Magnetostatics

Biot-Savart's Law¶

Sources

\begin{matrix} \vec{u} d Q = \frac{d \vec{ℓ}}{d t} Q = I d \vec{ℓ} \end{matrix}

Magnetic flux density:

\begin{matrix} d \vec{B} = \frac{μ_{o}}{4 π} \frac{I d \vec{ℓ} \times \hat{a_{r}}}{R^{2}} \end{matrix}

Magnetic field density:

\begin{matrix} d \vec{H} = \frac{d \vec{B}}{μ_{o}} = \frac{1}{4 π} \frac{I d \vec{ℓ} \times \hat{a_{r}}}{R^{2}} \end{matrix}

Magnetic force

\begin{aligned} d \vec{F} & = I d \vec{ℓ} \times d \vec{B} \\ = I d \vec{ℓ} \times (\frac{μ_{o}}{4 π} \frac{I d \vec{ℓ} \times \hat{a_{r}}}{R^{2}}) \end{aligned}

Comparison

\begin{matrix} d \vec{F} = q \vec{E} \end{matrix}

Ampere's Law¶

original Ampere's law 👎

\begin{matrix} \oint_{C} \vec{H} \cdot d \vec{ℓ} = \int_{S} \vec{J} \cdot d \vec{S} = I_{e n c} \\ \nabla \times \vec{H} = \vec{J} \end{matrix}

Ampere-Maxwell equation 👍

\begin{matrix} \oint_{C} \vec{H} \cdot d \vec{ℓ} = \int_{S} ({\vec{J}}_{f} + \frac{\partial \vec{D}}{\partial t}) \cdot d \vec{S} \\ \nabla \times \vec{H} = {\vec{J}}_{f} + \frac{\partial \vec{D}}{\partial t} \end{matrix}

displacement current

for the identity

\begin{matrix} \nabla \cdot (\nabla \times \vec{H}) = 0 \\ \nabla \cdot (\nabla \times \vec{H}) = \nabla \cdot \vec{J} \neq 0 when \nabla \cdot \vec{J} = - \frac{\partial ρ_{v}}{\partial t} \neq 0 \end{matrix}

thus, let's write

\begin{matrix} \nabla \times \vec{H} = \vec{J} + {\vec{J}}_{d} \end{matrix}

then

\begin{matrix} \nabla \cdot (\nabla \times \vec{H}) = \nabla \cdot \vec{J} + \nabla \cdot {\vec{J}}_{d} = 0 \\ ⟹ \nabla \cdot {\vec{J}}_{d} = \frac{\partial ρ_{v}}{\partial t} = \frac{\partial}{\partial t} (\nabla \cdot \vec{D}) \\ ⟹ {\vec{J}}_{d} = \frac{\partial \vec{D}}{\partial t} \end{matrix}

displacement current $I_{d}$ <br ${\vec{J}}_{d}$ is called displacement current density, and the displacement current $I_{d}$ id defined by <br

\begin{matrix} I_{d} = \int_{S} {\vec{J}}_{d} \cdot d \vec{S} \end{matrix}

for $I_{d} \neq 0$ whenever there is an accumulation of charges, or the imginary current inside capacitor. <br e.g. semi-infinite conductive wire.

Gauss' Law¶

\begin{matrix} \oint_{S} \vec{B} \cdot d \vec{S} = \int_{V} \nabla \cdot \vec{B} = 0 \end{matrix}

For it exists not isolated magnetic poles, Guass' law here in magnetism is pretty intuitive.

Faraday's law¶

static electric field $E_{e}$
transformer-induced electric field $E_{T}$
motion-induced electric field $E_{m}$
induced electric field $E_{i n d} \equiv E_{T} + E_{m}$

\begin{aligned} V_{e m f} & = \oint_{L} (E_{e} + E_{T} + E_{m}) \\ = 0 + (- \int_{S} \frac{\partial \vec{B}}{\partial t} \cdot d \vec{S}) + \oint_{L} (\vec{u} \times \vec{B}) \cdot d \vec{ℓ} \\ = - \frac{d}{d t} \int \vec{B} \cdot d \vec{S} \end{aligned}

differential form

\begin{matrix} \nabla \times \vec{E} = - \frac{\partial \vec{B}}{\partial t} + \nabla \times (\vec{u} \times \vec{B}) \\ ⟺ \nabla \times (\vec{E} - \vec{u} \times \vec{B}) = - \frac{\partial \vec{B}}{\partial t} \end{matrix}

