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Wyatt's Notes — University

Magnetostatics

3.1 The Biot-Savart Law

The magnetic field due to a steady current II in a wire element dld\mathbf{l}:

dB=μ0I4πdl×r^r2d\mathbf{B} = \frac{\mu_0 I}{4\pi} \frac{d\mathbf{l} \times \hat{\mathbf{r}}}{r^2}

For a complete circuit:

B(r)=μ0I4π∮dl×r^"∣r−r′∣2\mathbf{B}(\mathbf{r}) = \frac{\mu_0 I}{4\pi} \oint \frac{d\mathbf{l} \times \hat{\mathbf{r}}"}{|\mathbf{r} - \mathbf{r}'|^2}

3.2 Ampere’s Law

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

∮CB⋅dl=μ0Ienc\oint_C \mathbf{B} \cdot d\mathbf{l} = \mu_0 I_{\mathrm{enc}}

Example: Infinite straight wire carrying current II.

By cylindrical symmetry, BB is constant on circles centred on the wire. Choose an Amperian loop of Radius rr:

∮B⋅dl=B⋅2πr=μ0I  ⟹  B=μ0I2πr\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 nn turns per unit length carrying current II:

B=μ0nI(inside),B=0(outside)B = \mu_0 n I \quad \mathrm{(inside)}, \quad B = 0 \quad \mathrm{(outside)}

3.3 Magnetic Vector Potential

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

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

∇2A=−μ0J\nabla^2 \mathbf{A} = -\mu_0 \mathbf{J}

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

For a current loop, the solution is:

A(r)=μ04π∫J(r′)∣r−r′∣ d3r′\mathbf{A}(\mathbf{r}) = \frac{\mu_0}{4\pi} \int \frac{\mathbf{J}(\mathbf{r}')}{|\mathbf{r} - \mathbf{r}'|}\, d^3\mathbf{r}'

3.4 Additional Ampere’s Law Examples

Example: Toroid. A toroid with NN turns carrying current II has inner radius aa and outer Radius bb.

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

B⋅2πr=μ0NI  ⟹  B=μ0NI2πrB \cdot 2\pi r = \mu_0 N I \implies B = \frac{\mu_0 N I}{2\pi r}

For r<ar \lt a or r>br \gt b: B=0B = 0 (no enclosed current).

Unlike a solenoid, the field inside a toroid is not uniform --- it varies as 1/r1/r.

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

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

B⋅2L=μ0KL  ⟹  B=μ0K2B \cdot 2L = \mu_0 K L \implies B = \frac{\mu_0 K}{2}

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

B={+μ0K2 y^z>0−μ0K2 y^z<0\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}

3.5 Magnetic Dipole Moment

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

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

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

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

Bdip(r)=μ04π[3(m⋅r^)r^−mr3]\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:

τ=m×B\boldsymbol{\tau} = \mathbf{m} \times \mathbf{B}

Energy of a dipole in a field:

U=−m⋅BU = -\mathbf{m} \cdot \mathbf{B}

Force on a dipole in a non-uniform field:

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

Example: Field on the axis of a circular loop

A circular loop of radius RR carries current II. On the axis at distance zz from the centre, Every element dld\mathbf{l} is perpendicular to r^\hat{\mathbf{r}}So:

dB=μ0I4πdlR2+z2d\mathbf{B} = \frac{\mu_0 I}{4\pi}\frac{dl}{R^2 + z^2}

The component perpendicular to the axis cancels by symmetry. The axial component is:

Bz=∮dB sin⁡α=μ0I4π(R2+z2)RR2+z2∮dl=μ0IR22(R2+z2)3/2B_z = \oint dB\,\sin\alpha = \frac{\mu_0 I}{4\pi(R^2+z^2)}\frac{R}{\sqrt{R^2+z^2}}\oint dl = \frac{\mu_0 I R^2}{2(R^2+z^2)^{3/2}}

For z≫Rz \gg R: Bz≈μ0IR22z3=μ04π2mz3B_z \approx \frac{\mu_0 I R^2}{2z^3} = \frac{\mu_0}{4\pi}\frac{2\mathbf{m}}{z^3} Which matches the dipole formula with m=IπR2 z^\mathbf{m} = I\pi R^2\,\hat{\mathbf{z}}. ■\blacksquare

