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Fernando Garcia

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PHYS405 - Electricity & Magnetism I

Spring 2024

Problem 6.8

Let's start by calculating the volume and surface currents due to $\mathbf{M}$. Recall:

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

So

$$\begin{array}{rl}\text{Volume}{\mathbf{J}}_b& =\frac{1}{s}\frac{\partial }{\partial s}(s\cdot (k{s}^2))\hat{\mathbf{z}}\\ & =3ks\hat{\mathbf{z}}\end{array}$$

Similarly,

$$\begin{array}{rl}\text{Surface}{\mathbf{K}}_b& =k{s}^2{|}_{s=R}(\hat{\varphi }\times \hat{s})\\ & =-k{R}^2\hat{z}\end{array}$$

Where the negative sign comes from the orientation of that cross product.

We have cylindrical symmetry, so we can use Ampere's law. Inside, the current enclosed is:

$$\begin{array}{rl}{I}_{\text{enc}}^{(\text{inside})}& =\int {\mathbf{J}}_b\cdot d\mathbf{a}\\ & ={\int }_{\varphi =0}^{2\pi }{\int }_{s=0}^s{J}_{b}d{a}_{s,\varphi }\\ & ={\int }_{\varphi =0}^{2\pi }{\int }_{s=0}^s{J}_{b}rdrd\varphi \\ & ={\int }_{\varphi =0}^{2\pi }{\int }_{s=0}^{s}3kssdsd\varphi \\ & =2\pi k{\int }_{s=0}^{s}3s^{2}ds\\ & =2\pi k{s}^3\end{array}$$

So Ampere's law follows:

$$\begin{array}{rl}B\cdot \text{area}& ={\mu }_0{I}_{\text{enc}}^{(\text{inside})}\\ B2\pi s& ={\mu }_{0}2\pi k{s}^3\\ B& ={\mu }_{0}k{s}^2\hat{\varphi }\\ \text{That is...}\mathbf{B}& ={\mu }_{0}k{s}^2\hat{\varphi }\\ & =\mu \mathbf{M}\end{array}$$

Now, if we are inside, the charge enclosed will be the total from inside ($2\pi k{R}^3$) plus the one from the surface current density, but the surface charge turns out to be the exact opposite: $\int K_{b}dl=-2\pi k{R}^3$.

So the total current enclosed is 0, meaning that outside

$$\mathbf{B}=\mathbf{0}$$

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