Subject: Physics
Self-inductance is the property of a coil by virtue of which it opposes the growth or decay of the current flowing through it.
Consider a coil connected to a battery through a key(K) as shown in the figure. As the key is on, current in the coil starts increasing. Due to it, the magnetic field and hence flux linkage around the coil also increases. The direction of induced e.m.f. is such that it opposes the growth of current in the coil. This delays the current to acquire the maximum value.
When the key (K) is switched off, the current in the coil starts decreasing. So the magnetic flux linked with the coil decreases as a result. Due to change in magnetic flux e.m.f is also produced. According to lenz’s law, the direction of induced e.m.f is such that it opposes the decay of current in the coil. This delays the current to acquire minimum or zero value.
Such property of the coil which opposes the growth or decay of the current in the circuit.
Let I be the current flowing through a coil then the magnetic flux \(\phi \) linked with the coil is found to be proportional to the strength of current (I).
\begin{align*} \: \text {i.e} \: \phi &\propto I \\ \text {or,}\: \phi &= LI \\ \end{align*}
where L is coefficient of self-inductance. From Faraday’s law of electromagnetic induction, the induced e.m.f. in the coil is given by
\begin{align*} E &= -\frac {d\phi }{dt} \\ &= -\frac {d}{dt}(LI) = - L \frac {dI}{dt} \\ \therefore E &= -L \frac {dI}{dt} \dots (i) \\ \end{align*}
If \(\frac {dI}{dt} = I\), i.e. rate of decrease of current is unity then equation (i) becomes,
$$ E = L$$
Thus coefficient of self induction of a coil is the e.m.f. induced in the coil through which the rate of decrease of current is unity.
\begin{align*} E &= L \frac {dI}{dt} \\ \text {or,} \: L &= \frac {E}{dI/dt} = \frac {W/dq}{dI/dt}\:\:\:\:\: [\because \text {e.m.f.} = \frac {\text {work done}}{\text {charge}} ]\\ &= \frac {W}{\frac {dq}{dt} dI} = \frac {W}{IdI} \\ &= \frac {[ML^2T^{-2}]}{[A^2]} \\ &= \therefore [L] = [M^1L^2T{-2}A^{-2}] \end{align*}
An ideal inductance has high value of self inductance and zero ohmic resistance.
Let a long solenoid of length l, the area of cross-section A, the current flowing through solenoid I and a number of turns per unit length n.
The magnetic field inside the solenoid is uniform and given by
$$B = \mu _0nI$$
Total number of turns in the solenoid, N = nl. Now the magnetic flux linked with each turn of the solenoid\(= BA = \mu_0\: nIA\).
total magnetic flux with the whole solenoid, \(\phi \) = magnetic flux linked with each turn × number of turns in the solenoid.
\begin{align*} \therefore \: \phi = \mu_0 n IA\times nl = \mu_o n^2 IAl \\ \text {or,} \: LI &= \mu_0 n^2 IAl \:\:\: [\because \phi = Li ] \\ \therefore L = \mu_0\: n^2 Al\dots (i) \\ \text {Since,}\: n = \frac Nl. \text {So equation} \:(i) \: \text {becomes,} \\ L &= \mu _0 \frac {N^2}{l^2}Al \\ L &= \mu_0\frac {N^2}{l}A \\ \end{align*}
Thus, self inductance of an air cored solenoid (L) depends on
Reference
Manu Kumar Khatry, Manoj Kumar Thapa, et al.Principle of Physics. Kathmandu: Ayam publication PVT LTD, 2010.
S.K. Gautam, J.M. Pradhan. A text Book of Physics. Kathmandu: Surya Publication, 2003.
Self inductance is the property of a coil by virtue of which it opposes the growth or decay of the current flowing through it.
According lenz’s law, the direction of induced e.m.f is such that it opposes the decay of current in the coil.
Dimensional formula of self inductance is \( [L] = [M^1L^2T{-2}A^{-2}] \).
Self inductance of an air cored solenoid (L) depends on total number of turns (N) of the solenoid, length of the solenoid, and area f cross-section of the solenoid (A).
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