Question

the length of thin wire required to manufacture a solenoid of inductance L and length l,(if cross section diameter is considered less than its lenght)is

1) {(pie)Ll/2(nu not)}1/2

2) {4(pie)Ll/(nu not)}1/2

3) {2(pie)Ll/(nu not)}1/2

4) {(pie)Li/(nu not)}1/2

Answer 1

For solenoid , if N is the number of turns , A is the cross section area of solenoid , l is the length of wire then, $\bold{L=\frac{\mu_0N^2A}{l}}$
where L is inductance of solenoid .
cross section area of solenoid = A = πr² { where r is the radius of circular part }
length of wire used to form = circumference of circle × number of turns
L₁ = 2πrN
so, L = μ₀N²πr²/l
= μ₀(2πNr)²/4πl
= μ₀L₁²/4π
4πLl= μ₀L₁²⇒L₁²= (4πLl/ μ₀)
L₁ = $\bold{=\sqrt{\frac{4\pi Ll}{\mu_0}}}$

Answer 2

Answer:

The inductance of a solenoid is given by,

L=μ0N2Al

And the length of the wire needed to make it,

L1=2πNr

and A=πr2

So,

L=μ04π4π2N2r2l⇒L=μ04πL12l⇒L1=(4πμ0Ll)12