Past exam of the physics 2 paper of the University of Oxford / 2013 resit / 5

Faraday's law statesfor a stationary loop, with ${\Phi }_B=\int \mathbf{B}\cdot d\mathbf{S}$ and consistent contour and area orientations. The minus sign expresses Lenz's law: the induced response opposes the change of magnetic flux. An emf can exist across an open loop even when no conduction current flows.

Choose the loop normal along $+z$. Rotation about $x$ gives $B_z=B\cos \omega t$, so the linked magnetic flux is ${\Phi }_B=B{a}^2\cos \omega t$. ThereforeThe sign uses the contour orientation associated with $+z$. For the small open gap, this is the corresponding open-circuit terminal emf, with the stated terminal ordering. The tangential induced electric field is nonconservative, so its circulation is not the integral of a single-valued electrostatic potential gradient.

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.