Magnetostatics

The magnetic field due to a steady current $I$ in a wire element $d\mathbf{l}$:

$d\mathbf{B} = \frac{\mu_0 I}{4\pi} \frac{d\mathbf{l} \times \hat{\mathbf{r}}}{r^2}$

For a complete circuit:

$\mathbf{B}(\mathbf{r}) = \frac{\mu_0 I}{4\pi} \oint \frac{d\mathbf{l} \times \hat{\mathbf{r}}"}{|\mathbf{r} - \mathbf{r}'|^2}$

For steady currents ($\partial \mathbf{E} / \partial t = 0$):

$\oint_C \mathbf{B} \cdot d\mathbf{l} = \mu_0 I_{\mathrm{enc}}$

Example: Infinite straight wire carrying current $I$.

By cylindrical symmetry, $B$ is constant on circles centred on the wire. Choose an Amperian loop of Radius $r$:

$\oint \mathbf{B} \cdot d\mathbf{l} = B \cdot 2\pi r = \mu_0 I \implies B = \frac{\mu_0 I}{2\pi r}$

Example: Solenoid. For a long solenoid with $n$ turns per unit length carrying current $I$:

$B = \mu_0 n I \quad \mathrm{(inside)}, \quad B = 0 \quad \mathrm{(outside)}$

Since $\nabla \cdot \mathbf{B} = 0$We can write $\mathbf{B} = \nabla \times \mathbf{A}$Where $\mathbf{A}$ is the magnetic vector potential.

In the Coulomb gauge ($\nabla \cdot \mathbf{A} = 0$), the vector potential satisfies

$\nabla^2 \mathbf{A} = -\mu_0 \mathbf{J}$

This is Poisson’s equation for each component of $\mathbf{A}$.

For a current loop, the solution is:

$\mathbf{A}(\mathbf{r}) = \frac{\mu_0}{4\pi} \int \frac{\mathbf{J}(\mathbf{r}')}{|\mathbf{r} - \mathbf{r}'|}\, d^3\mathbf{r}'$

Example: Toroid. A toroid with $N$ turns carrying current $I$ has inner radius $a$ and outer Radius $b$.

By symmetry, $\mathbf{B}$ is tangential and constant on circular Amperian loops inside the Toroid. For a loop of radius $r$ ($a \lt r \lt b$):

$B \cdot 2\pi r = \mu_0 N I \implies B = \frac{\mu_0 N I}{2\pi r}$

For $r \lt a$ or $r \gt b$: $B = 0$ (no enclosed current).

Unlike a solenoid, the field inside a toroid is not uniform --- it varies as $1/r$.

Example: Infinite current sheet. A sheet in the $xy$-plane carries surface current density $\mathbf{K} = K\,\hat{\mathbf{x}}$.

By symmetry, $\mathbf{B}$ is parallel to $\pm\hat{\mathbf{y}}$ and depends only on $z$. Choose a rectangular Amperian loop straddling the sheet with sides parallel to $\hat{\mathbf{y}}$:

$B \cdot 2L = \mu_0 K L \implies B = \frac{\mu_0 K}{2}$

The field is uniform on each side, pointing in opposite directions:

$\mathbf{B} = \begin{cases} +\frac{\mu_0 K}{2}\,\hat{\mathbf{y}} & z \gt 0 \\[4pt] -\frac{\mu_0 K}{2}\,\hat{\mathbf{y}} & z \lt 0 \end{cases}$

A current loop carrying current $I$ enclosing area $\mathbf{a}$ has magnetic dipole moment:

$\mathbf{m} = I\mathbf{a}$

For a planar loop of $N$ turns: $\mathbf{m} = NIA\,\hat{\mathbf{n}}$Where $A$ is the area And $\hat{\mathbf{n}}$ is the unit normal given by the right-hand rule.

Field of a magnetic dipole (at position $\mathbf{r}$ from the dipole):

$\mathbf{B}_{\mathrm{dip}(\mathbf{r}) = \frac{\mu_0}{4\pi}\left[\frac{3(\mathbf{m} \cdot \hat{\mathbf{r}})\hat{\mathbf{r}} - \mathbf{m}}{r^3}\right]}$

This has the same angular structure as the electric dipole field.

