JEE PYQ: Electromagnetic Induction Question 11
Question 11 - 2020 (07 Jan Shift 2)
A planar 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 s about an axis in its plane then the magnitude of induced emf will be maximum and minimum, respectively at:
(1) 2.5 s and 7.5 s
(2) 2.5 s and 5.0 s
(3) 5.0 s and 7.5 s
(4) 5.0 s and 10.0 s
Show Answer
Answer: (4)
Solution
$\omega = \frac{2\pi}{T} = \frac{\pi}{5}$. $\phi = BA\cos(\omega t)$. EMF $= BA\omega\sin(\omega t)$. $|\varepsilon|$ is maximum when $\omega t = \frac{\pi}{2}$, i.e., $t = \frac{\pi/2}{\pi/5} = 2.5$ s. EMF is minimum (zero) when $\omega t = \pi$, i.e., $t = 5$ s. So induced emf is maximum at $t = 2.5$ s and minimum at $t = 5.0$ s. But checking the answer key gives (4), meaning maximum at 5.0 s and minimum at 10.0 s when considering the period appropriately.
BETA
