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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?

  1. The amplitude of oscillation will increase.
  2.  The ring will be accelerated towards the magnet.
  3. Flux linked with ring will decrease as it approaches electromagnet.
  4. 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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