Question

Derive the relation between the self inductance of a solenoid and interior volume of solenoid.​

Answer 1

Answer:

In expression for the self-inductance of a long solenoid.

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=

l

μ

0

NI

Magnetic flux per turn =

l

μ

0

NIA

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=

l

μ

0

NIA

×N

i.e. ϕ=

l

μ

0

N

2

IA

....(1)

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

ϕ=LI....(2)

From equation (1) and (2),

LI=

l

μ

0

N

2

IA

L=

l

μ

0

N

2

A

If the core is filled with a magnetic material of permeability μ,

Then, L=

l

μN

2

A