Q. A circular coil of 500 turns of wire has an enclosed area of $0.1\ m^{2}$ per turn. It kept perpendicular to a magnetic field of induction $0.2\ T$ and rotated by $180{^\circ}$ about a diameter perpendicular to the field in $0.1\ sec$. How much charge will pass when the coil is connected to a galvanometer with a combined resistance of $50\ ohms$
A) $0.2\ C$
B) $0.4\ C$
C) $2\ C$
D) $4\ C$
Explanation
(b) $\Delta Q = \frac{NBA}{R}\left( \cos\theta_{1} - \cos\theta_{2} \right)$
$= \frac{500 \times 0.2 \times 0.1(\cos{0 - \cos 180})}{50} = 0.4\ C$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.