnet. Hence, the magnetic flux will increase due to the induced magnetic field. So, the direction of induced current for such magnetic field must be anti-clockwise, as shown in the figure.

Hence, $\left(B\right)$ and $\left(C\right)$ are the correct answers.