Chapter 5 - Electrodynamics with Moving Charges
Section 5.1 - Currents in Steady-State Regime
We want to work in a steady-state system. Thus, we restrict ourselves to currents that do not change in time.
With math, we see that \(\nabla \cdot \mathbf{J}(\mathbf{r}) = -\frac{\partial \rho(\mathbf{r})}{\partial t}\). Since we are only considering a steady-state system, \(\nabla \cdot \mathbf{J}_e = \nabla \cdot \mathbf{J}_m = 0\).
Definition. The conductance of a material is \(G = \frac{1}{R}\), where \(R\) is the resistance of a material.
For a wire of uniform cross-sectional area, we see that \(G = \sigma \frac{A}{L}\), where \(A\) is the cross-sectional area, \(L\) is the length of the wire, and \(\sigma\) is the conductivity of a wire. Inverted, we see that \(R\) = \(\rho \frac{L}{A}\), where \(\rho = \frac{1}{\sigma}\) is the resistivity of the wire.
Definition. Ohm's Law can be written as \(I = G V\), or inverted, \(V = IR\). In a wire, we see that current density \(\mathbf{} = \frac{I}{A} = \sigma \frac{V}{L} = \sigma \mathbf{E}\)
Section 5.2 - Currents and Curling Fields
We know that \(\mathbf{J}_e = \nabla \times{\mathbf{H}}\) and \(\mathbf{J}_m = -\nabla \times{\mathbf{E}}\). That is, current densities cause the opposing field to curl.
For a wire with current \(I_e\), we see that applying Stoke's theorem to the first equation,
$ \int_S \nabla \times{\mathbf{H}} \cdot \hat{\mathbf{n}} d{S} = \int_{\partial S} = \mathbf{H} \cdot d{\mathbf{l}}$. Apply the identity \(\nabla \times{\mathbf{H}} = \mathbf{J}_e\) to the left side to see that \(\int_S \nabla \times{\mathbf{H}} \cdot \hat{\mathbf{n}} d{S} = \int_S \mathbf{J}_e \cdot \hat{\mathbf{n}} d{S} = (I_e)_S\), or the current passing through the cross-sectional area. By the original equation, we see that \((I_e)_S = \mathbf{H} \cdot d{\mathbf{l}}\).
If we assume cylindrical coordinates and that \(\mathbf{H}(vb{r}) = H_\varphi(s) \hat{\mathbf{\varphi}}\), then \(\mathbf{H} \cdot d{\mathbf{l}} = \int_0^{2\pi} H_\varphi(S) s d{\varphi}\), so then \((I_e)_S = \int_0^{2\pi} H_\varphi(S) s d{\varphi}\). Thus, for \(s > a\) (where \(a\) is the radius of the wire), \(2\pi s H_\varphi = I_e\), and for \(s < a\), \(2\pi s H_\varphi = I_e \frac{s^2}{a^2}\).
By Helmholtz Theorem, we know that \(\mathbf{H}(\mathbf{r}) = \nabla \times{\mathbf{A}(\mathbf{r})}\). For a current-carrying wire, \(\mathbf{A}(\mathbf{r}) = \frac{I_e}{4\pi} \int_{\text{wire}} \frac{d{\mathbf{l'}}}{|\mathbf{r}-\mathbf{r'}|}\). Applying identities, we see the *Law of Biot and Savart$, where
Consider a current loop instead, on the \(x-y\) plane and current \(I\). Then, \(r = z \hat{\mathbf{z}}\) and \(d{\mathbf{l'}} = R \hat{\mathbf{\varphi'}} d\phi'\), and the magnetic field collapses to \(\mathbf{H}(s = 0, z) = \frac{I_e R^2}{2(R^2 + z^2)^{\frac{3}{2}}} \hat{\mathbf{z}}\)
Consider some infinite bar magnet with height \(h\) and width \(w\). Then, the top and bottom surfaces will have a magnetic charge with density \(\mathbf{J}_m^+ = M_0 \mathbf{b} \delta(z - h)\) and \(\mathbf{J}_m^- = -M_0 \mathbf{v} \delta(z)\) respectively. By definition, \(I_m = M_0 w v\).
Now, consider a loop around only the top of the conductor. Then,
By definition,
Applying Stokes theorem,
Section 5.3 - Forces on Moving Charges and Current
Consider an electric charge moving with velocity \(\mathbf{v}\) in a magnetic parallel plate capacitor with charge densities \(\pm \sigma_m\). That is, \(\mu_0 \mathbf{H} = \sigma_m \hat{\mathbf{z}}\). Then, we can apply theorems to see the resulting force.
Theorem. Lorentz Force Law states that \(\mathbf{F} = q_e \mathbf{v} \times \mu_0 \mathbf{H}\) in the presence of a magnetic field. In the presence of both an electric and magnetic field, \(\mathbf{F} = q_e (\mathbf{E} + \mathbf{v} \times \mu_0 \mathbf{H})\).
Theorem. Ampere's Force Law states that generalizing the previous theorem, we can see that
Section 5.4 - Multipole Expansion of a Vector Potential
This is messy. Skipped.