4.4 KiB
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
\mathbf{H}(\mathbf{r}) = \int{I_e}{4\pi}\int_{\text{wire}} \frac{-(\mathbf{r}-\mathbf{r'}) \times d{\mathbf{l'}}}{|\mathbf{r}-\mathbf{r'}|^3}
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,
\int_S \mathbf{J}_m \cdot \hat{\mathbf{n}} d{S} = I_m = M_o w v
By definition,
\int_S \mathbf{J}_m \cdot \hat{\mathbf{n}} d{S} = -\int_S (\nabla \times{\mathbf{E}}) \cdot \hat{\mathbf{n}} d{S}
Applying Stokes theorem,
\int_S (\nabla \times{\mathbf{E}}) \cdot \hat{\mathbf{n}} d{S} = M_0 w v
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
d\mathbf{F} = I_e d{\mathbf{L}} \times \mu_0 \mathbf{H}(\mathbf{r})
Section 5.4 - Multipole Expansion of a Vector Potential
This is messy. Skipped.