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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{align*} \text{Volume} \;\;\; \mathbf{J}_b &= \nabla \times \mathbf{M} \\ \text{Surface} \;\;\; \mathbf{K}_b &= \mathbf{M}\times \hat{\mathbf{n}} \end{align*}

So

\begin{align*} \text{Volume} \;\;\; \mathbf{J}_b &= \frac{1}{s}\frac{\partial }{\partial s}(s\cdot (ks^2 )) \hat{\mathbf{z}} \\ &= 3ks \hat{\mathbf{z}} \end{align*}

Similarly,

\begin{align*} \text{Surface} \;\;\; \mathbf{K}_b &= ks^2 \Big|_{s=R} (\hat{\phi }\times \hat{s}) \\ &= -kR^2 \hat{z} \end{align*}

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{align*} I_{\text{enc} }^{(\text{inside} )} &= \int \mathbf{J}_b\cdot d\mathbf{a} \\ &= \int_{\phi =0}^{2\pi } \int_{s=0}^{s} J_b da_{s,\phi }\\ &= \int_{\phi =0}^{2\pi } \int_{s=0}^{s} J_b rdrd\phi \\ &= \int_{\phi =0}^{2\pi } \int_{s=0}^{s} 3ks sdsd\phi \\ &= 2\pi k\int_{s=0}^{s} 3s^2 ds \\ &= 2\pi ks^3 \end{align*}

So Ampere's law follows:

\begin{align*} B\cdot \text{area} &= \mu _0 I_{\text{enc} }^{(\text{inside} )} \\ B2\pi s &= \mu _0 2\pi ks^3 \\ B &= \mu _0 ks^2 \hat{\phi } \\ \text{That is...} \;\;\; \mathbf{B} &= \mu _0 ks^2 \hat{\phi } \\ &= \mu \mathbf{M} \end{align*}

Now, if we are inside, the charge enclosed will be the total from inside ($2\pi kR^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 kR^3 $.

So the total current enclosed is 0, meaning that outside

\[\mathbf{B}=\mathbf{0} \]

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