 Solution

The longitudinal Faraday effect produces circular birefringence: right- and left-circular components have different phase velocities. Linear input contains equal amplitudes of both circular eigenpolarizations. A relative phase ${\delta }_F$ rotates its linear polarization through ${\delta }_F/2$, preserving linearity in an ideal transparent sample. Conventionally the signed rotation is ${\vartheta }_F=V{B}_{\parallel }{d}_F$, with Verdet constant $V$.

A pure Faraday rotator cannot serve as the required switchable quarter-wave plate. It rotates a line into another line rather than producing the required relative phase between fixed orthogonal linear components. Applying a field that gives a nominal quarter-cycle circular-eigenmode retardation simply rotates linear polarization by $45^{\circ }$. Circular dichroism or a separate linear retarder would change the device, and is not the stated pure Faraday mechanism.