 Solution

ID: past-exam-of-the-physics-2-paper-of-the-university-of-oxford/2022/3/ii/solution

j (r) = Kr

. Normalize it using the total electric current:

I = \int_{0}^{a} Kr 2 π r d r = \frac{2 π K a ^{3}}{3}, K = \frac{3 I}{2 π a ^{3}} .

r

I_{enc} (r) = 2 π K r^{3} /3 = I (r / a)^{3}

B (r) = \frac{μ _{0} I r ^{2}}{2 π a ^{3}} (0 \leq r \leq a) .

Both profiles join the exterior magnetic flux density

μ_{0} I / (2 πa)

at the surface, but the second vanishes quadratically at the axis because less electric current is concentrated near it.

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