# Suggestions for practice problems in E&M

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1. Jan 5, 2017

### IxRxPhysicist

Hey all,
I am working my way through a couple of emag books (Griffiths, Jackson, and Schwinger) and I was wondering if any of y'all have suggestions for problems that you thought were particularly physically insightful or useful.

Cheers,
IR

2. Jan 5, 2017

Try these problems/posts that were recently on Physics Forums:
https://www.physicsforums.com/threa...field-of-a-uniformly-polarized-sphere.877891/
@IxRxPhysicist I have one more E&M problem that I don't think is included in the above that you might find useful. Begin with the equation $B=\mu_o H +M$ which is an equation that comes out of the "pole" model of magnetostatics. (J.D. Jackson emphasizes the "pole" model.) (Sometimes you will see this equation as $B=\mu_o H +\mu_o M'$ where $M'=M/\mu_o$.) Upon taking the divergence of both sides $\nabla \cdot B=\mu_o \nabla \cdot H +\nabla \cdot M$. But $\nabla \cdot B=0$ so that $\mu_o \nabla \cdot H=-\nabla \cdot M$. You might recognize the right side as $-\nabla \cdot M=\rho_m$ where $\rho_m$ is the magnetic charge density (fictitious). The problem is to solve this for $H$. $\\$ The result is that $H$ has an integral solution with the inverse square law $H(x)=\int \frac{1}{4 \pi \mu_o} \frac{\rho_m(x')(x-x')}{|x-x'|^3} \, d^3x'$. The question is, where is the current in conductor contribution to $H$ which is absent from this solution? I will give you a hint: The inhomogeneous differential equation $\nabla \cdot H =\frac{\rho_m}{\mu_o}$ can also have a solution to the homogeneous equation as the complete solution. (The current in conductor contribution to $H$ can be found using Biot-Savart's law. The Biot-Savart solution obeys $\nabla \cdot H=0$.) $\\$ Note: A similar problem is encountered if you take the curl of both sides of the above equation. $\nabla \times B=\mu_o J_{total}$ (in the steady-state case) where $J_{total}=J_{free}+J_m$ and $\nabla \times M=\mu_o J_m$ so that $\nabla \times H=J_{free}$. This has a Biot-Savart type integral for $H$, but the question is where did the magnetic "pole" contribution go that we found above with the $\nabla \cdot H$ equation? And the answer is again similar: This time, the homogeneous $\nabla \times H =0$ needs to be considered as having a contribution to the complete solution for $H$.