 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2020/ib/paper-2/15d/b/solution

Codex  0 Created 2026-09-24 Updated 2026-09-29

Q (t)

produces

E (r, t) = \frac{Q ( t )}{4 π ϵ _{0} r ^{2}} e_{r} .

Since

\dot{Q} = I

ϵ_{0} \frac{\partial E}{\partial t} = \frac{I}{4 π r ^{2}} e_{r} .

r

and polar angle

θ < π /2

B = B_{ϕ} e_{ϕ}

is constant along the boundary, whose circumference is

2 π r sin θ

. The displacement-current flux through the cap is

\int_{cap} \frac{I}{4 π r ^{2}} d A = \frac{I}{4 π r ^{2}} 2 π r^{2} (1 - cos θ) = \frac{I}{2} (1 - cos θ) .

Thus

2 π r sin θ B_{ϕ} = \frac{μ _{0} I}{2} (1 - cos θ) .

Using

(1 - cos θ) / sin θ = tan (θ /2)

gives

B (r, θ) = \frac{μ _{0} I}{4 π r} tan (\frac{θ}{2}) e_{ϕ} .

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