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Self-inductance

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 of inductance is the henry and it is defined as the inductance of a coil (or circuit) in which an e.m.f. of one volt is induced when the current changes at the rate of one ampere per second. The unit can be expressed as 1 H = 1 VsA^-1.

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.

Self Inductance of a solenoid

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 = μ_oN I/x

The flux (φ) through each turn is BA, and the flux linkage for the solenoid is
Nφ = BAN.
Therefore Nφ = [AN²I]/x

Let the current now change by an amount dI in a time dt, giving a change of flux linkage d(Nφ)/dt.
From Faraday's law
E.M.F generated is given by E = -d(Nφ)/dt = [μ_oAN²/x] dI/dt

Therefore, since E = -L dI/dt we have LdI/dt = [μ_oAN²/x] dI/dt

and so for a solenoid:

Self inductance (L) = [μ_oAN²/x]

Example problem
Calculate the inductance of a solenoid 0.5 m Iong of cross-sectional area 20 cm² and with 500 turns.

L = [4πx10^-7 x 20x10^-4 x 500²]/0.5 = 1.25x10^-3 H = 1.25 mH

Energy stored in an inductor

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.

Energy stored in inductor = ½LI²

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.

Example problem

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

Student investigation
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.

Student investigation
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?

Growth and decay of current in an inductor

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:

I = E/R(1 – e-t/(L/R))

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:

I = E/R(e-t/(L/R))

Example problem
A coil with a self-inductance of 1.25 mH (see previous examples) is connected to a resistor of 24 Ω 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^-t/(L/R)) 0.45 = 12/24(1 – e^{-t/(0.00125/24)})
Therefore 0.1 = e^{-t/(0.00125/24)} and so
10 = et/(0.00125/24) ln 10 = t/(0.00125/24)
and therefore:
t = (2.3x0.001250/24 = 1.1 x10^-4s

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