$$d\vec B={\mu_0\over 4\pi}{I\,d\vec s\times\vec d\over d^3}$$

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.

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.

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.