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ID: past-exam-of-the-optics-and-electrodynamics-course-of-the-university-of-cambridge/2012/paper-1/4/c/solution

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Past exam of the optics and electrodynamics course of the University of Cambridge / 2012 / Paper 1 / 4 / c / Solution by

Codex  0 2026-10-20

In

S

, adding the stationary opposite line cancels charge density and electric field but contributes no current. Hence

λ_{tot} = 0, I_{tot} = λ u, E_{tot} = 0, B_{tot} = \frac{μ _{0} λ u}{2 π r} φ .

The neutral current filament in a boosted frame is not neutral. Transform its line-integrated four-current:

λ_{tot}^{'} = - \frac{γ u I _{tot}}{c ^{2}} = - γλ \frac{u ^{2}}{c ^{2}}, I_{tot}^{'} = γ I_{tot} = γλ u .

The same result follows by adding the original resting line

(λ^{'}, 0)

to the new negative line, whose primed density and current are

(- γλ, + γλ u)

. In particular

λ^{'} - γλ = - γλ u^{2} / c^{2}

because

λ = γ λ^{'}

.

Transforming the total fields gives

E_{tot}^{'} = - \frac{γλ u ^{2}}{2 π ϵ _{0} c ^{2} r} r, B_{tot}^{'} = \frac{μ _{0} γλ u}{2 π r} φ .

For positive

λ

,

u \times B

points inward, explaining the electric-field sign. The primed Gauss and Ampère fields calculated from

λ_{tot}^{'}

and

I_{tot}^{'}

reproduce these expressions. Both a moving charge distribution and a moving current must be transformed; boosting just one constituent would give an inconsistent total.

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