Magnetic Materials

Step 1: Vector potential of a single magnetic dipole

A small volume element $d\tau'$ at position $\mathbf{r}'$ with magnetization $\mathbf{M}$ has dipole moment $d\mathbf{m} = \mathbf{M}(\mathbf{r}')\,d\tau'$. Its contribution to $\mathbf{A}$ is:

$$d\mathbf{A} = \frac{\mu_0}{4\pi}\frac{\mathbf{M}(\mathbf{r}') \times \hat{\mathscr{r}}}{\mathscr{r}^2}\,d\tau'$$

Step 2: Total vector potential of the magnetized body

Integrating over the entire volume $V$ and using $\hat{\mathscr{r}}/\mathscr{r}^2 = \nabla'(1/\mathscr{r})$:

$$\mathbf{A}(\mathbf{r}) = \frac{\mu_0}{4\pi}\int_V \mathbf{M}(\mathbf{r}') \times \nabla'\!\left(\frac{1}{\mathscr{r}}\right)d\tau'$$

Step 3: Integrate by parts using vector identity

Using the product rule $\mathbf{M} \times \nabla'(1/\mathscr{r}) = \frac{1}{\mathscr{r}}(\nabla' \times \mathbf{M}) - \nabla' \times (\mathbf{M}/\mathscr{r})$:

$$\mathbf{A} = \frac{\mu_0}{4\pi}\left[\int_V \frac{\nabla' \times \mathbf{M}(\mathbf{r}')}{\mathscr{r}}\,d\tau' - \int_V \nabla' \times \left(\frac{\mathbf{M}}{\mathscr{r}}\right)d\tau'\right]$$

Step 4: Convert the second integral to a surface integral

By the curl theorem (generalization of Stokes' theorem), $\int_V \nabla' \times \mathbf{F}\,d\tau' = -\oint_S \mathbf{F} \times d\mathbf{a}'$:

$$\int_V \nabla' \times \left(\frac{\mathbf{M}}{\mathscr{r}}\right)d\tau' = -\oint_S \frac{\mathbf{M} \times \hat{n}'}{\mathscr{r}}\,da'$$

Step 5: Combine into the final expression

$$\mathbf{A} = \frac{\mu_0}{4\pi}\int_V \frac{\nabla' \times \mathbf{M}}{\mathscr{r}}\,d\tau' + \frac{\mu_0}{4\pi}\oint_S \frac{\mathbf{M} \times \hat{n}'}{\mathscr{r}}\,da'$$

Step 6: Identify the bound currents

The first integral has the form of $\mathbf{A}$ due to a volume current, the second has the form due to a surface current. By comparison with $\mathbf{A} = \frac{\mu_0}{4\pi}\int \frac{\mathbf{J}}{\mathscr{r}}d\tau'$:

$$\boxed{\mathbf{J}_b = \nabla \times \mathbf{M}} \qquad \boxed{\mathbf{K}_b = \mathbf{M} \times \hat{n}}$$

Step 7: Physical interpretation

If $\mathbf{M}$ is uniform, $\nabla \times \mathbf{M} = 0$ and there is no bulk bound current -- only a surface current $\mathbf{K}_b = \mathbf{M} \times \hat{n}$. This makes physical sense: neighboring atomic current loops cancel in the interior of a uniformly magnetized material, leaving only a net surface current.

Step 1: Ampere's law with all currents

In the presence of magnetic materials, the total current density has free and bound parts:

$$\nabla \times \mathbf{B} = \mu_0 \mathbf{J}_{\text{total}} = \mu_0(\mathbf{J}_f + \mathbf{J}_b)$$

Step 2: Substitute the bound current expression

Using $\mathbf{J}_b = \nabla \times \mathbf{M}$:

$$\nabla \times \mathbf{B} = \mu_0(\mathbf{J}_f + \nabla \times \mathbf{M})$$

Step 3: Rearrange to isolate the free current

Move the magnetization term to the left side:

$$\nabla \times \mathbf{B} - \mu_0\nabla \times \mathbf{M} = \mu_0 \mathbf{J}_f$$

$$\nabla \times \left(\frac{\mathbf{B}}{\mu_0} - \mathbf{M}\right) = \mathbf{J}_f$$

Step 4: Define the H field

Define the auxiliary field:

$$\boxed{\mathbf{H} \equiv \frac{1}{\mu_0}\mathbf{B} - \mathbf{M}}$$

Step 5: Ampere's law in terms of H

The curl equation becomes elegantly simple:

$$\boxed{\nabla \times \mathbf{H} = \mathbf{J}_f}$$

Step 6: Integral form

Applying Stokes' theorem:

$$\oint \mathbf{H}\cdot d\boldsymbol{\ell} = I_{f,\text{enc}}$$

This is the same form as the original Ampere's law but with $\mathbf{H}$ replacing $\mathbf{B}/\mu_0$ and only free currents on the right side. The practical advantage: in many problems we know $\mathbf{J}_f$ but not $\mathbf{J}_b$.

