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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answer is 5.

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Detailed Solution

By Newton's law,Using KVL Blv=iR+qC ----(ii)Differentiating equation (ii) w.r.t. time, we getBldvdt=Rdidt+iC ----(iii)Eliminating dvdt from equations (i) and (iii), we getmg-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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