Physics · Electromagnetic Induction

JEE Main 2025 — 4 April, Morning Shift — Question 65

Conductor wire $A B C D E$ 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 $\qquad$ mV .

Answer: 10

Numerical answer — enter this value.

\begin{aligned} I & =10 \sqrt{2} \mathrm{~cm} \\ & =\frac{\sqrt{2}}{10} \mathrm{~m} \end{aligned}

$E M F=B v l$

\begin{aligned} & =\frac{1}{\sqrt{2}} \times(0.1)\left(\frac{\sqrt{2}}{10}\right) & =0.01 & =10 \mathrm{mV} \end{aligned}

Answer key and solution verified before publishing.

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