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A current generator $l_{g}$ sends current, $l = l_{0} (1 - \frac{3 t}{T}), 0 \leq t \leq \frac{T}{3}$ through the circuit. The wire $P Q$ has length $l$ and mass $m$ and can slide on the smooth rails connected to $l_{g}$ . The entire system lies in a vertical uniform magnetic field $B_{0}$ . If initial velocity of wire is zero, then the maximum velocity of wire, would be $\frac{l_{0} l B_{0} T}{x m}$ . Find x

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answer is 6.

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Detailed Solution

Based on force acting on current-carrying element.

$m \frac{d v}{d t} = l_{0} (1 - \frac{3 t}{T}) l B_{0} v = \frac{l_{0} l B_{0}}{m} (t - \frac{3 t^{2}}{2 T}) \frac{d v}{d t} = 0 at t = \frac{T}{3} v_{\max} = \frac{l_{0} l B_{0} T}{6 m}$

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