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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 centre O of the loop is:
Detailed Solution
$R_1:R_2=1:2$ and $I\propto\frac{1}{R}$, so $I_1=\frac{2}{3}I_0$ (through ABC) and $I_2=\frac{1}{3}I_0$ (through ADC). The two semicircles give opposite fields at O: $B_{net}=\frac{\mu_0I_1}{4r}-\frac{\mu_0I_2}{4r}=\frac{\mu_0}{4r}\left(\frac{2I_0}{3}-\frac{I_0}{3}\right)=\frac{\mu_0I_0}{12r}$.