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(D) Let the lengths of the two arcs be $l_1 = r	heta$ and $l_2 = r(2\pi - 	heta)$.Since the ring is uniform,the resistances are $R_1 = ho \frac{l_

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Zero for all values of $	heta$

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(D) Let the lengths of the two arcs be $l_1 = r	heta$ and $l_2 = r(2\pi - 	heta)$.Since the ring is uniform,the resistances are $R_1 = ho \frac{l_1}{A}$ and $R_2 = ho \frac{l_2}{A}$.When connected to a battery,the potential difference $V$ across both arcs is the same. Thus,$i_1 R_1 = i_2 R_2$,which implies $i_1 l_1 = i_2 l_2$.The magnetic field at the center due to an arc of length $l$ carrying current $i$ is $B = \frac{\mu_0 i}{4\pi r^2} l$.For the two arcs,$B_1 = \frac{\mu_0 i_1 l_1}{4\pi r^2}$ and $B_2 = \frac{\mu_0 i_2 l_2}{4\pi r^2}$.Since $i_1 l_1 = i_2 l_2$,we have $B_1 = B_2$.The currents flow in opposite directions around the center,so the magnetic fields produced by the two arcs are equal in magnitude and opposite in direction.Therefore,the resultant magnetic induction at the center is $B_{net} = B_1 - B_2 = 0$,which is independent of $	heta$.

$A$ battery is connected between two points $A$ and $B$ on the circumference of a uniform conducting ring of radius $r$ and resistance $R$. One of the arcs $AB$ of the ring subtends an angle $\theta$ at the centre. The value of the magnetic induction at the centre due to the current in the ring is

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