The magnetic field in a region is given by \(\vec{B} = B_o\left(\frac{x}{a}\right)\hat{k}\). 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 :
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- \(\frac{B_o v_o d^2}{2a}\)
- \(\frac{B_o v_o^2 d}{2a}\)
- \(\frac{B_o v_o d}{2a}\)
- \(\frac{B_o v_o d^2}{a}\)
The Correct Option is D
Solution and Explanation
Left side at \(x\): \(\varepsilon_1 = B(x)\,v_0\,d = \frac{B_o x}{a}v_0 d\). Right side at \(x + d\): \(\varepsilon_2 = \frac{B_o (x+d)}{a}v_0 d\).
The two emfs oppose in the loop: \(\varepsilon = \varepsilon_2 - \varepsilon_1 = \frac{B_o v_0 d}{a}\,[(x+d) - x] = \frac{B_o v_0 d^2}{a}\).
Answer: (4) \(\frac{B_o v_o d^2}{a}\)
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