Question: A big circular coil of 1000 turns and average radius $10 \mathrm{~m}$ is rotating about its horizontal diameter at $2 \mathrm{rad} \mathrm{s}^{-1}$. If the vertical component of earth’s magnetic field at that place is $2 \times 10^{-5} \mathrm{~T}$ and electrical resistance of the coil is $12.56 \Omega$, then the maximum induced current in the coil will be
A) $0.25 \mathrm{~A}$
B) $1.5 \mathrm{~A}$
C) $1 \mathrm{~A}$
D) $2 \mathrm{~A}$
Answer: $1 \mathrm{~A}$
Sol:
$\begin{aligned} & \phi_B=N B A \cos \omega t \ & \varepsilon=\frac{-d \phi_B}{d t}=-N B A \omega(-\sin \omega t) \ & \varepsilon=N B A \omega \sin \omega t \ & i_{\max }=\frac{\varepsilon_{\max }}{R}-\frac{N B A \omega}{R} \ & =\frac{1000 \times 2 \times 10^{-5} \times \pi(10)^2 \times 2}{12.56} \ & =1 \mathrm{~A} \end{aligned}$
