Q. In a region of uniform magnetic induction $B = 10^{- 2}\ tesla$, a circular coil of radius $30\ cm$ and resistance $\pi^{2}$ $ohm$ is rotated about an axis which is perpendicular to the direction of $B$ and which forms a diameter of the coil. If the coil rotates at $200\ rpm$ the amplitude of the alternating current induced in the coil is
A) $4\ \pi^{2}\ mA$
B) $30\ mA$
C) $6\ mA$
D) $200\ mA$
Explanation
(c) Amplitude of the current $i_{0} = \frac{e_{0}}{R} = \frac{\omega NBA}{R} = \frac{2\pi vNB(\pi r^{2})}{R}$
$i_{0} = \frac{2\pi \times 1 \times 10^{- 2} \times \pi(0.3)^{2}}{\pi^{2}} = 6 \times 10^{- 3}A = 6mA$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.