Pre-Quantum Electrodynamics
The Magnetic Field issuing from a Steady Currentems.ms.BS.sc
This is given experimentally (around 1820) by the
{\bf Biot-Savart law} \[ {\bf B} ({\bf r}) = \frac{\mu_0}{4\pi} \int dl' \frac{{\bf I} \times ({\bf r} - {\bf r}')}{|{\bf r} - {\bf r}'|^3} \label{eq:BiotSavart} \]
in which \(\mu_0\) is the vacuum permeability (or alternately permeability of free space, permeability of vacuum or magnetic constant), \[ \mu_0 = 1.25663706212(19)x10^{-6} H/m \] with the {\it henry} \(H = kg m^2 / s^2 A^2\) being the unit for inductance.
For surface and volume density currents:
\[ {\bf B} ({\bf r}) = \frac{\mu_0}{4\pi} \int da' \frac{{\bf K} ({\bf r}') \times ({\bf r} - {\bf r}')}{|{\bf r} - {\bf r}'|^3}, \hspace{2cm} {\bf B} ({\bf r}) = \frac{\mu_0}{4\pi} \int d\tau' \frac{{\bf J} ({\bf r}') \times ({\bf r} - {\bf r}')}{|{\bf r} - {\bf r}'|^3} \label{Gr(5.39)} \]
{\bf N.B.:} there is no Biot-Savart law for a single point charge. Not steady current !
The {\bf superposition principle} applies here as well: collection of currents generates a \({\bf B}\) field which is the vector sum of the fields generated by the individual currents.
\paragraph{Example 5.5:} find \({\bf B}\) a distance \(s\) from a long straight wire carrying steady current \(I\). \paragraph{Solution:} {\bf Gr Fig 5.18}: \(dl {\bf I} \times ({\bf r} - {\bf r}')\) points out of the page (blackboard !), and has magnitude \(dl' \sin \alpha = dl' \cos \theta\). But \(l' = s \tan \theta\) so \(dl' = \frac{s}{\cos^2 \theta} d\theta\), and \(s = |{\bf r} - {\bf r}'| \cos \theta\). Then, \[ B = \frac{\mu_0}{4\pi} I \int_{\theta_1}^{\theta_2} d\theta \cos \theta \frac{\cos^2 \theta}{s^2} \frac{s}{\cos^2 \theta} = \frac{\mu_0 I}{4\pi s} (\sin \theta_2 - \sin \theta_1) \label{Gr(5.35)} \] For infinite wire: \(\theta_1 = -\pi/2\), \(\theta_2 = \pi/2\), so \[ B = \frac{\mu_0 I}{2\pi s} \label{Gr(5.36)} \]
Immediate consequence: force per unit length between two wires with currents \(I_1\) and \(I_2\) separated by distance \(d\): \[ f = \frac{\mu_0}{2\pi} \frac{I_1 I_2}{d} \label{Gr(5.37)} \] (like currents attract).
\paragraph{Example 5.6:} find {\bf B} a distance \(z\) above the center of a circular loop of radius \(R\), carrying a steady counterclockwise current \(I\). \paragraph{Solution:} Gr Fig 5.21. By symmetry, only the vertical component doesn't cancel. \[ B(z) = \frac{\mu_0 I}{4\pi} \int dl' \frac{\cos \theta}{|{\bf r} - {\bf r}'|} = \frac{\mu_0 I}{4\pi} \frac{\cos \theta}{R^2 + z^2} \int dl' = \frac{\mu_0 I}{2} \frac{R^2}{(R^2 + z^2)^{3/2}} \label{Gr(5.38)} \] (since \(\cos \theta = R/\sqrt{R^2 + z^2}\)).
Created: 2022-02-08 Tue 06:55