Title | Magnetomotive Force |
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Author | Simonetta Goldsmith |
Course | Electromagnestism Optics |
Institution | Imperial College London |
Pages | 2 |
File Size | 131.3 KB |
File Type | |
Total Downloads | 50 |
Total Views | 147 |
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Magnetomotive Force - Electronics and Micros
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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
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system.
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The MMF resulting from passing an electrical current through a coil is given by the electrical current
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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
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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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