Past exam of the optics and electrodynamics course of the University of Cambridge / 2022 / Section B / 6 / b

Write $a_y=E_{A,y}^0$, $a_z=E_{A,z}^0$, $b_y=E_{B,y}^0$ and $b_z=E_{B,z}^0$, evaluated at the same transverse coordinates $(y,z)$. A superscript $0$ labels each filament's own rest frame, rather than a single common frame in all cases. The filaments are infinite and uniform along $x$, so the change of longitudinal coordinate under a Lorentz boost creates no extra dependence in these functions. After the acceleration transients have passed, the magnetic field of each filament vanishes in its own rest frame.

For a filament moving at $+v\hat{\mathbf{x}}$, use the inverse of part (a): its transverse electric field in the laboratory is $\gamma {\mathbf{E}}^0$ and its magnetic field is $\mathbf{v}\times \mathbf{E}/c^2$. Superposition for the two co-moving filaments therefore gives, in $O$,In the common co-moving frame $O^{\prime }$ both filaments are at rest, soFor example, $B_y^{\prime }=\gamma [-\gamma v(a_z+b_z)/c^2+v\gamma (a_z+b_z)/c^2]=0$, a useful sign check. The supplied $\eta$ is the proper linear charge density. Thus each laboratory density is $\gamma \eta$ and each laboratory electric current is $\gamma \eta v$, consistently with length contraction and the four-current transformation. In $O^{\prime }$ each density is $\eta$ and both currents vanish.

Retain $a_y,a_z,b_y,b_z$ as defined in Case 1. In the laboratory, A moves and B is stationary. Transform A's fields and add B's unchanged electrostatic field:The laboratory linear charge densities are ${\lambda }_A=\gamma \eta$ and ${\lambda }_B=\eta$, with electric currents $I_A=\gamma \eta v$ and $I_B=0$.

In $O^{\prime }$, A is stationary while B moves with velocity $-v\hat{\mathbf{x}}$. Thus A contributes its rest electrostatic field, and B contributes the boosted electric field and the opposite-sign convective magnetic field:The densities in this frame are ${\lambda }_A^{\prime }=\eta$, ${\lambda }_B^{\prime }=\gamma \eta$, with $I_A^{\prime }=0$ and $I_B^{\prime }=-\gamma \eta v$. Substituting the laboratory result into part (a) gives the same fields; for instance $E_y^{\prime }=\gamma [(\gamma a_y+b_y)-v\gamma v{a}_y/c^2]=a_y+\gamma b_y$. The magnetic contribution attributed to A disappears in its own frame, but B's contribution does not.