2.3 KiB
Chapter 5 - Electrodynamics with Moving Charges
Section 5.1 - Currents in Steady-State Regine
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
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-carying 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}