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(A) The total length of the wire is $L = 2 \pi r = 2 \pi (2) = 4 \pi \ m$.The total resistance of the wire is $R_{total} = (	ext{resistance per unit

Vedclass · Accepted Answer

$8$

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(A) The total length of the wire is $L = 2 \pi r = 2 \pi (2) = 4 \pi \ m$.The total resistance of the wire is $R_{total} = (	ext{resistance per unit length}) 	imes L = \frac{1}{\pi} 	imes 4 \pi = 4 \ \Omega$.Since $\angle AOB = 90^{\circ}$,the wire is divided into two arcs: a minor arc $AB$ and a major arc $AB$.The length of the minor arc is $L_1 = \frac{90^{\circ}}{360^{\circ}} 	imes (2 \pi r) = \frac{1}{4} 	imes 4 \pi = \pi \ m$.The resistance of the minor arc is $R_1 = \frac{1}{\pi} 	imes \pi = 1 \ \Omega$.The length of the major arc is $L_2 = L - L_1 = 4 \pi - \pi = 3 \pi \ m$.The resistance of the major arc is $R_2 = \frac{1}{\pi} 	imes 3 \pi = 3 \ \Omega$.These two resistances are connected in parallel between points $A$ and $B$.The equivalent resistance $R_p$ is given by $\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{1} + \frac{1}{3} = \frac{4}{3} \ \Omega^{-1}$,so $R_p = \frac{3}{4} \ \Omega$.The current flowing through the battery is $I = \frac{V}{R_p} = \frac{6}{3/4} = 6 	imes \frac{4}{3} = 8 \ A$.

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