---

Fernando Garcia

Home     Research     Blog     Other     About me

---

PHYS405 - Electricity & Magnetism I

Spring 2024

Problem 6.16

An amperian loop around the central axis will give us:

$$\mathbf{H}=\frac{I}{2\pi s}\hat{\varphi }$$

Now that we know $\mathbf{H}$, knowing $\mathbf{B}$ is straight forward. Recall

$$\begin{array}{}\text{(6.31)}& \mathbf{B}=\mu \mathbf{H}\end{array}$$

Where

$$\begin{array}{}\text{(6.32)}& \mu \equiv {\mu }_0(1+{\chi }_m)\end{array}$$

Then

$$\begin{array}{rl}\mathbf{B}& ={\mu }_0(1+{\chi }_m)\mathbf{H}\\ & ={\mu }_0(1+{\chi }_m)\frac{I}{2\pi s}\hat{\varphi }\end{array}$$

That's it, we have the field. To show that the old methods give the same answer, we need to know what $\mathbf{M}$ is. For linear media (as in this problem), we have:

$$\begin{array}{rl}\text{(Eq. 6.29)}& \mathbf{M}& ={\chi }_m\mathbf{H}& ={\chi }_m\frac{I}{2\pi s}\hat{\varphi }\end{array}$$

By recalling

$$\begin{array}{rl}\text{Volume}{\mathbf{J}}_b& =\mathrm{\nabla }\times \mathbf{M}\\ \text{Surface}{\mathbf{K}}_b& =\mathbf{M}\times \hat{\mathbf{n}}\end{array}$$

It is straightforward to show that the volume current density vanishes:

$$\begin{array}{rl}{\mathbf{J}}_b& =\mathrm{\nabla }\times \mathbf{M}\\ & =\frac{1}{s}\frac{\partial }{\partial s}\left(sM\right)\hat{z}\\ & =\mathbf{0}\end{array}$$

And we have 2 surface current densities to take care of: one at $s=a$:

$${\mathbf{K}}_b^{\text{at }s=a}=\frac{{\chi }_{m}I}{2\pi a}\hat{z}$$

And one at $s=b$:

$${\mathbf{K}}_b^{\text{at }s=b}=-\frac{{\chi }_{m}I}{2\pi b}\hat{z}$$

Notice, of course, the sign in the the last one. The total current enclosed is $I$ plus that from the surface current density of the smaller cylinder, so:

$$\begin{array}{rl}{I}_{\text{enc }}& =I+I\frac{{\chi }_m}{2\pi a}\cdot (\text{circumference})\\ & =I+I\frac{{\chi }_m}{2\pi a}2\pi a\\ & =I(1+{\chi }_m)\end{array}$$

Using an amperian loop co-centered with the cylinders:

$$\begin{array}{rl}B& =\frac{1}{2\pi s}{\mu }_0{I}_{\text{enc}}\\ & ={\mu }_0(1+{\chi }_m)\frac{I}{2\pi s}\end{array}$$

That is

$$\mathbf{B}={\mu }_0(1+{\chi }_m)\frac{I}{2\pi s}\hat{\varphi }$$

As before!

Back to PHYS405 site