 Solution

If $R$ closes the break, a conduction current flows and electrical energy is dissipated. When loop self-inductance is negligible,The externally imposed flux-change emf remains the same. If self-inductance $L_{\ell }$ matters, the current instead satisfies $L_{\ell }\dot{I}+RI=B{a}^2\omega \sin \omega t$; its steady amplitude is $B{a}^2\omega /\sqrt{R^2+{\omega }^2{L}_{\ell }^2}$ 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.