*Absolute Permeability* (*μ*) is the ratio of magnetic flux density (B) to magnetic field strength (H). Hence,

For a non magnetic material, the ratio of B/H is a constant.

The permeability of a *classical vacuum* (an ‘ideal’ vacuum, neglecting quantum mechanical effects) is called the *permeability of free space* (μ_{0}) and is given by μ_{0} = 4 π × 10^{−7} H/m. For other materials, the permeability is normally specified as a scale factor, relative to that of a vacuum, this being known as the *relative permeability* (μ_{r}) – and having no unit, being a ratio). The relative permeability of *air* is very similar to that of a vacuum, so it is normally assumed that μ_{r}=1 for air.

It follows that the absolute permeability (μ) may be expressed as the product of the permeability of free space (μ_{0}) and the relative permeability (μ_{r}).

Combining these two equations gives the full relationship between B and H:

#### Worked Example 1

A coil is required to generate a magnetic flux density of 1.2 T in an air gap of length 0.8 mm. Calculate a) the magnetic field strength and b) the MMF.

Show / hide answer

a) The relative permeability of the air gap is 1 (μ_{r} = 1), so B = μ_{0} H

H = B / μ_{0}

= 1.2 / (4 π × 10^{−7})

= 954,930 A/m

b) The MMF may be found using F_{m} = H l

= 954,930 × 0.8 × 10^{−3}

= 763.94 A (to 2 d.p.)

## Permeability of Magnetic Materials

The absolute permeability is constant for non magnetic materials, as discussed previously, so a graph of B against H would be a straight line. Permeability varies dynamically for magnetic materials such as iron or steel, based on a number of factors including the currently applied magnetic field strength, the previous magnetisation of the material, and on the current magnetic flux density. Magnetic materials demonstrate effects such as *hysteresis* and *saturation*, as illustrated by a B-H Curve.

#### Worked Example 2

A piece of silicon steel is magnetised and a value of magnetic flux density of 1.4 T is observed when the magnetic field strength is 1,500 A/m. Find the relative permeability.

Show / hide answer

B = μ_{0} μ_{r} H

From the question, B = 1.4 T and H = 1,500 A/m. The permeability of free space, μ_{0} is taken to be 4 π × 10^{−7} H/m.

Rearranging the above to make μ_{r} the subject gives:

μ_{r} = B / (μ_{0} H)

= 1.4 / (4 π × 10^{−7} × 1,500)

= 742.72 (to 2 d.p.)