3.6 Vector Potential: Detailed Derivation

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

B(r)=μ04π∫J(r′)×(r−r′)∣r−r′∣3 d3r′=−μ04π∫J(r′)×∇1∣r−r′∣ d3r′\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 J×(∇f)=∇×(fJ)−f(∇×J)\mathbf{J} \times (\nabla f) = \nabla \times (f\mathbf{J}) - f(\nabla \times \mathbf{J}) And noting that ∇×J(r′)=0\nabla \times \mathbf{J}(\mathbf{r}') = 0 (since J\mathbf{J} depends on r′\mathbf{r}'Not r\mathbf{r}):

B(r)=μ04π∇×∫J(r′)∣r−r′∣ d3r′\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 B=∇×A\mathbf{B} = \nabla \times \mathbf{A}:

A(r)=μ04π∫J(r′)∣r−r′∣ d3r′\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:

A(r)=μ0I4π∮dl′∣r−r′∣\mathbf{A}(\mathbf{r}) = \frac{\mu_0 I}{4\pi}\oint \frac{d\mathbf{l}'}{|\mathbf{r}-\mathbf{r}'|}

Example: Vector potential of an infinite wire

An infinite straight wire along the zz-axis carries current II. In cylindrical coordinates (s,ϕ,z)(s, \phi, z)The vector potential can only depend on ss by symmetry, and must point along z^\hat{\mathbf{z}}.

A(s)=μ0I4π∫−∞∞dz′s2+z′2 z^\mathbf{A}(s) = \frac{\mu_0 I}{4\pi}\int_{-\infty}^{\infty}\frac{dz'}{\sqrt{s^2 + z'^2}}\,\hat{\mathbf{z}}

This integral diverges logarithmically. Introduce a cutoff at z′=±Lz' = \pm L:

A(s)≈μ0I2πln⁡ ⁣(2Ls)z^+const\mathbf{A}(s) \approx \frac{\mu_0 I}{2\pi}\ln\!\left(\frac{2L}{s}\right)\hat{\mathbf{z}} + \mathrm{const}

Since A\mathbf{A} is defined only up to a gauge transformation, we write:

A(s)=−μ0I2πln⁡ ⁣(ss0)z^\mathbf{A}(s) = -\frac{\mu_0 I}{2\pi}\ln\!\left(\frac{s}{s_0}\right)\hat{\mathbf{z}}

Verify: B=∇×A=−∂Az∂s ϕ^=μ0I2πs ϕ^\mathbf{B} = \nabla \times \mathbf{A} = -\frac{\partial A_z}{\partial s}\,\hat{\boldsymbol{\phi}} = \frac{\mu_0 I}{2\pi s}\,\hat{\boldsymbol{\phi}}. This matches the Ampere’s law result. ■\blacksquare

3.7 Magnetization and the H Field

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

Jb=∇×M,Kb=M×n^\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:

H=1μ0B−M\mathbf{H} = \frac{1}{\mu_0}\mathbf{B} - \mathbf{M}

Ampere’s law for H\mathbf{H}:

∇×H=Jf\nabla \times \mathbf{H} = \mathbf{J}_f

∮H⋅dl=If,enc\oint \mathbf{H} \cdot d\mathbf{l} = I_{f,\mathrm{enc}}

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

Linear magnetic materials. For isotropic linear materials:

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

Where χm\chi_m is the magnetic susceptibility and μ=μ0(1+χm)\mu = \mu_0(1 + \chi_m) is the permeability. The relative permeability is μr=1+χm\mu_r = 1 + \chi_m.

3.8 Magnetic Materials

Diamagnetic materials (χm<0\chi_m \lt 0, ∣χm∣≪1\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 (χm>0\chi_m \gt 0, χm≪1\chi_m \ll 1): Weakly attracted by magnetic fields. Atomic dipoles align partially with the applied field. Examples: aluminium, platinum, oxygen.

Ferromagnetic materials (χm≫1\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 B\mathbf{B} vs H\mathbf{H} as the external field cycles. Key Features:

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

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

:::caution Common Pitfall The magnetic field B\mathbf{B} is the fundamental quantity; H\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, B\mathbf{B} is the magnetic field and H\mathbf{H} is the Auxiliary H field.

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