Question

A copper ring is suspended by a long, light rod pivoted at $X$ so that it may swing as a pendulum, as shown in the diagram below. An electromagnet is mounted so that the ring passes over it as it swings. The ring is set into oscillation with switch $K$ open. What happens to the motion after switch $K$ has been closed?

• A. The amplitude of oscillation will increase.
• B. The ring will be accelerated towards the magnet.
• C. Flux linked with ring will decrease as it approaches electromagnet.
• D. Oscillation will be damped.

Answer 1

The correct option is D Oscillation will be damped.

When the switch $K$ is closed, current flows through the electromagnet thus a magnetic field $B$ is produced due to electromagnet.
$\Rightarrow$When copper ring approaches towards electromagnet the flux $\varphi$ linked with the ring increases, because strength of $B\uparrow$ as ring comes closer.
$\therefore$Current will be induced in ring in such a way that ring is repelled by electromagnet, by Lenz's law.

$\Rightarrow$Hence, motion of ring will be decelerated $\&$ oscillation will be damped( decreasing amplitude), since a part of total energy is dissipated as heat produced in ring.