Test Index

Electromagnetic Induction

Section: Physics

© examsnet.com

Question : 3 of 20

Marks: +1, -0

A conducting square loop initially lies in the

X Z

plane with its lower edge hinged along the

X

-axis. Only in the region

y \geq 0

, there is a time dependent magnetic field pointing along the

Z

\vec{B}(t)=B_0(\cos \omega t) \hat{k}

B_0

is a constant. The magnetic field is zero everywhere else. At time

t=0

, the loop starts rotating with constant angular speed

\omega

X

axis in the clockwise direction as viewed from the

+X

axis (as shown in the figure). Ignoring self-inductance of the loop and gravity, which of the following plots correctly represents the induced e.m.f. (

V

) in the loop as a function of time:

[JEE Adv 2025 P1]

Solution:

\phi = B_0 \cos(\omega t) l^2 \sin(\omega t)

\phi = \frac{B_0 I^2}{2} \sin(2 \omega t)

\sum = \left| \frac{d\phi}{dt} \right| = B_0 \omega l^2 \cos(2 \omega t)

only for half rotation

© examsnet.com

Go to Question:

Prev Question

Next Question