Magnetomotive Force PDF

Title Magnetomotive Force
Author Simonetta Goldsmith
Course Electromagnestism Optics
Institution Imperial College London
Pages 2
File Size 131.3 KB
File Type PDF
Total Downloads 50
Total Views 147

Summary

electromotive force...


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Magnetomotive Force - Electronics and Micros

http://www.electronics-micros.com/electrical/magnetomotive-force

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Magnetomotive Force (MMF) performs a similar role in a magnetic circuit to the electromotive force

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(EMF) of a battery in a basic electrical circuit, thus acting as the 'prime mover' of an electromagnetic

Electrical Basic Concepts

system.

Measuring Instrume DC Circuits Resistor Circuits Capacitor Circui Inductor Circuits Circuit Theorems DC Transients Single Phase AC C Three Phase AC Ci Electric Field Conce

The MMF resulting from passing an electrical current through a coil is given by the electrical current

Magnetic Circuits

flowing through the coil multiplied by the number of turns, as shown in the following equation:

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

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

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

Fm = I × N = 0.25 × 400 = 100 A

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Magnetomotive Force - Electronics and Micros

http://www electronics-micros com/electrical/magnetomotive-force

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

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I = Fm / N = 20 / 100 = 0.2 A

 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:

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. Show / hide answer a) Fm = H l, where l = 2  r = 5,000 × 2 ×  × 0.01 = 314.16 A (to 2 d.p.) b) I = Fm / N = 5,000 × 2 ×  × 0.01 / 600 = 0.52 A (to 2 d.p.)

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