**Definition**: The magnetomotive force (MMF) is defined as the work done in moving a unit magnetic pole once around the magnetic circuit. It is measured in ampere-turns.

Magnetomotive force (MMF) is like the pressure that makes magnetic flux happen in a magnetic circuit. Without MMF, there can’t be any flux in the magnetic circuit. MMF in a magnetic circuit is similar to electromotive force in an electric circuit.

## What is Magnetomotive Force?

The product of current and turns, known as IN, denotes the **magnetizing force** or **magnetic potential**, termed as **magnetomotive force (mmf)**.

The magnetic field strength of a coil magnet is influenced by the amount of current flowing through its turns.

More current flowing through the coil leads to a stronger magnetic field. Additionally, having more turns in a specific length concentrates the magnetic field.

The coil, resembling a bar magnet, possesses opposite poles at its ends, resulting in a magnetic field proportional to the ampere-turns.

The magnetomotive force (MMF) drives or tends to drive flux through a magnetic circuit and corresponds to electromotive force (EMF) in an electric circuit. Magnetomotive force is equal to the work done in joules in carrying a unit magnetic pole once through the entire magnetic circuit.

In fact, as potential difference between any two points is measured by the work done in carrying a unit charge from one points to another, similarly, mmf between two points is measured by the work done in joules in carrying a unit magnetic pole from one point to another.

## MMF Equation

This relationship is expressed as:

**MMF = Current (I) × Number of Turns (N)**

where I is the current in amperes and N is number of turns.

#### Practical Unit: Ampere-turn

The practical unit for measuring magnetomotive force is the ampere-turn, abbreviated as AT. This unit reflects the combined effect of current and turns in a coil.

MMF = N x I = 10 x 1 = 10 AT

MMF = N x I = 5 x 2 = 10 AT

Two examples of equal ampere-turns for the same mmf are shown above. A solenoid with 5 turns and 2 amperes exhibits the same magnetizing force as one with 10 turns and 1 ampere, as the product of amperes and turns remains consistent.

Utilizing thinner wire allows for more turns to be accommodated within a given space, influencing the magnetic field strength. The amount of current flowing through the coil depends on the wire’s resistance and the source voltage.

The required number of ampere-turns depends on the desired magnetic field strength. Adjustments in current or turns can be made accordingly.

## Units of Magnetomotive Force

The ampere-turn, AT, is an SI unit. It is calculated as NI with the current in amperes.

The cgs unit of mmf is the **gilbert**, abbreviated **Gb**.

One ampere-turn equals 1.26 Gb. The number 1.26 is approximately 4π/10, derived from the surface area of a sphere, which is 4πr^{2}.

To convert IN to gilberts, multiply the ampere-turns by the constant conversion factor 1.26G/1 AT.

As an example, 1000 AT is the same mmf as 1260 Gb. The calculations are 1000 AT × 1.26 Gb/1 AT = 1260 Gb. Note that the units of AT cancel in the conversion.

Description | Information |
---|---|

Ampere-turn (AT) | SI unit for magnetomotive force |

Calculation | AT = I × N |

cgs unit of mmf | Gilbert (Gb) |

Conversion factor | 1 AT = 1.26 Gb |

Conversion to gilberts | Multiply AT by 1.26 Gb |

Example conversion | 1000 AT = 1260 Gb |

## Solved Example Problems

**Example 1**

Calculate the ampere-turns of mmf for a coil with 2000 turns and a 5-mA current.

**Solution**:

To calculate the ampere-turns (AT) of magnetomotive force (mmf) for a coil with 2000 turns and a 5-mA (0.005 A) current, we can use the formula:

$\text{AT} = I \times N $

Where:

- ( I ) = Current in amperes
- ( N ) = Number of turns

Given:

- I = 0.005 A
- N = 2000 turns

Substituting the values:

$\text{AT} = 0.005 \, \text{A} \times 2000 \, \text{turns} $

$ \text{AT} = 10 \, \text{A-turns} $

So, the ampere-turns of mmf for the given coil is 10 A-turns.

**Example 2**

To find the number of turns required for a coil to provide a magnetizing force of 600 A?t with a current of 4 A, we can rearrange the formula for ampere-turns:

$ \text{AT} = I \times N $

Given:

- AT = 600 ) AT
- I = 4 A

We rearrange the formula to solve for ( N ):

$ N = \frac{\text{AT}}{I} $

Substituting the given values:

$ N = \frac{600 \, \text{A?t}}{4 \, \text{A}} $

$ N = 150 \, \text{turns} $

Therefore, 150 turns are necessary for the coil to provide a magnetizing force of 600 AT with a current of 4 A.