#### Steady Currents: the Biot-Savart Law
ems.ms.BS

The magnetic field issuing from a steady surrent is given experimentally (around 1820) by the

Biot-Savart law

- FLS II (14.43)
- Gr (5.34)
- PM (6.49)

\[ {\bf B} ({\bf r}) = \frac{\mu_0}{4\pi} \int dl' \frac{{\bf I} \times ({\bf r} - {\bf r}')}{|{\bf r} - {\bf r}'|^3} \tag{BiotSavart}\label{BiotSavart} \]

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 henry \(H = kg ~m^2 / s^2 A^2\) being the unit for inductance.

For surface and volume density currents:

\[ {\bf B} ({\bf r}) = \frac{\mu_0}{4\pi} \int da' \frac{{\bf K} ({\bf r}') \times ({\bf r} - {\bf r}')}{|{\bf r} - {\bf r}'|^3}, \tag{BiotSavart_s}\label{BiotSavart_s} \]

\[ {\bf B} ({\bf r}) = \frac{\mu_0}{4\pi} \int d\tau' \frac{{\bf J} ({\bf r}') \times ({\bf r} - {\bf r}')}{|{\bf r} - {\bf r}'|^3} \tag{BiotSavart_v}\label{BiotSavart_v} \]

N.B.: there is no such thing as a Biot-Savart law for a point charge, since this cannot represent a steady current.

The superposition principle applies here as well: a collection of currents generates a \({\bf B}\) field which is the vector sum of the fields generated by the individual currents.

Example: \({\bf B}\) from long straight wire

Task: find \({\bf B}\) a distance \(r\) from a long straight wire carrying steady current \(I\).

Solution: \(dl {\bf I} \times ({\bf r} - {\bf r}')\) points out of the page, and has magnitude \(dl' \cos \theta\). But \(l' = r \tan \theta\) so \(dl' = \frac{r}{\cos^2 \theta} d\theta\), and \(r = |{\bf r} - {\bf r}'| \cos \theta\). Then,

\[ B = \frac{\mu_0}{4\pi} I \int_{\theta_1}^{\theta_2} d\theta \cos \theta \frac{\cos^2 \theta}{r^2} \frac{r}{\cos^2 \theta} = \frac{\mu_0 I}{4\pi r} (\sin \theta_2 - \sin \theta_1) \] For infinite wire: \(\theta_1 = -\pi/2\), \(\theta_2 = \pi/2\), so

- Gr (5.38)

\[ B = \frac{\mu_0 I}{2\pi r} \tag{Bwire1}\label{Bwire1} \]

As an immediate consequence, we see that the force per unit length between two wires with currents \(I_1\) and \(I_2\) separated by distance \(d\) is

\[ f = \frac{\mu_0}{2\pi} \frac{I_1 I_2}{d} \] (like currents attract).

Example: \({\bf B}\) above a circular loop

Task: find \({\bf B}\) a distance \(z\) above the center of a circular loop of radius \(R\), carrying a steady counterclockwise current \(I\).

Solution: By symmetry, only the vertical component doesn't cancel.

\[ B(z) = \frac{\mu_0 I}{4\pi} \int dl' \frac{\cos \theta}{|{\bf r} - {\bf r}'|^2} = \frac{\mu_0 I}{4\pi} \frac{\cos \theta}{R^2 + z^2} \int dl' = \frac{\mu_0 I}{2} \frac{R^2}{(R^2 + z^2)^{3/2}} \] (since \(\cos \theta = R/\sqrt{R^2 + z^2}\)).

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Author: Jean-Sébastien Caux

Created: 2022-03-24 Thu 08:42