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 (μ0) and is given by μ0 = 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 (μr) – 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 μr=1 for air.

It follows that the absolute permeability (μ) may be expressed as the product of the permeability of free space (μ0) and the relative permeability (μr). 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.
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a) The relative permeability of the air gap is 1 (μr = 1), so B = μ0 H

H = B / μ0

= 1.2 / (4 π × 10−7)

= 954,930 A/m

b) The MMF may be found using Fm = 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.
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B = μ0 μr H

From the question, B = 1.4 T and H = 1,500 A/m. The permeability of free space, μ0 is taken to be 4 π × 10−7 H/m.

Rearranging the above to make μr the subject gives:

μr = B / (μ0 H)

= 1.4 / (4 π × 10−7 × 1,500)

= 742.72 (to 2 d.p.)