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The magnetic field in a region is given by $\vec{B} = B_{0} (1 + \frac{x}{a}) \hat{k}$ . A square loop of edge - length d is placed with its edge along x & y axis. The loop is moved with constant velocity $\vec{V} = V_{0} \hat{i}$ . The emf induced in the loop at that instant is

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\frac{V_{0} B_{0} d^{2}}{a}

\frac{V_{0} B_{0} a^{2}}{d}

None

\frac{V_{0} B_{0} d^{2}}{2a}

answer is A.

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Detailed Solution

B = B_{0} (1 + \frac{x}{a}) \hat{k}; V = v_{0} \hat{i}

Emf between $AB = B_{0} {dv}_{0}$

Emf between $BC = B_{0} [1 + \frac{d}{a}] v_{0} (d)$

$= B_{0} v_{0} d [1 + \frac{d}{a}] - B_{0} {dv}_{0}$

$E_{net} = \frac{B_{0} d^{2} v_{0}}{a}$

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