Question

A square loop of side a is rotating about its diagonal with angular velocity $\omega$ in a perpendicular magnetic field $\overrightarrow{B}$. It has 10 turns. The emf induced is

• A. $B{a}^{2}sin\omega t$
• B. $B{a}^{2}cos\omega t$
• C. $5\sqrt{2}B{a}^2$
• D. $10B{a}^{2}sin\omega t$

Answer 1

Answer: (D)

$10B{a}^{2}sin\omega t$

$\varphi =nBAcos\theta =10B{a}^{2}cos\omega t$

$e=-\frac{d\varphi }{dt}$

$=-\frac{d}{dt}\left(10B{a}^{2}cos\omega t\right)$

$=10B{a}^{2}sin\omega t\left(\omega \right)$