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Magnetic flux and induced emf

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Question

Figure shows a conducting rod of length l = l0 cm, resistance R and mass m = 100 mg moving vertically downward due to gravity. Other parts are kept fixed. Magnetic field is B = 1 T. MN and PQ are vertical, smooth, conducting rails. The capacitance of the capacitor is C = 10 mF. The rod is released from rest. Find the maximum current (in mA) in the circuit.

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Solution

By Newton's law,

mg=ilB=mdvdt     -----(i)

Using KVL Blv=iR+qC    ----(ii)

Differentiating equation (ii) w.r.t. time, we get

Bldvdt=Rdidt+iC    ----(iii)

Eliminating dvdt from equations (i) and (iii), we get

mg-ilB=mBlRdidt+iC

mgBl-iB2l2=mRdidt+miC   ---(iv)

I will be maximum when didt=0. use this in equation (iv)

mgBlC=iB2l2C+m

imax=mgBlCm+B2l2C


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