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Question

A rectangular loop ABCD is being pulled out of a magnetic field B with uniform velocity v by applying an external force F. Length AB is l. Length AD is 3l and total resistance of the loop is R. The thermal power developed in the loop and the force F respectively is


A
B2v2l2R,B2l2vR
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B
9B2v2l2R,9B2l2vR
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C
8B2v2l2R,8B2l2vR
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D
3B2v2l2R,3B2l2vR
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Solution

The correct option is A B2v2l2R,B2l2vR
When the loop is passing out of the magnetic field, the flux through the loop decreases. Therefore, induced current will flow in such a direction as to oppose the change in flux. Since the flux into the plane is decreasing, current in the loop will flow in clockwise direction.


A rod of length l moving in a uniform perpendicular magnetic field with velocity v acts like a battery of emf
ϵ=Bvl
And current induced in the rod i=BvlR where R is the resistance of the loop.

Thermal power P=i2R=B2l2v2R

Segment AB is a current carrying rod moving in a magnetic field. Hence it will experience the force
FAB=i(l×B)
i.e FAB=ilB in the direction shown


Inorder to keep the loop moving with constant velocity,
F=FAB=ilB=(BvlR)lB=B2l2vR

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