Torque on a dipole in a uniform field:

$\boldsymbol{\tau} = \mathbf{m} \times \mathbf{B}$

Energy of a dipole in a field:

$U = -\mathbf{m} \cdot \mathbf{B}$

Force on a dipole in a non-uniform field:

$\mathbf{F} = \nabla(\mathbf{m} \cdot \mathbf{B})$

Starting from the Biot-Savart law and the identity $\frac{\mathbf{r} - \mathbf{r}'}{|\mathbf{r}-\mathbf{r}'|^3} = -\nabla\frac{1}{|\mathbf{r}-\mathbf{r}'|}$:

$\mathbf{B}(\mathbf{r}) = \frac{\mu_0}{4\pi}\int \mathbf{J}(\mathbf{r}') \times \frac{(\mathbf{r}-\mathbf{r}')}{|\mathbf{r}-\mathbf{r}'|^3}\,d^3\mathbf{r}' = -\frac{\mu_0}{4\pi}\int \mathbf{J}(\mathbf{r}') \times \nabla\frac{1}{|\mathbf{r}-\mathbf{r}'|}\,d^3\mathbf{r}'$

Using the product rule $\mathbf{J} \times (\nabla f) = \nabla \times (f\mathbf{J}) - f(\nabla \times \mathbf{J})$ And noting that $\nabla \times \mathbf{J}(\mathbf{r}') = 0$ (since $\mathbf{J}$ depends on $\mathbf{r}'$Not $\mathbf{r}$):

$\mathbf{B}(\mathbf{r}) = \frac{\mu_0}{4\pi}\nabla \times \int \frac{\mathbf{J}(\mathbf{r}')}{|\mathbf{r}-\mathbf{r}'|}\,d^3\mathbf{r}'$

Comparing with $\mathbf{B} = \nabla \times \mathbf{A}$:

$\mathbf{A}(\mathbf{r}) = \frac{\mu_0}{4\pi}\int \frac{\mathbf{J}(\mathbf{r}')}{|\mathbf{r}-\mathbf{r}'|}\,d^3\mathbf{r}'$

This is the general solution for the vector potential in the Coulomb gauge. For a line current:

$\mathbf{A}(\mathbf{r}) = \frac{\mu_0 I}{4\pi}\oint \frac{d\mathbf{l}'}{|\mathbf{r}-\mathbf{r}'|}$

Magnetization. The magnetization $\mathbf{M}$ is the magnetic dipole moment per unit volume. It produces bound currents:

$\mathbf{J}_b = \nabla \times \mathbf{M}, \quad \mathbf{K}_b = \mathbf{M} \times \hat{\mathbf{n}}$

The H field (magnetic field intensity) is defined as:

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

Ampere’s law for $\mathbf{H}$:

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

$\oint \mathbf{H} \cdot d\mathbf{l} = I_{f,\mathrm{enc}}$

This is simpler than Ampere’s law for $\mathbf{B}$ because only free currents appear.

Linear magnetic materials. For isotropic linear materials:

$\mathbf{M} = \chi_m \mathbf{H}, \quad \mathbf{B} = \mu \mathbf{H}$

Where $\chi_m$ is the magnetic susceptibility and $\mu = \mu_0(1 + \chi_m)$ is the permeability. The relative permeability is $\mu_r = 1 + \chi_m$.

Diamagnetic materials ($\chi_m \lt 0$, $\lvert\chi_m\rvert \ll 1$): Weakly repelled by Magnetic fields. The induced magnetization opposes the applied field (Lenz’s law at the Atomic level). Examples: bismuth, copper, water.

Paramagnetic materials ($\chi_m \gt 0$, $\chi_m \ll 1$): Weakly attracted by magnetic fields. Atomic dipoles align partially with the applied field. Examples: aluminium, platinum, oxygen.

Ferromagnetic materials ($\chi_m \gg 1$): Strongly attracted by magnetic fields. Exhibit hysteresis: the magnetization depends on the history of the applied field.

The hysteresis loop traces $\mathbf{B}$ vs $\mathbf{H}$ as the external field cycles. Key Features:

• Remanence $B_r$: the residual field when $H = 0$.
• Coercivity $H_c$: the field required to demagnetize the material.
• Saturation: the maximum magnetization achievable.

For soft ferromagnets (iron, nickel), $H_c$ is small and the hysteresis loop is narrow. For hard ferromagnets (permanent magnets), $H_c$ is large.

:::caution Common Pitfall The magnetic field $\mathbf{B}$ is the fundamental quantity; $\mathbf{H}$ is an auxiliary field Convenient for problems with free currents. The names “magnetic field” and “magnetic field Intensity” vary across textbooks --- always check which symbol a given text associates with Which name. In this document, $\mathbf{B}$ is the magnetic field and $\mathbf{H}$ is the Auxiliary H field.

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