JEE MAIN - Physics (2025 - 4th April Morning Shift - No. 24)
Conductor wire ABCDE with each arm 10 cm in length is placed in magnetic field of $\frac{1}{\sqrt{2}}$ Tesla, perpendicular to its plane. When conductor is pulled towards right with constant velocity of $10 \mathrm{~cm} / \mathrm{s}$, induced emf between points A and E is ________ mV .
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Answer
10
Explanation
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As field is uniform we can replace the bent wire with straight wire from A to B.
So EMF :
$$\begin{aligned} & \varepsilon=\operatorname{Bv} \ell_{\mathrm{AB}} \\ & =\frac{1}{\sqrt{2}} \times \frac{10 \mathrm{~cm}}{5} \times 2\left(10 \sin 45^{\circ}\right) \mathrm{cm} \\ & \varepsilon=10 \mathrm{mV} \end{aligned}$$
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