Question 28 of 153

The magnetic field in a region is given by $\vec{B}={B}_{0}\left(\dfrac{x}{a}\right)\hat{k}$. 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 $\vec{v}={v}_{0}\hat{i}$. The emf induced in the loop is :

A$\dfrac{{B}_{0}{v}_{0}^{2}d}{2a}$B$\dfrac{{B}_{0}{v}_{0}d}{2a}$C$\dfrac{{B}_{0}{v}_{0}{d}^{2}}{a}$D$\dfrac{{B}_{0}{v}_{0}{d}^{2}}{2a}$

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Answer: (C) $\dfrac{{B}_{0}{v}_{0}{d}^{2}}{a}$

${E}_{1}=\dfrac{{B}_{0}\left(x+d\right)}{a}{v}_{0}d$
${E}_{2}=\dfrac{{B}_{0}\left(x\right)}{a}{v}_{0}d$
${E}_{net}={E}_{1}-{E}_{2}$
${E}_{net}=\dfrac{{B}_{0}{v}_{0}{d}^{2}}{a}$