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? The amplitude of oscillation will increase. The ring will be accelerated towards the magnet.Flux linked with ring will decrease as it approaches electromagnet.Oscillation will be damped.

Solution

## 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.  ⇒When copper ring approaches towards electromagnet the flux ϕ linked with the ring increases, because strength of B↑ as ring comes closer. ∴Current will be induced in ring in such a way that ring is repelled by electromagnet, by Lenz's law. ⇒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.

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