obtain an expression for the self inductance of a solenoid
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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= lμ 0 NI
Magnetic flux per turn = lμ 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= lμ NIA×N
i.e. ϕ= lμ 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= lμ 0 N IA
L= lμ 0 N 2 A
If the core is filled with a magnetic material of permeability μ,
Then, L= lμN 2 A
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