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A square loop of side $a$ , resistance $R$ , mass $m$ is sliding as shown on a smooth horizontal table with speed $ν_{0}$ . It enters a uniform magnetic field of magnitude $B_{0}$ perpendicular to the table which exist everywhere towards right of dotted line as shown in figure. It is seen that the loop comes to rest after entering a distance $l$ inside the magnetic field. The value of $v_{0}$ can be

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$\frac{2 B_{0}^{2} a^{3}}{m R}$

$\frac{B_{0}^{2} a^{3}}{3 m R}$

$\frac{B_{0}^{2} a^{3}}{2 m R}$

$\frac{B_{0}^{2} a^{3}}{6 m R}$

answer is B, C, D.

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Detailed Solution

$i = \frac{B a v}{R}$

$\frac{m v d v}{d x} = F = - \frac{B^{2} a^{2} v}{R}$

$m v_{0} = \frac{B^{2} a^{2} l}{R}$

$l = \frac{R v_{0} m}{B^{2} a^{2}} \leq a$

$v_{0} \leq \frac{B^{2} a^{2}}{m R}$

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