A closed planar wire loop of area $A$ and arbitrary shape is placed in a uniform magnetic field of magnitude $B$ , with its plane perpendicular to the magnetic field. The resistance of the wire loop is $R$ . The loop is now turned upside down by $180^{o}$ so that its plane again becomes perpendicular to the magnetic field. The total charge that must have flowed through the wire ring in the process is:

\frac{A B}{2 R}

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\frac{A B}{R}

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\frac{2 A B}{R}

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Solution

The correct option is D $\frac{2 A B}{R}$
induced emf= $\frac{d (f l u x)}{d t} = R \frac{d q}{d t} \Rightarrow Δ q = \frac{Δ (f l u x)}{R} = \frac{2 B A}{R}$

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