Clever Manipulation of Maxwell's Equations

[Adoniram]

Homework Statement

This is a general question that applies to many homework problems (and real world problems), but I will provide an example to help guide the discussion.

I am hoping you all can give me some examples of particularly clever manipulations of Maxwell's equations to make a difficult problem easier. For example, Griffiths' E&M (4th Edition) problem 7.22b states (paraphrased):

A current I flows in a small circular loop of wire of radius a. This loop of wire sits above another loop of wire of radius b (where b > a). They are coaxial (the planes they span are parallel), and are separated by a distance z.

Find the flux through the big loop (of radius b).

Homework Equations

Φ = ∫B⋅da

The Attempt at a Solution

The magnetic field of the top loop can be written as the magnetic field of a dipole, where m=Iπa^2 z-hat

Now, naively, I would want to calculate the flux over the flat area spanned by the loop of radius b. However, the solution is much simpler if one uses a spherical cap of fixed radius R, which is bounded by the same loop of radius b. Hilarious. Because I never would have thought of that...

So, the answer is the same no matter how you calculate, but this basically uses the idea that the flux through any surface, bounded by the same line, is the same. Clever. I mean, I knew that, but I don't see many examples like this...

Can anyone provide another example of a clever use of math to help solve a difficult E&M problem?

 

[Charles Link]

E&M seems to have quite a number of times where an unexpected result appears which simplifies a calculation and/or puts some order to the system. One in particular that I find of interest is the equation $B=\mu_oH+M$ that essentially emerges out of some calculations with the magnetic pole theory which is a mathematical alternative to the magnetic surface current theory. Here $H$ consists of two sources: 1) magnetic poles (comes from $- \nabla \cdot M=\rho_m$) and gives an inverse square law result for $H$ analogous to the $E$ of electrostatics and 2) currents in conductors, where the $H$ is computed via Biot-Savart. A puzzle arises if you take divergence of both sides of the above equation and solve for $H$. Since $\nabla \cdot B=0$, an integral solution is found for $H$ consisting of magnetic charges. (fictitious magnetic charges, but still a valid calculation.) Question is though, what became of the currents in conductors contribution to $H$ if there are any in the problem? It appears to be absent from the solution. And the puzzle has a simple, but perhaps not obvious solution: $\nabla \cdot H=-\nabla \cdot M/\mu_o$ is an inhomogeneous differential equation for $H$. To get the complete $H$, we must consider solutions to the homogeneous equation $\nabla \cdot H=0$ and add them to the solution of the inhomogeneous equation . The currents in conductors (since they follow Biot-Savart) will have $\nabla \cdot H=0$. Thereby, the $H$ from the currents in conductors is part of the solution, and our calculations are in complete order. editing... I can think of a couple others, but one in particular, the simple solution that arises for a dielectric sphere in a uniform electric field is one that I might elaborate on in a subsequent post.

 

[Charles Link]

The problem of a dielectric sphere (of arbitrary radius) in a uniform electric field $E_o$ using Poisson's equation and Legendre polynomials is somewhat complex, but the solution is simple. The result is that the polarization is uniform inside the sphere and the electric field inside the sphere is also uniform inside the sphere and is given by $E_i=E_o+E_p$ where $E_p=-P/(3 \epsilon_o)$. $\$ Using $P=\epsilon_o \chi E_i$, this has the result that $E_i=E_o (1/(1+\chi/3))$. The -1/3 factor for this spherical problem can be shown by computing the result from the surface polarization charge density $\sigma_p=P \cdot \hat{n}$ on the surface of the sphere, and using Coulombs law $E=Q/(4 \pi \epsilon_o r^2)$ for the center of the sphere. This simple Coulomb's law calculation does not show that the $P$ or $E_i$ is indeed uniform inside the sphere, but it is a quick way to show the answer that results from the Legendre Polynomial solution. It would be helpful for E&M students to see this result before being presented the complete Legendre Polynomial solution. Otherwise, it is very easy to lose the physics in the mathematics.

 

[Charles Link]

The problem of a dielectric sphere (of arbitrary radius) in a uniform electric field $E_o$ using Poisson's equation and Legendre polynomials is somewhat complex, but the solution is simple. The result is that the polarization is uniform inside the sphere and the electric field inside the sphere is also uniform inside the sphere and is given by $E_i=E_o+E_p$ where $E_p=-P/(3 \epsilon_o)$. $\$ Using $P=\epsilon_o \chi E_i$, this has the result that $E_i=E_o (1/(1+\chi/3))$. The -1/3 factor for this spherical problem can be shown by computing the result from the surface polarization charge density $\sigma_p=P \cdot \hat{n}$ on the surface of the sphere, and using Coulombs law $E=Q/(4 \pi \epsilon_o r^2)$ for the center of the sphere. This simple Coulomb's law calculation does not show that the $P$ or $E_i$ is indeed uniform inside the sphere, but it is a quick way to show the answer that results from the Legendre Polynomial solution. It would be helpful for E&M students to see this result before being presented the complete Legendre Polynomial solution. Otherwise, it is very easy to lose the physics in the mathematics.

Just one additional comment on this one: In the E&M textbooks this one is so often presented with so much mathematical detail that the student with limited study time doesn't have the time to sift through it to figure out what all the mathematics is showing. If they would first present it in a simpler fashion, I think more students would pick up on the solution.