Project on to study the factors on which self inductance of a coil depends by observing the effect of this coil,when put in series with a bulb/resistor in a circuit fed by an ac source of adjustable frequency.

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factors affecting self inductance of a coil when bulb connected in series

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 = μoN 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 = [μoAN2/x] dI/dt

Therefore, since E = -L dI/dt we have LdI/dt = [μoAN2/x] dI/dt

and so for a solenoid:
Self inductance (L) = [μoAN2/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.