A rod of length l = 1m and resistance R = 10 $Ω$ is being pulled with a constant velocity of 4ms^-1 towards right. At time t = 0 the rod is 1m away from the left end. The magnetic field is $B = (1 - 3 t) T$ . Find the work done (in J) by the external agent to move the rod with constant velocity for first 5s. The rest of circuit is resistance-less. Neglect friction.

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answer is 1067.

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Detailed Solution

Area enclosed at time $t, A = (1 + 4 t)$
Flux associated to this enclosed area $= (1 + 4 t) (1 - 3 t)$
Induced emf, $E = \frac{d}{dt} (1 + t - 12 t^{2})$
$\begin{array}{l} E = 1 - 24 t \\ i = \frac{E}{R} = \frac{1 - 24 t}{10} \end{array}$
$F_{ext} = Bil$ to move rod with constant velocity
$= \frac{(1 - 3 t) (1 - 24 t)}{10} \times 1 = \frac{1}{10} [1 + 72 t^{2} - 27 t]$
Power, $\frac{dW}{dt} = Fv$
$\begin{array}{l} dW = \frac{4}{10} \int_{0}^{5} (1 + 72 t^{2} - 27 t) dt \\ W = 0 .4 {[t + \frac{72}{3} t^{3} - \frac{27}{2} t^{2}]}_{0}^{5} = 0 .4 [5 + 24 \times 125 - 13 .5 \times 25] = 1067 J \end{array}$