>>
Study Material >>
IIT JEE Physics >>
Electromagnetic Induction >>
Self Induction
INDUCTANCE
We have already discussed capacitors – devices that store energy using electric fields. Like a capacitor, an inductor is also quite a commonly used element in electric circuits. It stores magnetic energy. As we know that when current flows through a conductor a magnetic field is set-up in surrounding of it, and hence it is associated with magnetic flux. If magnetic flux associated with a coil is Φ and current flowing through it is I, then its inductance is given by the expression L=Φ ⁄ l. The quantity 'L' is called self-inductance of the coil. It does not depend on the current, but it depends on the permeability of the core and the dimensions of the coil.
S.I. unit of inductance is Henry.
Consider the circuit, in which a solenoid is connected across a cell through a resistor. When the switch is open, the current in the circuit is zero. When the switch is closed, a current flow in it. Since current in the circuit increases from zero to a certain value, magnetic field associated with it changes that causes induction of an emf across the solenoid.
Induction of an emf due to variation in current flowing through the coil itself is known as self induction.
Since ΦB = LI, and


Inductance of an ideal solenoid:
Let a current I flow through a solenoid. The magnetic field due to the current flowing within the solenoid is, B = μ0nI, where n is the number of turns per unit length.
If area of cross section of the solenoid is A then flux associated with length l is equal to
Φ = nlBA. (Assuming that the solenoid is ideal and long)
where l is the length of the solenoid
Now B = ΦnI

Self Inductance of a Coil
Consider a coil of N turns and area of crossection A carrying a current i. The length of the coil is l(l≥√A)

Comparing with Φ = L i, we get:
