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 \(s > R\), $$\vec{B}=\mu_0(0)=0$$