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Q. A circular coil of $500$ turns of wire has an enclosed area of $0.1\, m ^{2}$ per turn. It is kept perpendicular to a magnetic field of induction $0.2\, T$ and rotated by $180^{\circ}$ about a diameter perpendicular to the field in $0.1\, s$. How much charge will pass when the coil is connected to a galvanometer with a combined resistance of $50$ ohms?

Electromagnetic Induction

Solution:

$\Delta Q =\frac{N B A}{R}\left(\cos \theta_{1}-\cos \theta_{2}\right)$
$=\frac{500 \times 0.2 \times 0.1\left(\cos 0^{\circ}-\cos 180^{\circ}\right)}{50}=0.4\, C$