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Two metal bars are fixed vertically and are connected on the top by a capacitor C. A sliding conductor AB of length L slides without friction with its ends in contact with the bars. The arrangement is placed in a uniform horizontal magnetic field directed normal to the plane of the figure. The conductor is released from rest. Find the displacement (x) in meter of the conductor at time t = 2 sec.[Given $m = 100 g m, g = 10 m / s^{2}, B = 100 T e s l a, L = 1 a n d C = 10 μ F$ ]

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Detailed Solution

At any time t.
mg – BIL = ma …(i)
and Q = C B v L
$I = \frac{d Q}{d t} = B C L a ... (i i) m g = m a + B^{2} L^{2} C a a = \frac{m g}{m + B^{2} L^{2} C} = 5 S = \frac{1}{2} \times 5 \times 2^{2} = 10 m$

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