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A square-shaped wire loop of mass $\mathrm{m}$, resistance $R$ and side 'a' moving with speed $v_{0}$, parallel to the $\mathrm{x}$-axis, enters a region of uniform magnetic field $\mathrm{B}$, which is perpendicular to the plane of the loop. The speed of the loop changes with distance $x(x < a)$ in the filed, as
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The correct answer is:
$v_{0}-\frac{\mathrm{B}^{2} \mathrm{a}^{2}}{\mathrm{Rm}} \mathrm{x}$
$\mathrm{F}=\mathrm{i} \ell \mathrm{B}=\frac{\mathrm{vBa} \cdot \mathrm{Ba}}{\mathrm{R}}$
$\mathrm{ma}=\frac{\mathrm{vB}^{2} \mathrm{a}^{2}}{\mathrm{R}}$
$\frac{-\mathrm{mdv}}{\mathrm{dt}}=\frac{\mathrm{vB}^{2} \mathrm{a}^{2}}{\mathrm{R}}$
$\Rightarrow-\mathrm{m} \frac{\mathrm{vdv}}{\mathrm{dx}}=\mathrm{v} \frac{\mathrm{B}^{2} \mathrm{a}^{2}}{\mathrm{R}}$
$-\mathrm{m} \int_{\mathrm{v}_{0}} \mathrm{dv}=\frac{\mathrm{B}^{2} \mathrm{a}^{2}}{\mathrm{R}} \int_{0}^{\mathrm{x}} \mathrm{d} \mathrm{x}$
$-\mathrm{m}\left(\mathrm{v}-\mathrm{v}_{0}\right)=\frac{\mathrm{B}^{2} \mathrm{a}^{2} \mathrm{x}}{\mathrm{R}}$
$\Rightarrow \mathrm{v}=\mathrm{v}_{0}-\frac{\mathrm{B}^{2} \mathrm{a}^{2}}{\mathrm{mR}} \mathrm{x}$
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