Question

A light copper ring is freely suspended by a light string. A bar magnet is held horizontally with its length along the axis of the ring. The magnet is moved towards the ring with its N pole facing the loop. What will happen to the ring and its position? Explain.

Hint:

In the case of electromagnetic induction, when a magnet is moved near a conducting loop, a current is induced in the loop that creates a magnetic field opposing the magnet's motion, according to Lenz's Law.

Answer 1

This phenomenon can be explained using the principle of electromagnetic induction. When the bar magnet is moved towards the freely suspended copper ring, the magnetic field associated with the magnet changes the magnetic flux through the ring. According to Faraday’s law of induction, a change in magnetic flux through a loop of wire induces an electromotive force (EMF), which drives a current in the ring. As the magnet’s north pole approaches the ring, the flux through the ring increases because the magnetic field strength near the N pole of the magnet is stronger. The direction of the induced current is such that the ring tries to oppose the change in flux as per Lenz's Law. Lenz's Law states that the induced current will flow in a direction that creates a magnetic field opposing the change in the magnetic flux. Therefore, the copper ring will develop a magnetic field that repels the magnet. This repulsion will cause the ring to move away from the magnet, as the ring experiences a force due to the interaction between its induced magnetic field and the magnet’s field. The magnitude of the current in the ring depends on several factors such as the speed at which the magnet is moved, the strength of the magnet, and the material of the ring. The induced current will create a magnetic dipole moment, and the interaction between this dipole moment and the external magnetic field of the magnet results in a force that moves the ring.