Biot-Savart Law

With electric fields (Section 4.2), we began by talking about the electric field created by point charges: $$\vec E=\frc{4\pi\epsilon_0}{q_s\over d^3}\vec d$$ (Remember that \(\frc{4\pi\epsilon_0}=k\).) There's no such thing as a "point current", but we can imagine a very small length of current. It would be like a very small length of electric charge with charge density \(\lambda\) and length \(ds\). The electric field that little bit of line charge would create is $$d\vec E=\frc{4\pi\epsilon_0}{\lambda\,ds\,\vec d\over d^3}$$ With a few modifications, and adding a cross-product (because this is magnetism so of course there's a cross-product), we get the Biot-Savart Law:
$$d\vec B={\mu_0\over 4\pi}{I\,d\vec s\times\vec d\over d^3}$$
This is the magnetic field created by a tiny segment of current \(I\), which has length \(ds\), and points in the direction of \(d\vec s\). As before, the vector \(\vec d\) points from the segment (the "source") to the place you want to know the field (the "target"). This field is tiny, so we represent it as an infinitesimal \(d\vec B\).

The constant \(\mu_0\) is called the permeability of free space, and has the value $$\mu_0=4\pi\ten{-7}\u{Tm/A}$$ If you're wondering what π is doing in there, it's because this is an exact value, not an approximation. How can we know that? Simple: the ampere is defined in terms of this value of \(\mu_0\), and not the other way around. (The speed of light has the same relationship with the meter.)

At this point you're probably thinking "integration", and indeed we will be doing some integrals in the next section. But the Biot-Savart law can tell us something about the appearance of magnetic fields even without integration.

Long Straight Wire

Using the Biot-Savart law, find the direction of the magnetic field at the star due to the two small segments of current shown.
The magnetic field at the star is \(d\vec B=I{d\vec s\times\vec d\over d^3}\). Because \(I\) and \(d^3\) are positive, \(d\vec B\) points in the direction of \(d\vec s\times\vec d\): of the current direction crossed with the source-to-target vector.

For the segment on the left, \(d\vec s\) points to the right (in the direction of the current). The vector \(\vec d\) points up and to the right as shown. Because both vectors lie in the plane of the screen, the cross-product must point either into or out of the page. The cross-product right-hand rule shows us that \(d\vec s\times\vec d\) points out of the page, as does the magnetic field.

For the segment on the right, the vector \(d\vec s\) is the same. The vector \(\vec d\) is different, but their cross product is still out of the page. In fact, any small segment of wire along the dotted line will create a magnetic field that points out of the page at the star, and if you have a whole wire made up of these short segments, those magnetic fields will all add together, and the field of the wire would point out of the page.

Therefore, a wire which carries current to the right creates a magnetic field which points out of the page, above the wire; and we can similarly show that the field points into the page below the wire. Because the wire has rotational symmetry around itself, the magnetic field lines are circles around the wire, which fits with our finding in Section 12.2 that field lines never begin or end.


The "permeability" part of the name doesn't mean very much (and makes it easy to confuse with the "permittivity" \(\epsilon_0\)), but the "free space" part means that the equation above only works if the target is in vacuum. If the target is in some material like air or copper or iron, then we would replace the permeability of free space with just the permeability \(\mu\). For paramagnets, \(\mu>\mu_0\), and the field will be slightly stronger; for diamagnets, \(\mu<\mu_0\) and the field is slightly weaker. In most cases, however, the difference is so slight as to be negligible. Not so for ferromagnets. Iron's permeability is 5000 times stronger than \(\mu_0\): place iron in a magnetic field, and (due to magnetization) it acts as an amplifier.