derive an expression for coeffcient of self inductance of solenoid
Answers
Answer:
And,DA∮Bdℓ=0. (∴Boutside the solenoid in negligible Now, Img=(n×a)×is⇒B×a=μ_o(n×a×i)
Answer:
Explanation: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