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<strong>www</strong>.<strong>GOALias</strong>.<strong>blogspot</strong>.<strong>com</strong>Physics222Now, let us recollect Experiment 6.3 in Section 6.2. In that experiment,emf is induced in coil C 1wherever there was any change in current throughcoil C 2. Let Φ 1be the flux through coil C 1(say of N 1turns) when current incoil C 2is I 2.Then, from Eq. (6.9), we haveN 1Φ 1= MI 2For currents varrying with time,( N ) d( MI )d1Φ 1=2dtdtSince induced emf in coil C 1is given by( N Φ )d1 1ε 1= –dtWe get,dI2ε 1= – Md tIt shows that varying current in a coil can induce emf in a neighbouringcoil. The magnitude of the induced emf depends upon the rate of changeof current and mutual inductance of the two coils.6.9.2 Self-inductanceIn the previous sub-section, we considered the flux in one solenoid dueto the current in the other. It is also possible that emf is induced in asingle isolated coil due to change of flux through the coil by means ofvarying the current through the same coil. This phenomenon is calledself-induction. In this case, flux linkage through a coil of N turns isproportional to the current through the coil and is expressed asNΦ∝ IBNΦ B= L I(6.15)where constant of proportionality L is called self-inductance of the coil. Itis also called the coefficient of self-induction of the coil. When the currentis varied, the flux linked with the coil also changes and an emf is inducedin the coil. Using Eq. (6.15), the induced emf is given by( NΦ)dε = –dtBdIε = – L(6.16)dtThus, the self-induced emf always opposes any change (increase ordecrease) of current in the coil.It is possible to calculate the self-inductance for circuits with simplegeometries. Let us calculate the self-inductance of a long solenoid of crosssectionalarea A and length l, having n turns per unit length. The magneticfield due to a current I flowing in the solenoid is B = μ 0n I (neglecting edgeeffects, as before). The total flux linked with the solenoid isB( ) ( μ ) ( )NΦ= nl n I A0

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