Question

A planer loop of wire rotates in a uniform magnetic field. Initially, at $t=0$, the plane of the loop is perpendicular to the magnetic field. If it rotates with a period of $10\text{s}$ about an axis in its plane, then the magnitude of induced emf will be maximum and minimum, respectively at:

• A. $\text{2.5 s and 7.5 s}$
• B. $\text{2.5 s and 5.0 s}$
• C. $\text{5.0 s and 7.5 s}$
• D. $\text{5.0 s and 10.0 s}$

Answer 1

The correct option is B $\text{2.5 s and 5.0 s}$

Given that, the time period, $T=10\text{s}$

$\therefore \text{Angular velocity},\omega =\frac{2\pi }{10}=\frac{\pi }{5}$
$\text{Magnetic flux},\varphi \left(t\right)=BA\cos \omega t$
Emf induced,
$E=\frac{-d\varphi }{dt}=BA\omega \sin \omega t$
Induced emf, $|E|$ is maximum when $\sin \omega t=1\to \omega t=\frac{\pi }{2}$
$\Rightarrow t=\frac{\left(\frac{\pi }{2}\right)}{\left(\frac{\pi }{5}\right)}=2.5\text{s}$
$\therefore$ Induced emf is maximum at $t=2.5\text{s}$

For induced emf to be minimum i.e. zero
$\omega t=\pi \Rightarrow t=\frac{\pi }{\left(\frac{\pi }{5}\right)}=5\text{s}$
$\therefore$ Induced emf is minimum at $t=5\text{s}$

Hence, option $\left(B\right)$ is correct.