A straight conductor carrying current i splits into two parts as shown in the figure. The radius of the circular loop is R. The total magnetic field at the centre P of the loop is

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(1) Zero. Given: the current enters at one point of the loop and leaves at another, splitting into i₁ through the longer arc and i₂ through the shorter; the figure marks the two arcs as 3/4 and 1/4 of the circle. Idea: the two arcs are in parallel, so the current divides inversely as their resistances, which for a uniform wire means inversely as their lengths. (i₁)/(i₂)=(ℓ₂)/(ℓ₁)=(1/4)/(3/4)=1/3, so i₂=3i₁. Field of an arc at the centre: B=(μ₀i)/(2R)·θ/(2π), and the two arcs circle P in opposite senses. Longer arc: (μ₀i₁)/(2R)·(3π)/(2π)=(μ₀)/(2R)·(3i₁)/4. Shorter arc: (μ₀i₂)/(2R)·π/(2π)=(μ₀)/(2R)·(i₂)/4=(μ₀)/(2R)·(3i₁)/4. They are equal and opposite, so the net field is zero — true for any split point of a uniform loop.

A straight conductor carrying current i splits into two parts as shown in the figure. The radius of the circular loop is R. The total magnetic field at the centre P of the loop is

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