Magnetostatics
Biot-Savart's Law¶
- Sources
\[
\begin{gather}
\vec udQ = \frac{d\vec ℓ}{dt}Q = Id\vec ℓ
\end{gather}
\]
- Magnetic flux density:
\[
\begin{gather}
d\vec B = \frac{\mu_o}{4\pi}\frac{Id\vec ℓ \times \hat{a_r}}{R^{2}}
\end{gather}
\]
- Magnetic field density:
\[
\begin{gather}
d\vec H = \frac{d\vec B}{\mu_o}= \frac{1}{4\pi}\frac{Id\vec ℓ \times \hat{a_r}}{R^{2}}
\end{gather}
\]
- Magnetic force
\[
\begin{align}
d\vec F &=
Id\vec ℓ \times d\vec B
\\\\
&=
Id\vec ℓ \times \left(\frac{\mu_o}{4\pi}\frac{Id\vec ℓ \times \hat{a_r}}{R^{2}}\right)
\end{align}
\]
Comparison
\[
\begin{gather}
d\vec F= q \vec E
\end{gather}
\]
Ampere's Law¶
- original Ampere's law 👎
\[
\begin{gather}
\oint_C{\vec H \cdot d\vec ℓ}=\int_S{\vec J\cdot d\vec S}=I_{enc}
\\\\
\nabla \times \vec H=\vec J
\end{gather}
\]
- Ampere-Maxwell equation 👍
\[
\begin{gather}
\oint_C{\vec H \cdot d \vec ℓ} = \int_S{\left(\vec J_f+\frac{\partial \vec D}{\partial t}\right)\cdot d \vec S}
\\\\
\nabla \times \vec H =\vec J_f + \frac{\partial \vec D}{\partial t}
\end{gather}
\]
displacement current
for the identity
\[
\begin{gather}
\nabla \cdot (\nabla \times \vec H) =0
\\\\
\nabla \cdot (\nabla \times \vec H) = \nabla \cdot \vec J \neq 0 \quad\text{ when}\quad\nabla \cdot \vec J=-\frac{\partial \rho_v}{\partial t} \neq 0
\end{gather}
\]
thus, let's write
\[
\begin{gather}
\nabla \times \vec H = \vec J +\vec J_d
\end{gather}
\]
then
\[
\begin{gather}
\nabla \cdot (\nabla \times \vec H) = \nabla \cdot \vec J + \nabla \cdot \vec J_d= 0
\\\\
\implies \nabla \cdot \vec J_d = \frac{\partial \rho_v}{\partial t} = \frac{\partial }{\partial t}\left(\nabla \cdot \vec D\right)
\\\\
\implies\vec J_d= \frac{\partial \vec D}{\partial t}
\end{gather}
\]
- displacement current \(I_d\) <br \(\vec J_d\) is called displacement current density, and the displacement current \(I_d\) id defined by <br
\[
\begin{gather}
I_d = \int_S{\vec J_d \cdot d\vec S}
\end{gather}
\]
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¶
\[
\begin{gather}
\oint_S{\vec B \cdot d \vec S} = \int_V{\nabla \cdot \vec B}=0
\end{gather}
\]
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 \(E_{ind} \equiv E_T +E_m\)
\[
\begin{align}
V_{emf}&=\oint_L (E_e+E_T+E_m)
\\\\
&=0+\left(-\int_S{\frac{\partial \vec B}{\partial t}\cdot d\vec S}\right)+\oint_L{\left(\vec u \times \vec B\right)\cdot d\vec ℓ}
\\\\
&=-\frac{d}{dt}\int{\vec B \cdot d\vec S}
\end{align}
\]
- differential form
\[
\begin{gather}
\nabla \times \vec E=-\frac{\partial \vec B}{\partial t}+\nabla \times \left(\vec u \times \vec B\right)
\\\\
\iff
\nabla \times \left(\vec E- \vec u \times \vec B\right)=-\frac{\partial \vec B}{\partial t}
\end{gather}
\]
Magnetic Potentials¶
Magnetic Scalar Potential¶
- magnetic scalar potential \(V_m\) (in \(A\))
Ignoring displacement current \(\vec J_d\), if \(\vec J = 0\), then
\[
