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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.
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
ERMS = 4.44 N × f × Φpk voltsThis is called the transformer equation.
Calculating this 'flip side to Faraday' is also easy; we take
his law in its differential form
above and integrate:
where E is the externally applied voltage and N is the number of turns.
= ( E.dt ) / N webers
We'll call this the integral form of Faraday's law.
Inductor with AC applied
Let's apply a sinusoidal voltage, frequency f, RMS amplitude a -
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)]0t
pk = (√2)a/(2.f.N)Compare this with the transformer equation above.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?
pk = 230 / (4.44 × 50 × 200)One important general point: if your winding has to cope with a given signal amplitude then the core flux is proportional to the inverse of the frequency. This means, for example, that mains transformers operating at 50 or 60 Hz are larger than transformers in switching supplies (capable of handling the same power) working at 50 kHz.pk = 5.18 mWb
Using inductance
If the material permeability is constant then
the relation between flux and current is linear and, by the definition
of inductance :
= 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
Conclusion: why Faraday is cool
Satisfy yourself that this result above is consistent with our original formulation of Faraday's law: voltage is
proportional to the rate of change of flux.
Consider also how useful this integration method is in practical inductor
design; if you know the number of turns on a winding and the voltage
waveform on it then you integrate wrt time and voilá you have
found the amount of flux. What's more is that you found it without knowing
about the inductance or the core: its permeabilty, size, shape, or even whether there
was a core at all.
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.