Exam

JEE Main

Year

2025

Date

4 April

Shift

Morning Shift

Subject

Physics

Chapter

Electromagnetic Induction

JEE Main 2025 · 4 April · Morning Shift · Physics · Electromagnetic Induction

Conductor wire ABCDE with each arm 10 ~cm in length is placed in...

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.

Step-by-step solution

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