\begin{gather}
\nabla \times \vec H = 0
\\\\
\vec H = - \nabla V_m
\end{gather}
\]
comparison
\[
\begin{gather}
\frac{1}{\varepsilon}\vec D = - \nabla V
\end{gather}
\]
Magnetic Vector Potential¶
- magnetic vector potential \(\vec A\) (Wb/m)
since
\[
\begin{gather}
\nabla \cdot \vec B = 0
\end{gather}
\]
and recall the identity
\[
\begin{gather}
\nabla \cdot \nabla \times \vec A = 0
\end{gather}
\]
then we can write
\[
\begin{gather}
\vec B = \nabla \times \vec A
\end{gather}
\]
by Stokes' theorem
\[
\begin{gather}
\psi = \int_S{\vec B \cdot d \vec S}=\int_S{(\nabla \times \vec A)\cdot d\vec S}=\oint_L{\vec A\cdot d \vec ℓ}
\end{gather}
\]
and just remember that
\[
\begin{gather}
\vec A = \int_Q{\frac{\mu}{4\pi}\frac{\vec udq}{R}}
\end{gather}
\]
- line current
\[
\begin{gather}
\vec udq = I d\vec ℓ
\end{gather}
\]
- surface current
\[
\begin{gather}
\vec udq = \vec K dS
\end{gather}
\]
- volume current
\[
\begin{gather}
\vec udq = \vec J dv
\end{gather}
\]
Comparison
\[
\begin{gather}
V=\int_Q{\frac{1}{4\pi \varepsilon}\frac{dq}{R}}
\end{gather}
\]
Time-Varying Potentials¶
Easy but tedious derivation
- \(\nabla \times \vec E\)
\[
\begin{gather}
\nabla \times \vec E = -\frac{\partial \left(\nabla \times \vec A\right)}{\partial t}=-\nabla \times \frac{\partial \vec A}{\partial t}
\\\\
\implies
\nabla \times \left(\vec E + \frac{\partial \vec A }{\partial t}\right)=0
\iff
\nabla \times \left(-\nabla V\right)=0
\\\\
\implies \vec E = -\frac{\partial \vec A}{\partial t}-\nabla V
\end{gather}
\]
- \(\nabla \cdot \vec D\)
\[
\begin{gather}
\nabla \cdot \vec E= \frac{\rho_v}{\varepsilon}
\\\\
\implies
\nabla^{2}V +\frac{\partial }{\partial t}\left(\nabla \cdot \vec A\right) =-\frac{\rho_v}{\varepsilon}
\end{gather}
\]
- \(\nabla \times \vec H\)
\[
\begin{gather}
\nabla \times \vec H= \vec J + \varepsilon\frac{\partial \vec E}{\partial t}
\\\\
\nabla \times \left(\frac{\nabla \times A}{\mu}\right)= \vec J + \varepsilon\frac{\partial \vec E}{\partial t}
\\\\
\nabla^{2}\vec A=\nabla \left(\nabla \cdot \vec A\right)-\nabla \times \nabla \times \vec A
\\\\
\nabla^{2}\vec A = \nabla \left(\nabla \cdot \vec A\right)-\mu \vec J - \mu\varepsilon \frac{\partial \vec E}{\partial t}
\\\\
\nabla^{2}\vec A - \nabla \left(\nabla \cdot \vec A\right)=-\mu \vec J + \mu\varepsilon\nabla \left(\frac{\partial V}{\partial t}\right) +\mu\varepsilon \frac{\partial ^{2}\vec A}{\partial t^{2}}
\end{gather}
\]
- coupled equations for \(V\) and \(\vec A\), (normally given \(\rho_v\) and \(\vec J\))
\[
\left\{
\begin{align}
&\nabla^{2}V +\frac{\partial }{\partial t}\left(\nabla \cdot \vec A\right) =-\frac{\rho_v}{\varepsilon}
\\\\
&\nabla^{2}\vec A - \nabla \left(\nabla \cdot \vec A\right)=-\mu \vec J + \mu\varepsilon\nabla \left(\frac{\partial V}{\partial t}\right) +\mu\varepsilon \frac{\partial ^{2}\vec A}{\partial t^{2}}
\end{align}\right.
