[Magnetism] Tricky task on self-inductance of solenoid

[wetback]

Homework Statement

Problem:
A Solenoid has N turns of Copper wire (Cross-section area of the wire is 10^-6 m²) The Length of the solenoid is 25*10^-2 m and its Resistance is 0.5 Ω. What's the Self-Inductance of the solenoid?

Homework Equations

Given answer: 5.5*10^-5 H

Please help me with the solution.

The Attempt at a Solution

How do you relate Resistance with Self-inductance to eliminate the need for (or find) the number of turns? All I could find by using R was the length of the whole wire(an unrolled solenoid). But this gives me nothing on the number of turns cause the diameter of the solenoid is not given.

 

[NascentOxygen]

Hi wetback! http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif

I reckon you should assume the turns in the winding are single-layer, tightly-wound and close-packed. This allows you to determine the diameter of the coil.

What formula will you then use to calculate inductance?

 

[wetback]

Hey! As it often is, when you are desperate enough to look for help in a forum, you miraculously find the solution yourself! Eureka! And as always, it turns out to be pretty simple. Just some algebra.

If anyone ever stumbles upon this post looking for help, I'll explain what I did:

I used the inductance equation for a solenoid L=(\muN²A)/l
where \mu - permeability constant(4\pi*10^-7), N - number of turns, A - the cross-section area of the solenoid cylinder shape and l - the length of the solenoid
I also used the resistance equation for a conductor(this case the copper wire): R=(\rho*l1)/A1 where \rho-the resistivity of copper(1.7*10^-8), l1 - the length of the wire, A1- the cross-section area of the wire.
I expressed l1: l1=R*A1/\rho, then expressed N using l1: N=l1/(2\pi*r) where r - the radius of the solenoid cylinder shape. Then r=l1/(2\piN) Now I expressed A: A=\pi*r². When you insert this expression in the L equation, N disappears and we are left with just the given numbers and L really is 5.5*10^-5 H.