If the current through a coil is altered then the flux through that coil
also changes, and this will induce an e.m.f. in the coil itself. This effect is known self-induction and the
property of the coil is the self-inductance (L) of the coil, usually abbreviated as the inductance.

The self inductance can be defined in two ways:

(a) Nf = LI
or (b) Using the equation for the e.m.f. generated: E = - L(dI/dt)

So:

The unit can be expressed as 1 H = 1 VsA

A very simple
demonstration uses the apparatus shown in the diagram.

(Figure 1).

An air-cored inductor is
connected in series with a d.c. supply and a 12 V bulb. The resistance of the solenoid will be low so
that it barely affects the light emitted by the bulb, and placing an iron core inside the inductor will make
no difference to the bulb's brightness.

If the experiment is repeated using a.c. with an air core,
the inductance will probably prevent the lamp from reaching its full brightness. If an iron core is placed
inside the solenoid, however, its inductance is increased considerably and the lamp goes out due to
the increased self-inductance and resulting back e.m.f. in the coil. The coil and iron rod are called a
choke.

Consider an air-cored solenoid of length x, cross-
sectional area A and N turns carrying a current I (Figure 2)

The field B in the solenoid is B = m_{o}N
I/x

The flux (f) through each turn is BA, and the flux linkage
for the solenoid is

Nf = BAN.

Therefore Nf =
[AN^{2}I]/x

Let the current now change by an amount dI in a time dt, giving a change of
flux linkage d(Nf)/dt.

From Faraday's law

E.M.F generated is given
by E = -d(Nf)/dt = [m_{o}AN^{2}/x] dI/dt

Therefore, since
E = -L dI/dt we have LdI/dt = [m_{o}AN^{2}/x] dI/dt

and so for a solenoid:

Calculate the inductance of a solenoid 0.5 m Iong of cross-sectional area 20 cm

L = 4px10

Since a changing current in an inductor causes an e.m.f. if the source
supplying the current is to maintain a p.d. between its terminals the inductor must gain energy.

Let
the inductor carry an instantaneous current I which is changing at the rate of dI/dt. The induced e.rn.f.
is L dI/dt and the power P supplied to the inductor is

P = EI = LIdI/dt

The energy dW supplied in time dt is Pdt, or dW = LI dI/dt. Therefore:

The energy is used to produce the magnetic field in and around the
coil. If the current is suddenly interrupted a spark may occur as the energy is dissipated. Self-
inductance can be a problem in circuits, where the breaking of the circuit can induce a large e.m.f.,
and so the switches maybe immersed in oil to quench the arc. Alternatively a capacitor may be
connected across the terminals to slow down the decay of current and so reduce the induced
e.m.f.

The solenoid plays a rather similar role with relation to magnetic fields as the capacitor
does to electric fields - the ability to store energy.

Calculate the energy stored in the coil given in the previous example (L = 1.25 mH) if the current through it is 0.5 A. Energy stored = ½ LI2 = 1.25x10-3x0.52 = 0.31x10-3 J

When drilling a wall to put up a shelf, it is most important not to drill through a mains power cable carrying power to a light or to a power socket. Devise and test a simple detector based on electromagnetic induction that could be used to trace the path of a cable carrying a.c. mains.

Study the transmission of sound from one coil to another using the output from a cassette recorder fed into the primary coil The primary should be of about five turns of insulated copper wire wrapped round the edges of a laboratory table and the secondary coil should be placed within it. Would you consider this to be a suitable method for the transmission of messages to workers in a noisy room? What problems would there be?

When a battery of
e.m.f E is connected across a resistor and an inductor in series the current does not rise to its final
value instantaneously. There is a rise time that is due to the back e.m.f (e) in the inductor and the
resistance and inductance of the circuit.

The equation is:

E – e = E – L dI/dt = IR and this can be shown to have the solution

and this shows that the growth of current
is exponential towards a final value E/R.

A similar argument
can be applied to the decay of current when the cell is disconnected, the equation in this case being -L dI/dt = IR with
solution:

A coil with a self-inductance of 1.25 mH (see previous examples) is connected to a resistor of 24 W and a supply of 12 V. Calculate the time taken for the current to rise to 0.45A. (Final d.c current would be 0.5A in this circuit).

Using the equation: I = E/R(1 – e

Therefore 0.1 = e

10 = e