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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 $v_{0}$ . It enters a uniform magnetic field of magnitude $B_{0}$ perpendicular to the table. 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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$\frac{2 {B_{0}}^{2} a^{3}}{mR}$

$\frac{{B_{0}}^{2} a^{3}}{2 mR}$

$\frac{{B_{0}}^{2} a^{3}}{3 mR}$

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

answer is B, C, D.

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Detailed Solution

$\begin{array}{l} i = \frac{B_{0} av}{R} \\ \frac{mvdv}{dx} = F = - \frac{{B_{0}}^{2} d^{2} v}{R} \\ {mv}_{0} = \frac{{B_{0}}^{2} a^{2} 1}{R} \\ 1 = \frac{{Rv}_{0} m}{{B_{0}}^{2} a^{2}} \leq a \\ v_{0} \leq \frac{{B_{0}}^{2} a^{3}}{mR} \end{array}$

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