 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/ib/paper-3/15d/c/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

Take the wire along the

x

-axis and let

r

E = \frac{λ}{2 π ε _{0} r} e_{r}, B = 0.

For an observer boosted parallel to the wire,

E^{'} = γ \frac{λ}{2 π ε _{0} r} e_{r}, B^{'} = - \frac{γ}{c ^{2}} v \times E .

The moving observer sees length contraction and hence line density

λ^{'} = γλ

, together with current

I^{'} = - γλ v

. The transformed magnetic field is exactly

B^{'} = \frac{μ _{0} ∣ I ^{'} ∣}{2 π r},

with the direction prescribed by the current. This is Ampère's law for the current seen in the moving frame.

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