= 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 $\vartheta_F=VB_\parallel d_F$, with <Verdet constant> $V$.
\b[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.
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