# Inductance of Solenoid

1. Oct 28, 2007

### jiten827

1. The problem statement, all variables and given/known data
A solenoid is wound with a single layer of copper wire of radius 0.38 cm. The solenoid is 4.5 cm in diameter and 3 m long. What is the inductance of the solenoid per unit length? Assume that adjacent wires touch each other and that insulation thickness is negligible.

radius of copper wire = 0.0038 m
length of solenoid = 3 m
diameter of solenoid = 0.045 m

2. Relevant equations
L = ($$\mu$$ * N^2 * A)/l

where L = inductance
$$\mu$$ = 4*PI*e-7 (perm. of free space)
N = number of turns
A = cross sectional area
l = length of the solenoid

3. The attempt at a solution

by working backwards, i figured the number of turns = 3/2*$$\Pi$$*0.0038, but its not working out any help?

Last edited: Oct 28, 2007
2. Oct 28, 2007

### Jim L

I'm wondering why pi is in your formula for calculating the number of turns?

3. Oct 28, 2007

### jiten827

i was thinking that taking the length of the solenoid and dividing that by the circumference of the wire would give me the number of turns

4. Oct 28, 2007

### Jim L

I was assuming that the length of the solenoid is not the length of wire needed to construct the solenoid. It is the length of the "form" the wire would be wound on to construct the solenoid.

5. Oct 29, 2007

### rl.bhat

Number of turns = length of the solenoid/ diameter of the wire

6. Jul 21, 2010

### ckcc

Ok so number of turns in the solenoid=length of the solenoid/ average diameter of the WIRE used in the solenoid?? do we need to include pi?

7. Jul 21, 2010

### rl.bhat

No. number of turn N = L/d where d is the diameter of the wire.