Answer
$v_L=−5.85 V$
Work Step by Step
The inductor voltage phasor is $90^{\circ}$ more than that of the current. Therefore, at $t=$ $4.17 \mathrm{~ms}$, we find
$
\begin{aligned}
v_L & =I \sin \left(90^{\circ}-\left(-24.3^{\circ}\right)+90^{\circ}\right) X_L=-L X_L \sin \left(24.3^{\circ}\right)=-(0.164 \mathrm{~A})(86.7 \Omega) \sin \left(24.3^{\circ}\right) \\
& =-5.85 \mathrm{~V} .
\end{aligned}
$
