*Magnetomotive Force* (MMF) performs a similar role in a magnetic circuit to the electromotive force (EMF) of a battery in a basic electrical circuit, thus acting as the ‘prime mover’ of an electromagnetic system.

The MMF resulting from passing an electrical current through a coil is given by the electrical current flowing through the coil multiplied by the number of turns, as shown in the following equation:

MMF is measured in *amperes* (A) rather than ampere-turns, since ‘turns’ is not an SI unit.

The word ‘force’ in magnetomotive

forceis not directly equivalent to a conventional Newtonian force – a push or pull – which would of course be measured in newtons, rather than amperes.

#### Worked Example 1

Find the magnetomotive force produced by a coil of 400 turns if the current flowing through the coil is 0.25 A.

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F_{m} = I × N

= 0.25 × 400

= 100 A

#### Worked Example 2

Find the current flowing through a coil of 100 turns if the magnetomotive force is 20 A.

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I = F_{m} / N

= 20 / 100

= 0.2 A

## Alternative MMF Expression

Magnetomotive force is also used in the derivation of magnetic field strength (H), as shown below.

Rearranging this expression to make MMF the subject, gives:

#### Worked Example 3

A circular coil of mean radius 10 cm has a magnetic field strength of 5,000 A/m. a) Find the magnetomotive force. b) Calculate the current if the coil is wound with 600 turns.

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a) F_{m} = H l, where l = 2 π r

= 5,000 × 2 × π × 0.01

= 314.16 A (to 2 d.p.)

b) I = F_{m} / N

= 5,000 × 2 × π × 0.01 / 600

= 0.52 A (to 2 d.p.)