In a region of uniform magnetic induction $B = 5 imes 10^{-2} \mathrm{~T}$, a single-turn circular

Question

In a region of uniform magnetic induction $B = 5 	imes 10^{-2} \mathrm{~T}$, a single-turn circular coil of radius $20 \mathrm{~cm}$ and resistance $2\pi^2 \Omega$ is rotated about a diametric axis perpendicular to the magnetic field. If the coil rotates at a constant speed of $300 \mathrm{rpm}$, the amplitude of the alternating current induced is

Professor PC Thomas Classes · Answer

(d) : The peak induced emf is $E_0 = NBA\omega$. Here $N=1$, $B=5 	imes 10^{-2} \mathrm{T}$, area $A = \pi (0.2)^2 = 0.04\pi \mathrm{~m}^2$. The angular velocity $\omega = 2\pi f = 2\pi 	imes (300/60) = 10\pi \mathrm{~rad/s}$. Substituting values: $E_0 = 1 	imes (5 	imes 10^{-2}) 	imes (0.04\pi) 	imes (10\pi) = 0.02\pi^2 \mathrm{~V}$. The amplitude of the alternating current is $I_0 = \frac{E_0}{R} = \frac{0.02\pi^2}{2\pi^2} = 0.01 \mathrm{~A} = 10 \mathrm{~mA}$.