# Magnetostatics - Biot-Savart

The source of magnetic fields is current flow.

We first considered electro**statics** where stationary charges gave
rise to stationary fields. The sources of electric fields are charges.

We should like to consider now **stationary magnetic fields**. These arise when
the sources of the magnetic fields are stationary. What are those sources??

The sources of magnetic fields are **currents**.

Stationary currents means that the magnitudes and directions of all the kinds of current densities we've been considering are constant. There's one additional proviso: that there's no charge pileup ($\frac{\del \rho}{\del t}=0$), and as we've just seen the continuity equation, this means that... $$\myv \grad \cdot \myv J = 0.$$

When this is all true, the **Biot-Savart Law** allows us to calculate the
magnetic field in the same way that Coulomb's Law let us calculate
the electric field

--that is to say, ponderously, and with much integration:

$$\myv B(\myv r) = \frac{\mu_0}{4 \pi} \int \frac{\myv I \times \uv{\rr}}{\rr^2} dl' =\frac{\mu_0}{4 \pi} I \int \frac{d \myv l' \times \uv{\rr}}{\rr^2} .$$

The constant $\mu_0$ is the permeability of free space, $$\mu_0 = 4\pi \times 10^{-7} N / A^2.$$

In this scheme, this results in the magnetic field being measured in units
of **tesla (T)**, where, 1 T = 1 N / (A $\cdot$ m) .

This should work out so that we get a force in N from the Lorentz force law.

Earth's magnetic field ~ $0.5 \times 10^{-4}$ T = 0.5 "gauss".

Strong lab magnet ~ 1 T.

### Field of a long straight wire

We'd like to know the magnitude of the magnetic field around a very long (~ infinite) straight wire.

Use our Biot-Savart equation... $$\myv B= \frac{\mu_0 I}{4 \pi}\int \frac{d\myv l' \times \uv\rr}{\rr^2}.$$

Consider a (rather artificial) current segment:

Convince yourself that, no matter what $\myv l$ is, $d\myv l \times \uv\r$ is always a vector pointing in to the board ($-\uv z$ direction).

The magnitude of the integral above (substituting dy=dl) is then: $$\int_{-a}{+a}\frac{ dy\, \sin(\phi)}{(x^2+y^2)}=\frac{2a}{x\sqrt{x^2+a^2}}.$$

Field at a distance $x$ from the center of a current-segment of length $2a$: $$B=\frac{\mu_0 I}{4\pi}\frac{2a}{x\sqrt{x^2+a^2}}=\frac{\mu_0 I}{2\pi x}\sin \theta_\text{max}$$where you have to remember how to come up with the direction...

Field at a distance $x$ from an* infinitely
long* straight wire: Let $\theta_\text{max}\to 90^o$ in the expression above, then
$$B=\frac{\mu_0 I}{2\pi x}$$

The cross product $d \myv l' \times \uv{\rr}$ is coming out of the page, and has a magnitude...

$$|d \myv l' \times \uv{\rr}| = dl' \sin \alpha = dl' \cos \theta$$

Differentiate $l'/s = \tan \theta \Rightarrow$ $$dl' =\frac{s}{\cos^2 \theta} d\theta.$$

Since $s=\rr \cos \theta$, $$\frac{1}{\rr^2} = \frac{\cos^2 \theta}{s^2}.$$

#### Force between two wires

If two wires carrying currents $I_1$ and $I_2$ are a distance $d$ apart, what is the force per unit length between the two currents?The field due to $I_1$ at the position of $I_2$ is pointing into the page, and has magnitude $$B = \frac{\mu_0 I_1}{2\pi d}$$

The magnetic force on $I_2$ is: $$ F = I_2 \int d \myv l \times \myv B = I_2 B \int d l .$$

So, the force per unit length is: $$ F/l = I_2 B = \frac{\mu_0 I_1 I_2}{2 \pi d}.$$