Step 7: The divergence of H

Note that $\nabla \cdot \mathbf{H} = \frac{1}{\mu_0}\nabla \cdot \mathbf{B} - \nabla \cdot \mathbf{M} = -\nabla \cdot \mathbf{M}$. Unlike $\mathbf{B}$, the field $\mathbf{H}$ is generally not divergence-free. This means $\mathbf{H}$ field lines can begin and end at points where $\nabla \cdot \mathbf{M} \neq 0$.

Step 1: Define the magnetic susceptibility

For a linear, isotropic magnetic material, the magnetization is proportional to $\mathbf{H}$:

$$\mathbf{M} = \chi_m \mathbf{H}$$

where $\chi_m$ is the dimensionless magnetic susceptibility ($\chi_m < 0$ for diamagnets, $\chi_m > 0$ for paramagnets).

Step 2: Substitute into the definition of H

From $\mathbf{H} = \mathbf{B}/\mu_0 - \mathbf{M}$, solve for $\mathbf{B}$:

$$\mathbf{B} = \mu_0(\mathbf{H} + \mathbf{M}) = \mu_0(\mathbf{H} + \chi_m\mathbf{H})$$

Step 3: Factor out H

$$\mathbf{B} = \mu_0(1 + \chi_m)\mathbf{H}$$

Step 4: Define the relative permeability

The relative permeability is $\mu_r = 1 + \chi_m$, and the permeability is $\mu = \mu_0\mu_r = \mu_0(1 + \chi_m)$:

$$\boxed{\mathbf{B} = \mu\mathbf{H} = \mu_0\mu_r\mathbf{H} = \mu_0(1+\chi_m)\mathbf{H}}$$

Step 5: Express M in terms of B

From $\mathbf{M} = \chi_m\mathbf{H} = \chi_m\mathbf{B}/\mu$, or equivalently:

$$\mathbf{M} = \frac{\chi_m}{\mu_0(1+\chi_m)}\mathbf{B} = \frac{\chi_m}{\mu_0\mu_r}\mathbf{B}$$

Step 6: Summary of material classifications

Vacuum: $\chi_m = 0$, $\mu_r = 1$, $\mu = \mu_0$. Diamagnets: $\chi_m \sim -10^{-5}$, $\mu_r \lesssim 1$. Paramagnets: $\chi_m \sim 10^{-4}$, $\mu_r \gtrsim 1$. Ferromagnets: $\chi_m \sim 10^3$ to $10^5$, $\mu_r \gg 1$. Note: ferromagnets are nonlinear, so these linear relations are only approximate.

Step 1: Normal component of B -- set up a Gaussian pillbox

Consider a thin pillbox of cross-sectional area $A$ and infinitesimal height $\epsilon$ straddling the interface. Apply the divergence theorem to $\nabla \cdot \mathbf{B} = 0$:

$$\oint_S \mathbf{B}\cdot d\mathbf{a} = 0$$

Step 2: Evaluate the surface integral

As $\epsilon \to 0$, the curved side contributes nothing. The top face gives $B_{\text{above}}^\perp \cdot A$ and the bottom face gives $-B_{\text{below}}^\perp \cdot A$ (outward normals):

$$B_{\text{above}}^\perp A - B_{\text{below}}^\perp A = 0$$

Step 3: Normal B boundary condition

$$\boxed{B_{\text{above}}^\perp = B_{\text{below}}^\perp}$$

The normal component of $\mathbf{B}$ is always continuous across any interface.

