A conducting frame ABCD is kept in vertical plane. A conducting rod EF of mass 1 kg and length 50 cm can slide smoothly on it remaining horizontal always. The resistance of the frame is negligible and inductance is constant having value 1H. The rod left from rest and allowed to fall under gravity with no current in the inductor. A magnetic field of constant magnitude 2T is present throughout the loop pointing inwards. Then (Assume length AB and CD of frame is sufficiency large): $(g = 10 m s^{- 2})$

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Position of the rod as a function of time is (assuming initial position of the rod to be x=0 and vertically downward as the positive X-axis) $x = 10 [1 - \cos t] metre .$

Maximum current in the circuit is 20A

Maximum velocity of the rod is 20 m/s

After some time, velocity of the rod becomes constant

answer is A, B.

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Detailed Solution

$B v l = L \frac{d i}{d t} \Rightarrow \frac{B l}{L} \times v d t = d i \Rightarrow \frac{B l}{L} \int_{0}^{x} d x = \int_{0}^{i} d i$

$\Rightarrow \frac{B l x}{L} = i$ ----------------------------(i)
Now F=ma=mg-Bil ------------------------(ii)
From equations (i) and (ii)
$\frac{d^{2} x}{d t^{2}} = g - \frac{B^{2} l^{2}}{m L} x \propto x \Rightarrow S H M$ ........