There are two square loops A and B. When A mo | vedclass

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(A) According to Faraday's law of electromagnetic induction,an induced current flows in a loop when the magnetic flux linked with it changes. In this

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There is a constant current in clockwise direction in $A$.

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(A) According to Faraday's law of electromagnetic induction,an induced current flows in a loop when the magnetic flux linked with it changes. In this case,loop $B$ is at rest,and loop $A$ is moving towards it. For an induced current to flow in $B$,there must be a magnetic field produced by $A$ that changes as $A$ moves. If $A$ carried a constant current,the magnetic field produced by it would change as it moves closer to $B$,thereby changing the magnetic flux through $B$ and inducing a current. However,the problem states that the current in $B$ stops when $A$ stops moving. If $A$ had a constant current,the magnetic field would still exist even when $A$ is stationary,but the flux would be constant,resulting in zero induced current. This matches the observation. Looking at the figure,the induced current in $B$ is counter-clockwise. By Lenz's law,this induced current creates a magnetic field to oppose the change in flux. If $A$ carries a constant clockwise current,it produces a magnetic field pointing into the page. As $A$ moves towards $B$,the flux into $B$ increases. To oppose this,$B$ induces a counter-clockwise current. Therefore,$A$ must carry a constant current.

There are two square loops $A$ and $B$. When $A$ moves towards $B$,a current starts flowing in $B$ as shown in the figure,and the current in $B$ stops when $A$ stops moving. From this,we can infer that (Assume loop $B$ is at rest):

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