A current I₀ flows through a metallic circular loop of radius r. The resistance of segment ABC is half that of ADC. The magnitude of the magnetic field at the centre O of the loop is:

Solution & Explanation

Principle: the centre field due to an arc of radius r carrying current i is B = (µ₀·i·φ)/(4π·r), where φ is the angle the arc subtends at O. The two arcs here are semicircular (φ = π each), so each gives B = µ₀·i/(4r). Current splits inversely with resistance. Given R_ABC = (1/2)·R_ADC. At the parallel junction the two branches have the same voltage: I₁·R_ABC = I₂·R_ADC. So I₁·(1/2·R_ADC) = I₂·R_ADC → I₁ = 2·I₂. With I₁ + I₂ = I₀: 2I₂ + I₂ = I₀ → I₂ = I₀/3 and I₁ = 2I₀/3. The two arcs carry current in opposite rotational senses around O, so their central fields point oppositely and subtract. B_O = (µ₀/4r)·(I₁ − I₂) = (µ₀/4r)·(2I₀/3 − I₀/3) = (µ₀/4r)·(I₀/3) = µ₀·I₀/(12r). This matches option A, µ₀·I₀/(12r). Note: more current flows through the lower-resistance arc, but each arc's field scales with its own current, and the difference (not sum) is taken because the senses oppose.