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A wire of mass m and length l can slide freely on a pair of smooth, vertical rails. A magnetic field B exists in region in the direction perpendicular to the plane of the rails. The rails are connected at the top end by a capacitor of capacitance C. Find the acceleration of the wire neglecting any electric resistance.

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\frac{mg}{m - {CB}^{2} l^{2}}

\frac{2mg}{m - {CB}^{2} l^{2}}

\frac{5mg}{m + {CB}^{2} l^{2}}

\frac{mg}{m + {CB}^{2} l^{2}}

answer is A.

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Detailed Solution

Force acting on the conductor

F = mg - B i l \to (1)

The charge Q on the capacitor is given by

$Q = CV = C (B l v) = (B I C) v$

$i = \frac{dQ}{dt} = B l C (\frac{dv}{dt}) \to (2)$

From (1) $m \frac{dv}{dt} = mg - B^{2} l^{2} C \frac{dv}{dt}$

$a = \frac{dv}{dt} = \frac{mg}{m + B^{2} l^{2} C}$

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