electromagnetism - Faraday's law in open circuits

Faraday's law needs a closed loop $\partial\Sigma$ in order to be applied (so as to be able to calculate the magnetic flux across $\Sigma$). However, $\partial\Sigma$ is just, at least according to my present comprehension, a mathematical object; nonetheless, in textbooks it often coincides with a physical object like a closed loop of wire. So my question is: does the physical wire have to be a closed loop or can it have discontinuities?

For example, if we consider a ring with a small discontinuity at some point, can we compute $-\partial\phi_B/\partial t$ across the surface described by the "convex envelope" of the wire (let's assume the ring lies on a single plane) in order to get the $\Delta V$ between the two open ends of the ring?

Answer

Faraday's Law relates the EMF in a closed loop to the flux through any surface having that loop as its boundary, but one can still use it to compute the EMF in, say, a wire segment by including that segment as part of a mathematical loop that is closed.

Consider, for example, a wire segment of length $\ell$ moving in the $x$-$y$ plane in the positive $x$-direction through a uniform field $-B\hat{\mathbf z}$. One can directly compute the EMF in this loop using the Lorentz force Law. The EMF is, by definition, the work per unit charge performed by the electromagnetic fields on a charge that is stationary relative to the segment, moving with the segment $$ \mathscr E = \frac{\text{Lorentz Force}\cdot\text{path vector}}{q} = \frac{[q(v\hat{\mathbf x})\times (-B\hat{\mathbf z})]\cdot(\ell\hat{\mathbf y})}{q} = \frac{qvB\ell}{q} = vB\ell $$ On the other hand, suppose that this segment is considered as one side of a mathematical rectangle in the $x$-$y$ plane with dimensions $\ell\times(\ell_0 + vt)$ and therefore with area $A_t = \ell(\ell_0 + vt)$, then the EMF in the loop computed via Faraday's Law is $$ \mathscr E = -\frac{d\Phi}{dt} = -\frac{d}{dt}[(-B\hat{\mathbf z})\cdot (A_t\hat{\mathbf z})] = -\frac{d}{dt}(-B\ell(\ell_0 + vt)) = vB\ell $$ We obtain the same result! This is often called motional EMF.