\]
- decoupling: Lorenz condition
eliminate \(\mu\varepsilon\nabla (\frac{\partial }{\partial t}V)\) term, by let
\[
\begin{gather}
\nabla \cdot \vec A=-\mu\varepsilon \frac{\partial V}{\partial t}
\end{gather}
\]
- Coulomb condition
let
\[
\begin{gather}
\nabla \cdot \vec A = 0
\end{gather}
\]
Magnetic Moment¶
- magnetic dipole
\[
\begin{gather}
\vec m = I_b\, \vec S
\end{gather}
\]
- magnetization vector
\[
\begin{gather}
\vec M = \lim_{\Delta v \to 0}{\frac{\sum \vec m_k}{\Delta v}}
\\\\
\implies
d\vec m = \vec M dv
\end{gather}
\]
Magnetization Current¶
- bound volume current density
\[
\begin{gather}
\vec J_b = \nabla \times \vec M
\end{gather}
\]
Comparison
- In a space \(\vec M = 0\) (with free current only)
\[
\begin{gather}
\nabla \times \vec H = \vec J_f
\end{gather}
\]
- bound surface current density
\[
\begin{gather}
\vec K_b = \vec M \times \hat{a_n}
\end{gather}
\]
the equation can be intuitively come up with by
Permeability¶
In material \(\vec M \neq 0\)
\[
\begin{gather}
\nabla \times \frac{\vec B}{\mu_0} &= &\vec J_f &+ \vec J_b &= \vec J
\\\\
&=&\nabla \times \vec H &+ \nabla \times \vec M &
\end{gather}
\]
\[
\begin{gather}
\\
\implies \vec B = \mu_0(\vec H + \vec M)
\end{gather}
\]
For linear Model
\[
\begin{gather}
\vec M = \chi_m\vec H
\end{gather}
\]
\[
\begin{align}
\\
\implies \vec B &= \mu_0(1+\chi_m)\vec H
\\\\
&= \mu_0\mu_r\,\vec H
\\\\
&= \mu\vec H
\end{align}
\]
Comparison
\[
\begin{gather}
\vec D = \varepsilon_0 \vec E + \vec P
\end{gather}
\]
Boundary Condition¶
by Guass' law
\[
\begin{gather}
\nabla \cdot \vec B = 0
\\\\
\implies \vec B_{1n}= \vec B_{2n}
\end{gather}
\]
by Amp.'s law
\[
\begin{gather}
\vec H_{1t\bot S} - \vec H_{2t\bot S}=\vec K
\end{gather}
\]
Inductance¶
- Definition in electric circuit
\[
\begin{gather}
v=L\frac{di}{dt}
\end{gather}
\]
Magnetic Flux Linkage¶
- magnetic flux linkage \(\lambda\)
\[
\begin{gather}
d\lambda = Nd \psi = \frac{I_{int}}{I_{ext}}d\psi
\end{gather}
\]
Self Inductance¶
\[
\begin{gather}
L= \frac{\lambda }{I}=\frac{\displaystyle {N\int_S \vec B \cdot d \vec S}}{I}
\end{gather}
\]
Mutual Inductance¶
\[
\begin{gather}
M_{12}=\frac{\lambda_{12}}{I_2}=\frac{\displaystyle{N_1\int_{S_1}{\vec B_2\cdot d\vec S}}}{I_2}
\\\\
M_{21}=\frac{\lambda_{21}}{I_1}=\frac{\displaystyle{N_2\int_{S_2}{\vec B_1\cdot d\vec S}}}{I_1}
\\\\
M_{12} = M_{21} = M
\end{gather}
\]
Magnetic Energy¶
- Magnetic energy in circuit theory
\[
\begin{align}
W_m &= \int{Pdt}=\int{VI\,dt}
\\\\
&= \int_0^{t}{L\frac{dI}{dt}I\,dt}
\\\\
&= \int_0^{I}{LI\,dI}=\frac{1}{2}LI^{2}
\end{align}
\]
Comparison
\[
\begin{gather}
W_e=\frac{1}{2}CV^{2}
\end{gather}
\]
- magnetostatic energy density (in general)
\[
\begin{gather}
w_m=\frac{dW_m}{dv}=\frac{1}{2}\vec B \cdot \vec H
\end{gather}
\]
Magnetic Circuits¶
| Electric | Magnetic |
|---|---|
| \(I\) | \(\psi\) |
| emf \(V\) | mmf \(ℱ\) |
| resistance \(R\) | reluctance \(ℛ\) |
- magnetomotive force (mmf)
\[
\begin{gather}
ℱ=\oint{\vec H \cdot d\vec ℓ}=NI
\end{gather}
\]
- Ohm's law
\[
\begin{gather}
ℛ=\frac{ℱ}{\psi}=\frac{Hℓ}{BS}=\frac{ℓ}{\mu S}
\end{gather}
\]