Induced Voltage: Faraday's Law
If you need to design a practical inductor then knowledge of Faraday's law
leads to understanding of how voltage is induced in a coil and how you can
relate that to the magnetic flux which its
core must be able to cope with.
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See also ...
[
Producing wound components]
[ Air coils]
[A guide to the terminology of magnetism]
[ Power loss in wound components]
The basis of Faraday's law
Michael Faraday's
greatest contribution to physics was to show that a
voltage, E, is generated by a coil of wire when the magnetic flux, 
, enclosed by it changes
E = N×d/dt volts
where N is the number of turns.
(Vector quantities express this more rigorously, but Faraday was even less
of a maths fan than me.) The flux may change because
- 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.
In electric motors and generators you will usually have more than one of
these causes at the same time. It doesn't matter what causes the change;
the result is an induced voltage, and the faster the flux changes the
greater the voltage.
Take an example of an inductor having 7 turns and in which the
flux varies according to
Φ = 0.008 sin(300t) webers
Substituting this into Faraday's law
E = 7 × d(0.008 sin(300t))/dt
E= 7 × 0.008 × 300 cos(300t)
E= 16.8 cos(300t) volts
So 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 show
ERMS = 4.44 N × f ×
Φpk volts
This is called the transformer equation.
The effect of coil current
OK, that's all clear enough, but there is one other reason for alteration
to the flux: current flow in the coil.
Hans Christian Oersted discovered that an electric current can produce a
magnetic field. The more current you have the more flux you generate.
That, too, is easy enough to grasp. What needs a firm intellectual grip is
to appreciate that Faraday's Law does not stop operating just because you
have current flowing in the coil. When the coil current varies then that
will alter the flux and, says Faraday, if the flux changes then you get an
induced voltage. This merry-go-round between current, flux and voltage
lies at the heart of electromagnetism. Calculating the sequence of operation just described is quite
easy.
The 'flip side to Faraday'
Now let's spin our fairground ride in the other direction. Instead of
getting an induced voltage by putting in a current we'll put a voltage
across the coil and see what happens. Normally, if you put several volts
across any randomly arranged bit of wire then what will happen will be a
flash and a bang; the current will follow
Ohm's law and (unless the wire is very long and thin) there won't be
enough resistance to prevent fireworks. It's a different story when the
wire is wound into a coil. If the current increases then we get flux build
up which induces a voltage of its own. The sign of this induced voltage is
always such that the voltage will be positive if the current into the coil
increases. We say that the induced voltage will oppose the
externally applied voltage which made the current change (Lenz's law).
This creates a limit to the rate of rise of the current and prevents (at
least temporarily) the melt-down we get without coiling.
Calculating this 'flip side to Faraday' is also easy; we take
his law in its differential form
above and integrate:
= ( 
E.dt ) / N webers
where E is the externally applied voltage and N is the number of turns.
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 ) / N
= ((√2)a/N) sin(2.f.t) .dt
= (-(√2)a/(2.f.N))
[cos(2..f.t)]0t
The expression with the limits of integration will always be between
-1 and +1 so that the peak value of flux is given by
pk = (√2)a/(2.f.N)
pk = a / (4.44 f.N) Wb
Compare this with the transformer equation above.
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)
pk = 5.18 mWb
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.
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 amps
where L is the inductance of the coil.
Substituting into the integral form of the law:
L×I/N = ( E.dt ) / N
I = ( E.dt ) / L amps
If E is a constant then this formula for the current simplifies to:
I = E×t/L amps
Example: 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.