+1 vote
in Class 12 by kratos

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

+4 votes
by kratos
 
Best answer

(i) Current in the loop PQRS,

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

F = BIl

= B(Blv/r)l

F = (B2l2v)/r

(iii) Power required to move the arm PQ

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