Step 4: Tangential component of H -- set up an Amperian loop

Consider a thin rectangular loop of width $\ell$ and infinitesimal height $\epsilon$ straddling the interface. Apply Stokes' theorem to $\nabla \times \mathbf{H} = \mathbf{J}_f$:

$$\oint_C \mathbf{H}\cdot d\boldsymbol{\ell} = \int_S \mathbf{J}_f \cdot d\mathbf{a} = I_{f,\text{enc}}$$

Step 5: Evaluate the line integral

As $\epsilon \to 0$, the vertical sides contribute nothing. The top side gives $H_{\text{above}}^\parallel \cdot \ell$ and the bottom gives $-H_{\text{below}}^\parallel \cdot \ell$. The enclosed free current is $K_f \cdot \ell$ if there is a free surface current $\mathbf{K}_f$:

$$H_{\text{above}}^\parallel \ell - H_{\text{below}}^\parallel \ell = K_f \ell$$

Step 6: Tangential H boundary condition

In vector form, where $\hat{n}$ points from below to above:

$$\boxed{H_{\text{above}}^\parallel - H_{\text{below}}^\parallel = K_f} \qquad \text{or equivalently} \qquad \hat{n} \times (\mathbf{H}_{\text{above}} - \mathbf{H}_{\text{below}}) = \mathbf{K}_f$$

If there is no free surface current ($K_f = 0$), the tangential component of $\mathbf{H}$ is continuous.

Step 7: Consequences for linear media

At an interface between two linear media ($\mu_1$ and $\mu_2$) with no free surface current:

$$\frac{B_1^\perp}{\mu_1}\tan\theta_1 = H_1^\parallel = H_2^\parallel = \frac{B_2^\perp}{\mu_2}\tan\theta_2$$

Since $B_1^\perp = B_2^\perp$, this gives the refraction law for magnetic field lines: $\frac{\tan\theta_1}{\tan\theta_2} = \frac{\mu_1}{\mu_2}$.

Step 1: Work to establish current in an inductor

The power delivered against the back-EMF $\mathcal{E} = -L\,dI/dt$ of an inductor is $P = -\mathcal{E}I = LI\,dI/dt$. The total work to build up current from 0 to $I$:

$$W = \int_0^I LI'\,dI' = \frac{1}{2}LI^2$$

Step 2: Express in terms of flux and A

Since $LI = \Phi = \int \mathbf{B}\cdot d\mathbf{a} = \int (\nabla \times \mathbf{A})\cdot d\mathbf{a} = \oint \mathbf{A}\cdot d\boldsymbol{\ell}$:

$$W = \frac{1}{2}I\oint \mathbf{A}\cdot d\boldsymbol{\ell}$$

Step 3: Generalize to a volume current distribution

Replace $I\,d\boldsymbol{\ell} \to \mathbf{J}\,d\tau$:

$$\boxed{W = \frac{1}{2}\int \mathbf{A}\cdot\mathbf{J}\,d\tau}$$

Step 4: Substitute Ampere's law for J

Use $\mathbf{J} = \frac{1}{\mu_0}\nabla \times \mathbf{B}$ (in free space):

$$W = \frac{1}{2\mu_0}\int \mathbf{A}\cdot(\nabla \times \mathbf{B})\,d\tau$$

Step 5: Integrate by parts using vector identity

Apply the product rule $\mathbf{A}\cdot(\nabla \times \mathbf{B}) = \mathbf{B}\cdot(\nabla \times \mathbf{A}) - \nabla \cdot (\mathbf{A} \times \mathbf{B})$:

$$W = \frac{1}{2\mu_0}\left[\int \mathbf{B}\cdot(\nabla \times \mathbf{A})\,d\tau - \int \nabla \cdot (\mathbf{A}\times\mathbf{B})\,d\tau\right]$$

Step 6: The surface term vanishes

By the divergence theorem, the second integral becomes a surface integral at infinity. Since $\mathbf{A} \sim 1/r^2$ and $\mathbf{B} \sim 1/r^3$ for a localized source, while $da \sim r^2$, the integrand falls off as $1/r^3 \to 0$:

$$\oint_S (\mathbf{A}\times\mathbf{B})\cdot d\mathbf{a} \to 0 \quad \text{as } S \to \infty$$

Step 7: Use B = curl A to get the field energy

Since $\nabla \times \mathbf{A} = \mathbf{B}$:

$$\boxed{W = \frac{1}{2\mu_0}\int B^2\,d\tau}$$

The energy density in the magnetic field is $u_B = \frac{1}{2\mu_0}B^2$. In a linear medium with permeability $\mu$, this generalizes to $u_B = \frac{1}{2}\mathbf{B}\cdot\mathbf{H} = \frac{B^2}{2\mu}$.