= Solution
If $R$ closes the break, a conduction current flows and electrical energy is dissipated. When loop self-<inductance> is negligible,
$$
\boxed{I(t)=\frac{Ba^2\omega}{R}\sin\omega t.}
$$
The externally imposed flux-change emf remains the same. If self-inductance $L_\ell$ matters, the current instead satisfies $L_\ell\dot I+RI=Ba^2\omega\sin\omega t$; its steady amplitude is $Ba^2\omega/\sqrt{R^2+\omega^2L_\ell^2}$ and it lags the emf by $\tan^{-1}(\omega L_\ell/R)$. If the break remains open after inserting a resistor elsewhere in series, the loop is still open and the conduction current remains zero.
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