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Question
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.
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Solution
Let the length RQ = x and RS = l.
Let the magnitude of the uniform magnetic field be B.
(a) The magnetic flux Φ enclosed by the loop PQRS is given by,Φ=BIx
As, x is changing with time, the rate of change of flux Φ will induce an emf given by:
`E = - (dphi)/dt = -(d(BIx))/dt`
`-BI dx/dt = Blv [as,dx/dt = -v]`
Current in the loop is given by,
`a = E/r = (BIv)/r`
(b) The magnetic force on the PQ is,
`F = BII = B((BIv)/r) xx I (B^2I^2v)/r`
(c) Power required to move the arm PQ is, `p = Fxx V = (B^2I^2V)/r xx v =(B^2I^2V^2)/r`
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