Question

$\huge\mathcal\color{red} QuEStiON$


Derive an expression for the self-inductance of a long air-cored solenoid of length l and number of turns N​

Answer 1

Consider a current(I) is passed through the inductor (L).

A: Cross-sectional area

N : Number of turns

L : Length

Magnetic field generated at inside the solenoid is ,

B = μ•NI/L

So, the flux through the coil is a obtained:

ϕ B =N( B . A ) ( B and A are along same direction)

ϕB = μ•N²IA/L ....( 1 )

Now, flux (ϕ B ) is related to inductance (L) as

ϕB=LI

L = ϕB/I

From (1) we get,

L = μ•N²A

I

Answer 2

Answer:

$\sf\bold\red{AnswEr}$

★ Solution :

• φ² I
• φ = LI

φ = FLUX

L = SELF INDUCTANCE

I = CURRENT

$L \: = \frac{ φ}{I}$

$L = \frac{ BA}{I} \: \: \: \: ( \: ∴ φ = BA \: )$

$\boxed{ l = \frac{ron \: a}{l} }$