JEE Mains · Physics · STD 12 -Electromagnetic induction
- JEE Mains 2025
- Hard
- Numerical
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.

Pick an option to check your answer.
Answer & Solution
Correct Answer
(A) 10
Step-by-step Solution
Detailed explanation

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