Is the closed contour integral of the Lorentz's Force equal to charge * EMF of the circuit?

Question

I have a situation as pictured

That is, a rotating rectangular spire subject to an uniform magnetic field $\vec{B}$ (The rods to which it is attached and make the system spin is non-conductive, so it isn't included in the circuit). I am asked to calculate the current that passes through the wire at a given instant, given that I know the resistance $R$ and side lengths of the spire. To do this I equated the e.m.f. $\mathcal{E}$ on the circuit to minus the time derivative of the magnetic flux. \begin{equation} \mathcal{E} = - \frac{\partial}{\partial t} \phi_{B} \end{equation} And set $$ I = \frac{1}{R}\mathcal{E} $$ The thing is, I know this relation to be derived from Faraday-Maxwell's equation $$ \nabla \times \vec{E} = - \frac{\partial}{\partial t}\vec{B} $$ where we integrate both sides et cetera. The problem is, since $\vec{B}$ is constant, $\frac{\partial}{\partial t}\vec{B} = 0$ and we can't to the integral as usual(changing the order of the derivative and integral). I feel the equation relating the e.m.f. and the change in magnetic flux should still count. The only way I can see this holding up is to note that since we have moving electrons on the spire, and they are subject to an external magnetic field, they will experience a Lorentz's force $$ \vec{F} = q \, \vec{v} \times \vec{B} $$ and we would set $\mathcal{E} = \frac{1}{q} \oint_{Wire} \vec{F} \cdot d\vec{l}$. Does this work? If so, how can we prove it?

Answer 1

First of all, there is no problem if $\vec B$ is constant, it is still possible to interchange integral and derivative, because, if $A$ is some area, $$ \int_A d\vec A ~ \underbrace{\partial_t\vec B}_{=0} = 0 = \partial_t \underbrace{\int_A d\vec A ~ \vec B}_{\text{independent of $t$}}~. $$ This is just a special case of the general theorem. In general, you may interchange integral and derivative whenever Leibniz's integral rule allows it.

Secondly, $\vec B$ is not constant at all in your problem. You are doing the integration in the reference frame of the rotating conductor loop, which means that $\vec B$ is actually rotating. This is also the reason for the current not being 0.

Thirdly, for your explicit situation, it is of course possible to use $$ \mathcal E = \frac 1 q \oint_{\partial A} \vec F d\vec l~, $$ where $\partial A$ is the boundary of the surface $A$ enclosed by the wire. Let $a$ be the length of the sides of the conductor loop not connected to the rods and $b$ the length of other two, then the flux is $\phi_B = -abB \sin(\omega t)$, and $$ \mathcal E = -\partial_t \phi_B = -\partial_t(-ab B \sin(\omega t)) = abB\cos(\omega t)~. $$ The same result can be obtained by using $$ \mathcal E = - \frac 1q \oint_{\partial A} \vec F d\vec l = - \frac 1q \oint_{\partial A} (q \vec v \times \vec B) d\vec l = - \oint_{\partial A} d\vec l(\vec v \times \vec B)~, $$ by parametrising $$ \vec v = \vec r \times \vec \omega = \frac a2 \omega \vec e_{\omega}~, \qquad \vec e_{\omega} = \begin{pmatrix} \cos(\omega t) \\ 0 \\ \sin(\omega t) \end{pmatrix}~, \qquad \vec B = \begin{pmatrix} 0 \\ 0 \\ B \end{pmatrix}~. $$ Then all there remains to do is splitting up the integral into the parts for the single sides of the conductor loop, noticing which of those cancel out and calculating the rest with the above parameterisation.