RE-NEET 2026 Physics Magnetic Field MCQ Question

Resistance of ABC is half that of ADC, so the current divides in inverse proportion:

$I_{ABC} = \frac{2I_0}{3}, \qquad I_{ADC} = \frac{I_0}{3}$

Each path is a semicircle, and a semicircular arc contributes $B = \frac{\mu_0 I}{4r}$ at the centre. The two halves carry current in opposite senses around O, so their fields subtract:

$B = \frac{\mu_0}{4r}\left(\frac{2I_0}{3} - \frac{I_0}{3}\right) = \frac{\mu_0 I_0}{12r}$

$\frac{\mu_0 I_0}{2r}$ is the full-loop result, and $\frac{\mu_0 I_0}{4r}$ is one semicircle on its own — both ignore that the second half cancels part of the first.