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Figure shows a rectangular loop conducting PQRS in which the arm PQ is free to move. A uniform magnetic field acts in the direction perpendicular to the plane of the loop. Arm PQ is moved with a velocity v towards the arm RS. Assuming that the arms QR, RS and SP have negligible resistances and the moving arm PQ has the resistance r, obtain the expression for 

(i) The current in the loop 

(ii) The force and 

(iii) The power required to move the arm PQ.

1 Answer

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Current in the loop PQRS,

I = \(\frac{ε}{r}\) 

Since ε = \(\frac{dφ}{dt}\) BIv  So, I = \(\frac{BIv}{r}\)

(ii) The force required to keep the arm PQ in 

Constant motion

F = BI =l = B\(\Big(\frac{BIv}{r}\Big)l\) = \(\frac{B^2l^2v}{r}\)

(iii) Power required to move the arm PQ

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