The Unapologetic Mathematician

Mathematics for the interested outsider

Faraday’s Law

Okay, so let’s say we have a closed circuit composed of a simple loop of wire following a closed path C. There’s no battery or anything that might normally induce an electromotive force around the circuit by chemical or other means. And, as we saw when discussing Gauss’ law, Coulomb’s law gives rise to an electric field that looks like

\displaystyle E(r)=\frac{1}{4\pi\epsilon_0}\int\rho(s)\frac{r-s}{\lvert r-s\rvert^3}\,d^3s

As we saw when discussing Gauss’ law for magnetism, we can rewrite the fraction in the integrand:

\displaystyle\begin{aligned}E(r)&=-\frac{1}{4\pi\epsilon_0}\int\rho(s)\nabla\left(\frac{1}{\lvert r-s\rvert}\right)\,d^3s\\&=-\nabla\left(\frac{1}{4\pi\epsilon_0}\int\rho(s)\frac{1}{\lvert r-s\rvert}\,d^3s\right)\end{aligned}

So this electric field is conservative, and so its integral around the closed circuit is automatically zero. Thus there is no electromotive force around the circuit, and no current flows.

And yet, that’s not actually what we see. Specifically, if we wave a magnet around near such a circuit, a current will indeed flow! Indeed, this is exactly how the simplest electric generators and motors work.

To put some quantitative meat on these qualitative observational bones, we have Faraday’s law of induction. This says that the electromotive force around a circuit is equal to the rate of change of the magnetic flux through any surface bounded by that circuit. What? maybe a formula will help:

\displaystyle\mathcal{E}=\frac{\partial}{\partial t}\int\limits_\Sigma B\cdot dS

where \Sigma is any surface with \partial\Sigma=C. Why can we pick any such surface? Because if \Sigma' is another one then:

\displaystyle\int\limits_\Sigma B\cdot dS-\int\limits_{\Sigma'}B\cdot dS=\int\limits_{\Sigma-\Sigma'}B\cdot dS

We can calculate the boundary of this combined surface:


Since our space is contractible, this means that our surface is itself the boundary of some region E.

\displaystyle\int\limits_{\partial E}B\cdot dS=\int\limits_E\nabla\cdot B\,dV

But Gauss’ law for magnetism tells us that this is automatically zero. That is, every surface has the same flux, and so it doesn’t matter which one we use in Faraday’s law.

Now, we can couple this with our original definition of electromotive force:

\displaystyle\begin{aligned}\int\limits_\Sigma\frac{\partial B}{\partial t}\cdot dS&=-\int\limits_{\partial\Sigma}E\cdot dr\\&=-\int\limits_\Sigma\nabla\times E\cdot dS\end{aligned}

But this works no matter what surface \Sigma we consider, so we come up with the differential form of Faraday’s law:

\displaystyle\nabla\times E=-\frac{\partial B}{\partial t}

January 14, 2012 Posted by | Electromagnetism, Mathematical Physics | 8 Comments