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(B) The total resistance of the wire is $R$. When the wire is bent into a circular loop,the resistance is distributed uniformly along the circumferenc

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$\frac{16 E^2}{3 R}$

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(B) The total resistance of the wire is $R$. When the wire is bent into a circular loop,the resistance is distributed uniformly along the circumference.Two points separated by a quarter circumference divide the loop into two arcs: one with resistance $R_1 = \frac{1}{4}R$ and the other with resistance $R_2 = \frac{3}{4}R$.These two segments are connected in parallel across the battery of emf $E$.The equivalent resistance $R_{eq}$ of the parallel combination is given by $\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{R/4} + \frac{1}{3R/4} = \frac{4}{R} + \frac{4}{3R} = \frac{12+4}{3R} = \frac{16}{3R}$.Thus,$R_{eq} = \frac{3R}{16}$.The heat generated per second is the power dissipated,given by $P = \frac{E^2}{R_{eq}}$.Substituting the value of $R_{eq}$,we get $P = \frac{E^2}{3R/16} = \frac{16 E^2}{3 R}$.

$A$ wire of resistance $R$ is bent in the form of a circular loop. Two points on the circle separated by a quarter circumference are connected to a battery of emf $E$ and negligible internal resistance. The heat generated in the wire per second is

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