Magnetic Potentials¶

Magnetic Scalar Potential¶

magnetic scalar potential $V_{m}$ (in $A$ )

Ignoring displacement current ${\vec{J}}_{d}$ , if $\vec{J} = 0$ , then

\begin{matrix} \nabla \times \vec{H} = 0 \\ \vec{H} = - \nabla V_{m} \end{matrix}

comparison

\begin{matrix} \frac{1}{ε} \vec{D} = - \nabla V \end{matrix}

Magnetic Vector Potential¶

magnetic vector potential $\vec{A}$ (Wb/m)

since

\begin{matrix} \nabla \cdot \vec{B} = 0 \end{matrix}

and recall the identity

\begin{matrix} \nabla \cdot \nabla \times \vec{A} = 0 \end{matrix}

then we can write

\begin{matrix} \vec{B} = \nabla \times \vec{A} \end{matrix}

by Stokes' theorem

\begin{matrix} ψ = \int_{S} \vec{B} \cdot d \vec{S} = \int_{S} (\nabla \times \vec{A}) \cdot d \vec{S} = \oint_{L} \vec{A} \cdot d \vec{ℓ} \end{matrix}

and just remember that

\begin{matrix} \vec{A} = \int_{Q} \frac{μ}{4 π} \frac{\vec{u} d q}{R} \end{matrix}

line current

\begin{matrix} \vec{u} d q = I d \vec{ℓ} \end{matrix}

surface current

\begin{matrix} \vec{u} d q = \vec{K} d S \end{matrix}

volume current

\begin{matrix} \vec{u} d q = \vec{J} d v \end{matrix}

Comparison

\begin{matrix} V = \int_{Q} \frac{1}{4 π ε} \frac{d q}{R} \end{matrix}

Time-Varying Potentials¶

Easy but tedious derivation

$\nabla \times \vec{E}$

\begin{matrix} \nabla \times \vec{E} = - \frac{\partial (\nabla \times \vec{A})}{\partial t} = - \nabla \times \frac{\partial \vec{A}}{\partial t} \\ ⟹ \nabla \times (\vec{E} + \frac{\partial \vec{A}}{\partial t}) = 0 ⟺ \nabla \times (- \nabla V) = 0 \\ ⟹ \vec{E} = - \frac{\partial \vec{A}}{\partial t} - \nabla V \end{matrix}

$\nabla \cdot \vec{D}$

\begin{matrix} \nabla \cdot \vec{E} = \frac{ρ_{v}}{ε} \\ ⟹ \nabla^{2} V + \frac{\partial}{\partial t} (\nabla \cdot \vec{A}) = - \frac{ρ_{v}}{ε} \end{matrix}

$\nabla \times \vec{H}$

\begin{matrix} \nabla \times \vec{H} = \vec{J} + ε \frac{\partial \vec{E}}{\partial t} \\ \nabla \times (\frac{\nabla \times A}{μ}) = \vec{J} + ε \frac{\partial \vec{E}}{\partial t} \\ \nabla^{2} \vec{A} = \nabla (\nabla \cdot \vec{A}) - \nabla \times \nabla \times \vec{A} \\ \nabla^{2} \vec{A} = \nabla (\nabla \cdot \vec{A}) - μ \vec{J} - μ ε \frac{\partial \vec{E}}{\partial t} \\ \nabla^{2} \vec{A} - \nabla (\nabla \cdot \vec{A}) = - μ \vec{J} + μ ε \nabla (\frac{\partial V}{\partial t}) + μ ε \frac{\partial^{2} \vec{A}}{\partial t^{2}} \end{matrix}

coupled equations for $V$ and $\vec{A}$ , (normally given $ρ_{v}$ and $\vec{J}$ )

{\begin{aligned} \nabla^{2} V + \frac{\partial}{\partial t} (\nabla \cdot \vec{A}) = - \frac{ρ_{v}}{ε} \\ \nabla^{2} \vec{A} - \nabla (\nabla \cdot \vec{A}) = - μ \vec{J} + μ ε \nabla (\frac{\partial V}{\partial t}) + μ ε \frac{\partial^{2} \vec{A}}{\partial t^{2}} \end{aligned}

decoupling: Lorenz condition
eliminate $μ ε \nabla (\frac{\partial}{\partial t} V)$ term, by let

\begin{matrix} \nabla \cdot \vec{A} = - μ ε \frac{\partial V}{\partial t} \end{matrix}

