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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 \div \vb{J}(\vb{r}) = -\frac{\partial \rho(\vb{r})}{\partial t}. Since we are only considering a steady-state system, \div \vb{J}_e = \div \vb{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 \vb{} = \frac{I}{A} = \sigma \frac{V}{L} = \sigma \vb{E}
Section 5.2 - Currents and Curling Fields
We know that \vb{J}_e = \curl{\vb{H}} and \vb{J}_m = -\curl{\vb{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 \curl{\vb{H}} \vdot \vu{n} \dd{S} = \int_{\partial S} = \vb{H} \vdot \dd{\vb{l}}. Apply the identity \curl{\vb{H}} = \vb{J}_e to the left side to see that \int_S \curl{\vb{H}} \vdot \vu{n} \dd{S} = \int_S \vb{J}_e \vdot \vu{n} \dd{S} = (I_e)_S, or the current passing through the cross-sectional area. By the original equation, we see that (I_e)_S = \vb{H} \vdot \dd{\vb{l}}.
If we assume cylindrical coordinates and that \vb{H}(vb{r}) = H_\varphi(s) \vu{\varphi}, then \vb{H} \vdot \dd{\vb{l}} = \int_0^{2\pi} H_\varphi(S) s \dd{\varphi}, so then (I_e)_S = \int_0^{2\pi} H_\varphi(S) s \dd{\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 \vb{H}(\vb{r}) = \curl{\vb{A}(\vb{r})}. For a current-carrying wire, \vb{A}(\vb{r}) = \frac{I_e}{4\pi} \int_{\text{wire}} \frac{\dd{\vb{l'}}}{|\vb{r}-\vb{r'}|}. Applying identities, we see the *Law of Biot and Savart$, where
\vb{H}(\vb{r}) = \int{I_e}{4\pi}\int_{\text{wire}} \frac{-(\vb{r}-\vb{r'}) \cross \dd{\vb{l'}}}{|\vb{r}-\vb{r'}|^3}
Consider a current loop instead, on the x-y plane and current I. Then, r = z \vu{z} and \dd{\vb{l'}} = R \vu{\varphi'} \dd{\phi'}, and the magnetic field collapses to \vb{H}(s = 0, z) = \frac{I_e R^2}{2(R^2 + z^2)^{\frac{3}{2}}} \vu{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 \vb{J}_m^+ = M_0 \vb{b} \delta(z - h) and \vb{J}_m^- = -M_0 \vb{v} \delta(z) respectively. By definition, I_m = M_0 w v.
Now, consider a loop around only the top of the conductor. Then,
\int_S \vb{J}_m \vdot \vu{n} \dd{S} = I_m = M_o w v
By definition,
\int_S \vb{J}_m \vdot \vu{n} \dd{S} = -\int_S (\curl{\vb{E}}) \vdot \vu{n} \dd{S}
Applying Stokes theorem,
\int_S (\curl{\vb{E}}) \vdot \vu{n} \dd{S} = M_0 w v
Section 5.3 - Forces on Moving Charges and Current
Consider an electric charge moving with velocity \vb{v} in a magnetic parallel plate capacitor with charge densities \plusminus \sigma_m. That is, \mu_0 \vb{H} = \sigma_m \vu{z}. Then, we can apply theorems to see the resulting force.
Theorem. Lorentz Force Law states that \vb{F} = q_e \vb{v} \cross \u_0 \vb{H} in the presence of a magnetic field. In the presence of both an electric and magnetic field, \vb{F} = q_e (\vb{E} + \vb{v} \cross \u_0 \vb{H}).
Theorem. Ampere's Force Law states that generalizing the previous theorem, we can see that
\dd{\vb{F}} = I_e \dd{\vb{L}} \cross \u_0 \vb{H}(\vb{r})
Section 5.4 - Multipole Expansion of a Vector Potential
This is messy. Skipped.