Describe Self Inductance of a Long Solenoid - QS Study
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Let consider a solenoid of N turns with length l and area of cross section A. It carries a current I. If B is the magnetic field at any point inside the solenoid, then

Magnetic flux per turn = B × area of each turn

But, B = μ0NI / l

Magnetic flux per turn = μ0NIA / l

Hence, the total magnetic flux (φ) linked with the solenoid is given by the product of flux through each turn and the total number of turns.

φ = (μ0NIA / l) × N

i.e φ = (μ0N2IA / l) … … (1)

If L is the coefficient of self induction of the solenoid, then

φ = LI … … (2)

From equations (1) and (2)

LI = (μ0N2IA / l)

L = (μ0N2A / l)

If the core is filled with a magnetic material of permeability μ, then, L = (μN2A / l)