The resistances of three parts of a circular | vedclass

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(D) The current $I$ enters at point $A$ and splits into two paths: one through the arc $AB$ (resistance $R$) and the other through the arc $AC$ (resis

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(D) The current $I$ enters at point $A$ and splits into two paths: one through the arc $AB$ (resistance $R$) and the other through the arc $AC$ (resistance $R$).Since the resistances of the two paths are equal,the current divides equally: $I_1 = I_2 = I/2$.The arc $BC$ has resistance $2R$ and is connected between points $B$ and $C$. However,since the potential at $B$ and $C$ is the same due to the symmetry of the circuit,no current flows through the arc $BC$.The magnetic field at the center $O$ due to arc $AB$ is $B_1 = \frac{\mu_0 (I/2)}{2a} 	imes \frac{120^\circ}{360^\circ} = \frac{\mu_0 I}{12a}$ (directed inwards).The magnetic field at the center $O$ due to arc $AC$ is $B_2 = \frac{\mu_0 (I/2)}{2a} 	imes \frac{120^\circ}{360^\circ} = \frac{\mu_0 I}{12a}$ (directed outwards).Since these two fields are equal in magnitude and opposite in direction,the net magnetic field at the center $O$ is $B_{net} = B_1 - B_2 = 0$.

The resistances of three parts of a circular loop are as shown in the figure. The magnetic field at the centre $O$ is:

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