Derive the expression for the self inductance of a long solenoid carring current
Answers
Let us 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
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.
Answer:
Hey mate here is your answer
Total number of turns in solenoid N = nlN=nlMagnetic field inside the long solenoid B = \mu_o ni B=μ
o
ni
Flux through one turn \phi_1 = BA = \mu_o ni Aϕ
1
=BA=μ
o
niA
Thus total flux through N turns \phi_t = N \phi_1 = nl \times \mu_o niA =\mu_o n^2 lAiϕ
t
=Nϕ
1
=nl×μ
o
niA =μ
o
n
2
lAi
Using \phi_t = Liϕ
t
=Li where LL is the self inductance of the coil
\therefore∴ \mu_o n^2 lA i = Liμ
o
n
2
lAi =Li \implies L = \mu_o n^2 Al⟹L=μ
o
n
2
Al
Hope it helps you