The magnetic field in a region is given by $\vec{B} = B_{0} (\frac{x}{a}) \hat{k} .$ A square loop of side d is placed with its edges along the x and y axes. The loop is moved with a constant velocity $\vec{v} = v_{o} \overset{\land}{i}$ . The emf induced in the loop is:

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

$\frac{B_{0} v_{0} {^{2}}_{} d}{2 a}$

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

$\frac{B_{0} v_{0} d}{2 a}$

answer is C.

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Detailed Solution

$ε_{1} = \frac{B_{0} (x + d)}{a} v_{0} d ε_{2} = \frac{B_{0} x}{a} v_{0} d ε_{n e t} = ε_{1} - ε_{2} = \frac{B_{0} v_{0} d^{2}}{a}$

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