A rigid wire loop of square shape having side of length L and resistance R is moving along the x -axis with a constant velocity $v_{0}$ in the plane of the paper. At $t = 0$ , the right edge of the loop enters a region of length 3L where there is a uniform magnetic field $B_{0}$ into the plane of the paper; as shown in the figure. For sufficiently large $v_{0}$ , the loop eventually crosses the region. Let x be the location of the right edge of the loop. Let $v (x), I (x)$ and $F (x)$ represent the velocity of the loop, current in the loop, and force on the loop, respectively, as a function of x. Counter-clockwise current is taken as positive.

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Detailed Solution

While entering i.e. $x < L$

$f = i l B = \frac{B^{2} l^{2} v}{R}$

$a_{2} = \frac{f}{m} = \frac{B^{2} l^{2} v}{m R} = k v = - \frac{v d v}{d x} [k = \frac{B^{2} l^{2}}{m}] \int_{v_{0}}^{v} d v = - k \int_{0}^{x} d x \Rightarrow v = v_{0} - k x f = \frac{B^{2} l^{2}}{R} (V_{0} - k x) = α - β x i = (v_{0} - k x) \frac{B l}{R} = i_{0} - γ x$

For $3 L > x > L f = 0 i = 0 v =$ constant.
For $4 L > x > 3 L$
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$f = i l B = \frac{B^{2} l^{2} v}{R} a = \frac{B^{2} l^{2}}{m R} v = k v = - \frac{v d v}{d x} v = {v^{'}}_{0} - k x f = α^{'} - β^{'} . x i = {i^{'}}_{0} - γ^{'} x$