Coulomb condition
let

\begin{matrix} \nabla \cdot \vec{A} = 0 \end{matrix}

Magnetic Moment¶

magnetic dipole

\begin{matrix} \vec{m} = I_{b} \vec{S} \end{matrix}

magnetization vector

\begin{matrix} \vec{M} = lim_{Δ v \to 0} \frac{\sum {\vec{m}}_{k}}{Δ v} \\ ⟹ d \vec{m} = \vec{M} d v \end{matrix}

Magnetization Current¶

bound volume current density

\begin{matrix} {\vec{J}}_{b} = \nabla \times \vec{M} \end{matrix}

Comparison

In a space $\vec{M} = 0$ (with free current only)

\begin{matrix} \nabla \times \vec{H} = {\vec{J}}_{f} \end{matrix}

bound surface current density

\begin{matrix} {\vec{K}}_{b} = \vec{M} \times \hat{a_{n}} \end{matrix}

the equation can be intuitively come up with by

Permeability¶

In material $\vec{M} \neq 0$

\begin{matrix} \nabla \times \frac{\vec{B}}{μ_{0}} & = & {\vec{J}}_{f} & + {\vec{J}}_{b} & = \vec{J} \\ = & \nabla \times \vec{H} & + \nabla \times \vec{M} \end{matrix}

\begin{matrix} ⟹ \vec{B} = μ_{0} (\vec{H} + \vec{M}) \end{matrix}

For linear Model

\begin{matrix} \vec{M} = χ_{m} \vec{H} \end{matrix}

\begin{aligned} ⟹ \vec{B} & = μ_{0} (1 + χ_{m}) \vec{H} \\ = μ_{0} μ_{r} \vec{H} \\ = μ \vec{H} \end{aligned}

Comparison

\begin{matrix} \vec{D} = ε_{0} \vec{E} + \vec{P} \end{matrix}

Boundary Condition¶

by Guass' law

\begin{matrix} \nabla \cdot \vec{B} = 0 \\ ⟹ {\vec{B}}_{1 n} = {\vec{B}}_{2 n} \end{matrix}

by Amp.'s law

\begin{matrix} {\vec{H}}_{1 t ⊥ S} - {\vec{H}}_{2 t ⊥ S} = \vec{K} \end{matrix}

Inductance¶

Definition in electric circuit

\begin{matrix} v = L \frac{d i}{d t} \end{matrix}

Magnetic Flux Linkage¶

magnetic flux linkage $λ$

\begin{matrix} d λ = N d ψ = \frac{I_{i n t}}{I_{e x t}} d ψ \end{matrix}

Self Inductance¶

\begin{matrix} L = \frac{λ}{I} = \frac{N \int_{S} \vec{B} \cdot d \vec{S}}{I} \end{matrix}

Mutual Inductance¶

\begin{matrix} M_{12} = \frac{λ_{12}}{I_{2}} = \frac{N_{1} \int_{S_{1}} {\vec{B}}_{2} \cdot d \vec{S}}{I_{2}} \\ M_{21} = \frac{λ_{21}}{I_{1}} = \frac{N_{2} \int_{S_{2}} {\vec{B}}_{1} \cdot d \vec{S}}{I_{1}} \\ M_{12} = M_{21} = M \end{matrix}

Magnetic Energy¶

Magnetic energy in circuit theory

\begin{aligned} W_{m} & = \int P d t = \int V I d t \\ = \int_{0}^{t} L \frac{d I}{d t} I d t \\ = \int_{0}^{I} L I d I = \frac{1}{2} L I^{2} \end{aligned}

Comparison

\begin{matrix} W_{e} = \frac{1}{2} C V^{2} \end{matrix}

magnetostatic energy density (in general)

\begin{matrix} w_{m} = \frac{d W_{m}}{d v} = \frac{1}{2} \vec{B} \cdot \vec{H} \end{matrix}

Magnetic Circuits¶

Electric	Magnetic
$I$	$ψ$
emf $V$	mmf $ℱ$
resistance $R$	reluctance $ℛ$

magnetomotive force (mmf)

\begin{matrix} ℱ = \oint \vec{H} \cdot d \vec{ℓ} = N I \end{matrix}

Ohm's law

\begin{matrix} ℛ = \frac{ℱ}{ψ} = \frac{H ℓ}{B S} = \frac{ℓ}{μ S} \end{matrix}