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:

\displaystyle\partial(\Sigma-\Sigma')=\partial\Sigma-\partial\Sigma'=C-C=0

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 »

  1. What’s the theorem that you used at the end that makes two functions be equal if their integrals are equal over any domain? I know how to prove that statement for continuous functions, but I’d like to know its name. I took two mechanics courses in the past year where this proof method was used extensively and only the professor who taught the first course had a name for it – “Lagrange’s lemma”, but I don’t know if that’s a canonical name.

    Comment by Andrei | January 14, 2012 | Reply

  2. I’m not sure what the name is, actually. It’s pretty straightforward for continuous functions, of course, and physics generally assumes everything is continuous — even smooth — unless specifically stated otherwise.

    Comment by John Armstrong | January 15, 2012 | Reply

  3. […] One unexpected source of electromotive force comes from our fourth and final experimentally-justified axiom: Faraday’s law of induction […]

    Pingback by Maxwell’s Equations « The Unapologetic Mathematician | February 1, 2012 | Reply

  4. […] the case of Faraday’s law, we’re already done, since it’s exactly the third of Maxwell’s equations in […]

    Pingback by Deriving Physics from Maxwell’s Equations « The Unapologetic Mathematician | February 3, 2012 | Reply

  5. […] charge” at a point in an electric field it experiences a force . As we saw when discussing Faraday’s law, for a static electric field we can write for some “electric potential” function . […]

    Pingback by Energy and the Electric Field « The Unapologetic Mathematician | February 14, 2012 | Reply

  6. […] Faraday’s law tells us about the electromotive force induced on the […]

    Pingback by Energy and the Magnetic Field « The Unapologetic Mathematician | February 14, 2012 | Reply

  7. […] Faraday’s law tells us […]

    Pingback by Conservation of Electromagnetic Energy « The Unapologetic Mathematician | February 17, 2012 | Reply

  8. […] about the existence of potentials, and the argument usually goes like this: as Faraday’s law tells us, for a static electric field we have ; therefore for some potential function because the […]

    Pingback by A Short Rant about Electromagnetism Texts « The Unapologetic Mathematician | February 18, 2012 | Reply


Leave a Reply

Fill in your details below or click an icon to log in:

Gravatar
WordPress.com Logo

You are commenting using your WordPress.com account. ( Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. ( Log Out /  Change )

Cancel

Connecting to %s

« Previous | Next »

%d bloggers like this: