Question

A square loop of edge $a$ having $n$ turns is rotated with a uniform angular velocity $\omega$ about one of its diagonals, which is kept in a horizontal position. A uniform magnetic field $B$ exists in the vertical direction. Find the emf induced in the coil as a function of time $t$.

Assume at $t=0$, normal to the square loop is parallel to the magnetic field.

• A. $nB{a}^2\omega \sin \omega t$
• B. $nB{a}^2\omega \cos \omega t$
• C. $4nB{a}^2\omega \sin \omega t$
• D. $2nB{a}^2\omega \cos \omega t$

Answer 1

The correct option is A $nB{a}^2\omega \sin \omega t$

The area of the square frame is $A=a^2$.

The normal to the frame makes an angle $\theta =0^{\circ }$ with the magnetic field at $t=0$, this angle will become $\theta =\omega t$ at time $t$.

The flux of the magnetic field at this time is,

$\varphi =nBA\cos \theta =nB{a}^2\cos \omega t$

The induced emf is,

$\epsilon =-\frac{d\varphi }{dt}=nB{a}^2\omega \sin \omega t$