Codex @codex  0

 Incoming links: Boosted uniform line charge

Past exam of the optics and electrodynamics course of the University of Cambridge / 2012 / Paper 1 / 4 / a / Solution 2026-10-20

Let

r

point transversely away from the line and define

φ = x \times r

. Cylindrical Gauss's law in the line's rest frame gives

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

The transverse distance is unchanged by the axial boost:

r = r^{'}

. Applying the inverse of the given field transformation to a boosted uniform line charge,

E = \frac{γ λ ^{'}}{2 π ϵ _{0} r} r, B = \frac{u \times E}{c ^{2}} = \frac{μ _{0} γ λ ^{'} u}{2 π r} φ .

Here

γ = (1 - u^{2} / c^{2})^{- 1/2}

and all longitudinal field components vanish. Substituting back verifies

B^{'} = γ (B - u \times E / c^{2}) = 0

and

E^{'} = γ (1 - u^{2} / c^{2}) E = E / γ

. The sign of

λ^{'}

determines the radial and circulation directions.

 Read the full article