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

Let $\hat{\mathbf{r}}$ point transversely away from the line and define $\hat{\mathbf{\phi }}=\hat{\mathbf{x}}\times \hat{\mathbf{r}}$. Cylindrical Gauss's law in the line's rest frame givesThe transverse distance is unchanged by the axial boost: $r=r^{\prime }$. Applying the inverse of the given field transformation to a boosted uniform line charge,Here $\gamma =(1-u^2/c^2{)}^{-1/2}$ and all longitudinal field components vanish. Substituting back verifies ${\mathbf{B}}^{\prime }=\gamma (\mathbf{B}-\mathbf{u}\times \mathbf{E}/c^2)=0$ and ${\mathbf{E}}^{\prime }=\gamma (1-u^2/c^2)\mathbf{E}=\mathbf{E}/\gamma$. The sign of ${\lambda }^{\prime }$ determines the radial and circulation directions.