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[
Producing wound components]
[ Air coils]
[A guide to the terminology of magnetism]
[ Power loss in wound components]

E = N×d/dt volts(Vector quantities express this more rigorously, but Faraday was even less of a maths fan than me.) The flux may change becausewhere N is the number of turns.

- A nearby permanent magnet is moving about.
- The coil rotates with respect to the magnetic field.
- The coil is wound on a core whose effective permeabilty changes.
- The coil is the secondary winding on a transformer where the primary current is changing.

Take an example of an inductor having 7 turns and in which the flux varies according to

Φ = 0.008 sin(300t) webersSubstituting this into Faraday's law

E = 7 × d(0.008 sin(300t))/dtSo our winding will have a peak voltage of 16.8 volts. You can easily generalize this for any flux varying sinusoidally at a frequency f to showE= 7 × 0.008 × 300 cos(300t)

E= 16.8 cos(300t) volts

EThis is called the_{RMS}= 4.44 N × f × Φ_{pk}volts

Calculating this 'flip side to Faraday' is also easy; we take his law in its differential form above and integrate:

= ( E.dt ) / N webersWe'll call this the integral form of Faraday's law.where E is the externally applied voltage and N is the number of turns.

E = (√2)a.sin(2.f.t)Substituting this into the integral form of Faraday:

= ( (√2)a.sin(2.f.t) .dt ) / NThe expression with the limits of integration will always be between -1 and +1 so that the peak value of flux is given by= ((√2)a/N) sin(2.f.t) .dt

= (-(√2)a/(2.f.N)) [cos(2..f.t)]

_{0}^{t}

Compare this with the transformer equation above._{pk}= (√2)a/(2.f.N)

_{pk}= a / (4.44 f.N) Wb

Example: If 230 volts at 50 Hz is applied to an inductor having 200 turns then what is the peak value of magnetic flux?

One important general point: if your winding has to cope with a given signal amplitude then the core flux is proportional to the_{pk}= 230 / (4.44 × 50 × 200)

_{pk}= 5.18 mWb

= L×I/N ampsSubstituting into the integral form of the law:where L is the inductance of the coil.

L×I/N = ( E.dt ) / NIf E is a constant then this formula for the current simplifies to:I = ( E.dt ) / L amps

I = E×t/L ampsExample: If a 820 mH coil has 2 volts applied then find the current at the end of three seconds.

I = 2×3/0.82 = 7.32 amps

Gee - thanks Mike!

Caveat (there's always one): The coils decribed here are idealised. See the paragraph on flux linkage for a more complete picture.

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E-mail:
R.Clarke@surrey.ac.uk

Last modified: 2001 April 29th.