 Solution

ID: past-exam-of-the-physics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2010/paper-1/section-a/3/solution

Apply Ampère's law on a concentric circle. Uniform current density is

J = I / (π a^{2})

, so the enclosed current is

I r^{2} / a^{2}

inside and

I

outside. With permeability

μ_{0}

B (r) = {μ_{0} I r / (2 π a^{2}), μ_{0} I / (2 π r), 0 \leq r \leq a, r \geq a .

The field circulates azimuthally by the right-hand rule. Its magnitude rises linearly from zero, is continuous and maximal at the surface, then falls as

1/ r

. End effects are neglected.

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