Bound Current

By definition, magnetization is defined as the magnetic dipole moment density $$\textbf{m} = \int \textbf{M} d\tau$$

Bound Current

Vector potential due to magnetic dipole is $$\textbf{A}(\textbf{r}) = \frac{\mu_0}{4\pi}\frac{\textbf{m}\times\hat{d}}{d^2}$$ where $\textbf{d} = \textbf{r} - \textbf{r}'$. If the magnetic dipole is expressed in magnetization, the vector potential can be written as $$\begin{align} \textbf{A}(\textbf{r}) &= \frac{\mu_0}{4\pi}\int\frac{\textbf{M}(\textbf{r}')\times\hat{d}}{d^2} d\tau \\ &= \frac{\mu_0}{4\pi}\int \left[\textbf{M}(\textbf{r}')\times\left(\nabla\frac{1}{d}\right)\right]d\tau \\ &= \frac{\mu_0}{4\pi}\int \left[ \frac{\nabla\times\textbf{M}(\textbf{r}')}{d} - \nabla\times\left(\frac{\textbf{M}(\textbf{r}')}{d}\right) \right] d\tau \\ &= \frac{\mu_0}{4\pi}\int \frac{1}{d}\left(\nabla\times\textbf{M}(\textbf{r}')\right)d\tau + \frac{\mu_0}{4\pi}\oint\frac{1}{d}\textbf{M}(\textbf{r}')\times d\textbf{a} \end{align}$$ Define bound surface current to be $$\bbox[5px,border:2px solid #666] { \textbf{J}_b = \nabla\times\textbf{M} }$$ and bound volume current to be $$\bbox[5px,border:2px solid #666] { \textbf{K}_b = \textbf{M}\times\hat{\textbf{n}} }$$ The potential can be written as $$\textbf{A}(\textbf{r}) = \frac{\mu_0}{4\pi}\int \frac{\textbf{J}_b(\textbf{r}')}{d}d\tau + \frac{\mu_0}{4\pi}\oint\frac{\textbf{K}_b(\textbf{r}')}{d} d\textbf{a}$$

Example (Griffiths Third Edition Ex 6.1)

Find the magnetic field of a uniformly magnetized sphere. Given that the magnetic field inside a spherical shell of radius $R$ carrying a uniform surface charge $\sigma$ spinning at angular velocity $\omega$ is $\frac{2}{3}\mu_0\sigma R\omega$.

Solution: Choose $z$ axis along the direction of $\vec{M}$, so $$\vec{M}=M\hat{z}.$$ The bound volume current is $$\vec{J}_b=\nabla\times\vec{M}=0$$ and the bound surface current is $$\vec{K}_b=\vec{M}\times\hat{n}=M\sin\theta\hat{\phi}$$ This is equivalent to a spherical shell of radius $R$ carrying a uniform surface charge $\sigma$ spinning at angular velocity $\omega$, where $\vec{M}=\sigma R\vec{\omega}$. So, the magnetic field inside the sphere is $$\vec{B}=\frac{2}{3}\mu_0\vec{M}$$ Outside the sphere, the field is the same as a pure dipole with dipole moment $$\vec{m}=\frac{4}{3}\pi R^3\vec{M}$$

Checkpoint (Griffiths Third Edition Q6.7)

An infinitely long circular cylinder carries a uniform magnetization $\vec{M}$ parallel to its axis. Find the magnetic field due to $\vec{M}$ inside and outside the cylinder.

The bound volume current is $$\vec{J}_b=\nabla\times\vec{M}=0$$ The bound surface current is $$\vec{K}_b=\vec{M}\times\hat{n}=M\hat{z}\times\hat{n}=M\hat{\phi}$$ This is equivalent to a solenoid, so the field outside is zero, and inside is $$B=\mu_0 K_b=\mu_0 M$$ in which the field points the same direction as $\vec{M}$, so $$\vec{B}=\mu_0\vec{M}$$

H-Field

By Ampere's law, $$\frac{1}{\mu_0}(\nabla\times\textbf{B})=\textbf{J} = \textbf{J}_f + \textbf{J}_b$$ where $\textbf{J}_b$ is the charge density due to bound current while $\textbf{J}_f$ is the charge density due to free current. Thus, $\textbf{B}$ here is the B-field due to both free current and bound current, i.e. magnetization. Express in terms of magnetization, we have $$\begin{align} \frac{1}{\mu_0}(\nabla\times\textbf{B}) &= \textbf{J}_f + \textbf{J}_b \\ &= \textbf{J}_f + (\nabla \times \textbf{M}) \\ \nabla \times \left(\frac{1}{\mu_0}\textbf{B} - \textbf{M}\right) &= \textbf{J}_f \end{align}$$ Define $\textbf{H}$ as $$\bbox[5px,border:2px solid #666] { \textbf{H} = \frac{1}{\mu_0}\textbf{B} - \textbf{M} }$$ Thus, Gauss's law can be written as $$\bbox[5px,border:2px solid #666] { \nabla\times\textbf{H} = \textbf{J}_f }$$ or $$\bbox[5px,border:2px solid #666] { \oint\textbf{H}\cdot d\textbf{l} = I_{f_{enc}} }$$ Note that Ampere's law is not modified, but just express in terms of $\textbf{H}$. By Ampere's law, we know that $$\textbf{H}_{above}^{\parallel} - \textbf{H}_{below}^{\parallel} = \textbf{K}_f \times \hat{\textbf{n}}$$ The divergence of $\textbf{H}$ is $$\nabla \cdot \textbf{H} = \frac{1}{\mu_0}\nabla\cdot\textbf{B} - \nabla\cdot\textbf{M} = - \nabla\cdot\textbf{M}$$ as divergence of magnetization is not necessarily zero, divergence of $\textbf{H}$ is not necessarily zero. Since $\textbf{H}$ is not divergence free, unlike magnetic field, its perpendicular component boundary condition is $$H^{\bot}_{above} - H^{\bot}_{below} = -(M^{\bot}_{above} - M^{\bot}_{below})$$

Example (Griffiths Third Edition Ex 6.2)

A long copper rod of radius $R$ carries a uniformly distributed free current $I$. Find $H$ inside and outside the rod.

Solution: For $s < R$, $$H(2\pi s) = I_{f_{\text{enc}}}=I\frac{\pi s^2}{\pi R^2}$$ so, $$\vec{H}=\frac{I}{2\pi R^2}s\hat{\phi}$$ For $s > R$, $$\vec{B}=\mu_0\vec{H}=\frac{I}{2\pi s}\hat{\phi}$$

Checkpoint (Griffiths Third Edition Q6.12)

An infinitely long cylinder, of radius $R$, carres a "frozen-in" magnetization parallel to the axis, $$\vec{M}=ks\hat{z},$$ where $k$ is a constant and $s$ is the distance from the axis; there is no free current anywhere. Find the magnetic field.

By symmetry, $\vec{H}$ points in the $z$ direction. By Ampere's law, $$\oint \vec{H}\cdot d\vec{l} = Hl = \mu_0 I_{f_{\text{enc}}}=0$$ So, $$\vec{H}=0$$ and hence $$\vec{B}=\mu_0\vec{M}$$ For $s < R$, $$\vec{B}=\mu_0 ks\hat{z}.$$ For $s > R$, $$\vec{B}=\mu_0(0)=0$$