What is meant by self-inductance of a coil? Obtain an expression for the self-inductance of a long solenoid.

Explanation

Suppose a sinusoidally alternating emf is applied to a pure inductor of large self-inductance L. Let the current in the inductor be
`I=I_(0) sin omega t " " ` …(1)
Where `I_(0)` is the peak current and frequency, `f=omega//2pi.`
By Faraday-Lenz's law, the back emf `(E_("back"))` induced in the coil is
`E_("back") = -L(dI)/(dt)(I_(0) sin omega t) " " ` ...(2)
`therefore E_("back") = - omega LI_(0) cos omega t= - omega LI_(0) sin (omega t+(pi)/(2)) " " ` ...(3)
Now, assuming the resistance of the coil to be negligible, Kirchhoff's voltage law applied around the closed circuit gives
`E+E_("back")=0`
`therefore` The applied emf, `E=-E_("back")`
`=omega LI_(0) sin(omega t+(pi)/(2)) " " ` ...(4)
That is, the applied emf must at every instant be (almost) equal and opposite to the back emf to maintain the alternating current.
The current and the applied emf are thus `90^(@)` out of phase (they are said to be in quadrature), and the emf leads the current (or the current lags behind the emf). This is shown in the phasor diagram, Fig. (b).
The peak emf from Eq. (4) is
`E_(0)=omega LI_(0) " "` ...(5)
`therefore (E_(0))/(I_(0))=omega L " " ` ...(6)

The quantity `omega L` is known as the inductive reactance of the coil and has the same dimensions as resistance. It is denoted by `X_(L)`,
and represents the resistance offered by the inductor to the alternating current through it.
Since `E_(0)=sqrt(2) E_("rms") and I_(0) =sqrt(2)I_("rms")`, from Eq. (6)
we get,
`X_(L) =omega L=(E_("rms"))/(I_("rms")) " " ` ...(8)
The inductive reactance is defined as the ratio of the rms value of the emf across the inductor to the rms value of the current in it.
SI unit : The ohm `(Omega)`.