A current $I_0$ flows through a metallic circular loop of radius $r$ as shown in the figure. Resistance of the segment ABC is half that of ADC. Magnitude of magnetic field at the center O of the loop is :
Answer: (A) $\dfrac{\mu_0 I_0}{12r}$
The current enters at A and leaves at C, which are diametrically opposite. The straight lead wires lie along the line through O, so they give no field at O.
The two semicircles ABC and ADC are in parallel. Since $R_{ABC} = \tfrac{1}{2}R_{ADC}$, the current divides as
$$I_{ABC} = \frac{2I_0}{3}, \qquad I_{ADC} = \frac{I_0}{3}$$
A semicircle carrying current $I$ gives $B = \dfrac{\mu_0 I}{4r}$ at the centre. The two halves carry current in opposite senses around O, so their fields oppose:
$$B = \frac{\mu_0}{4r}\left(\frac{2I_0}{3} - \frac{I_0}{3}\right) = \frac{\mu_0 I_0}{12r}$$
Solution by Sreeraj P, M.Sc Physics