# 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,Using KVL Differentiating equation (ii) w.r.t. time, we getEliminating $\frac{\mathrm{dv}}{\mathrm{dt}}$ from equations (i) and (iii), we get$\mathrm{mg}-\mathrm{ilB}=\frac{\mathrm{m}}{\mathrm{Bl}}\left[\mathrm{R}\frac{\mathrm{di}}{\mathrm{dt}}+\frac{\mathrm{i}}{\mathrm{C}}\right]$I will be maximum when $\frac{\mathrm{di}}{\mathrm{dt}}=0$. use this in equation (iv)$⇒\mathrm{mgB}\mathcal{l}\mathrm{C}=\mathrm{i}\left({\mathrm{B}}^{2}{\mathcal{l}}^{2}\mathrm{C}+\mathrm{m}\right)$$⇒{\mathrm{i}}_{\mathrm{max}}=\frac{\mathrm{mgB}\mathcal{l}\mathrm{C}}{\mathrm{m}+{\mathrm{B}}^{2}{\mathcal{l}}^{2}\mathrm{C}}$

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