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Magnetostatics | Physics - Wyatt's Notes

3.1 The Biot-Savart Law

Section titled “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

Section titled “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

Section titled “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

Section titled “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

Section titled “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

Section titled “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

Section titled “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

Section titled “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.

flowchart TD
    A[3_Magnetostatics] --> B[Key Concepts]
    A --> C[Core Principles]
    A --> D[Practical Applications]
    B --> E[Fundamental definitions]
    C --> F[Design patterns]
    D --> G[Real-world usage]

Intuition

Section titled “Intuition”

Magnetic fields are the universe’s way of pushing things sideways. Unlike electric fields that push along the field line, magnetic forces always act perpendicular to motion, which means they can change the direction of a moving charge but never its speed. The Biot-Savart law is like measuring the magnetic footprint of every tiny current element and adding them up. Ampere’s law is the magnetic equivalent of Gauss’s law: draw a loop around a current, and the magnetic field integrated along that loop tells you how much current passes through. The vector potential is a mathematical shortcut that simplifies calculations, even though it is not directly measurable. Ferromagnetism is like a crowd of tiny compass needles that all want to point the same way: once aligned, they stay aligned even after the external field is removed, which is why permanent magnets exist.

3.9 Common Mistakes

Section titled “3.9 Common Mistakes”

Mistake 1: Confusing the Biot-Savart law with Ampere’s law The Biot-Savart law gives the magnetic field from any current distribution by direct integration, while Ampere’s law relates the line integral of B\mathbf{B} to the enclosed current. Ampere’s law is easier to use when high symmetry exists (infinite wire, solenoid), but the Biot-Savart law is needed for finite or asymmetric configurations. Do not apply Ampere’s law without verifying cylindrical or planar symmetry.

Mistake 2: Misapplying the right-hand rule The right-hand rule for the magnetic field of a current element states that dBd\mathbf{B} is in the direction of dl×r^d\mathbf{l} \times \hat{\mathbf{r}}. Students often reverse the direction by curling the fingers in the wrong direction or using the left hand. For a straight wire, curl your right-hand fingers around the wire with your thumb pointing in the current direction; your fingers point in the direction of B\mathbf{B}.

Mistake 3: Assuming B\mathbf{B} is always parallel to H\mathbf{H} In linear magnetic materials, B=μH\mathbf{B} = \mu \mathbf{H}, so they are parallel. But in ferromagnetic materials, the relationship is nonlinear and hysteretic: B\mathbf{B} depends on the history of H\mathbf{H}. The H\mathbf{H} field is defined as H=B/μ0−M\mathbf{H} = \mathbf{B}/\mu_0 - \mathbf{M}, and it is the auxiliary field that simplifies problems with free currents, not a fundamental field.

Cross-References

Section titled “Cross-References”