Question 74 of 153

Conductor wire ABCDE with each $\operatorname{arm} 10 \mathrm{~cm}$ 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

As field is uniform we can replace the bent wire with straight wire from $A$ to $B$.
So EMF :
$\varepsilon=\mathrm{Bv} \ell_{\mathrm{AB}}$
$\begin{aligned} & =\frac{1}{\sqrt{2}} \times \frac{10 \mathrm{~cm}}{5} \times 2\left(10 \sin 45^{\circ}\right) \mathrm{cm} \\ & \varepsilon=10 \mathrm{mV}\end{aligned}$