= Solution
The regular hexagon has area $A=6(\sqrt3a^2/4)=3\sqrt3a^2/2$. Choose $t=0$ with its positive normal parallel to the imposed field. The external <magnetic flux> is $\Phi=B_0A\cos\omega t$, so <Faraday's law of induction> gives
$$
\mathcal E=-\frac{d\Phi}{dt}=B_0A\omega\sin\omega t.
$$
Taking the stated resistance to determine the loop response,
$$
\boxed{I(t)=\frac{3\sqrt3B_0a^2\omega}{2R}\sin\omega t.}
$$
A different initial orientation adds a <phase>. The sign is with respect to the chosen loop orientation and obeys <Lenz's law>; reversal of that orientation reverses the signed current. This expression neglects <self-inductance>, as in the resistive-loop model.
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