# Homework Help: Length of a Solenoid

1. Nov 30, 2013

### yaro99

1. The problem statement, all variables and given/known data

(a) What would have to be the self-inductance of a solenoid for it to store 9.3 J of energy when a 1.56-A current runs throught it?

(b) If this solenoid's cross-sectional diameter is 3.85 cm, and if you could wrap its coils to a density of 10 coils/mm, how long would the solenoid be?

2. Relevant equations

l=D*N
D=π*d

3. The attempt at a solution

I solved part a, I am only having trouble with part b.
The problem is, I only have the diameter, but I don't have N, the number of coils.

Here's how I tried to get N:

coils/m = 10/1*10^-3 = 10000 coils/m

then,
N = (10000 coils/m) * (0.0385 m) = 385 coils

then,
l = pi*(0.0385 m) * 385 coils = 46.57m

2. Nov 30, 2013

### Simon Bridge

0.0385m is the diameter - are there 10000coils per meter across the diameter?! Usually the coils are across the length aren't they?

You are close:
If n=#turns per unit length, and N is the number of turns in the solenoid, and L is the length of the solenoid, then n=N/L

You have a formula for the inductance in terms of the L and N don't you?
So use some algebra to express it in terms of n and L.

3. Nov 30, 2013

### yaro99

I got 7.643H for the inductance from part a

So if n=N/L, then N=n*L:
N = (10/(1*10^-3)) * 7.643 = 76430 turns

then,
l = pi*0.0385 * 76430 = 9244 m

It this correct?

4. Dec 1, 2013

### Simon Bridge

... um... no: not what I wrote.
Hmmm... L was not a good choice for a variable-name for length was it?
You pick one.

5. Dec 1, 2013

### yaro99

Actually I don't. I couldn't find one in my book, but searching online I believe I found it:
L = μ0*A*l*n2

rearranging it gives:
l = L/(μ0*A*n2)

So,
l = (7.643)/([4*π*10^-7]*[π*(0.0385/2)2]*100002) = 52.24m

Is this right?

6. Dec 1, 2013

### Simon Bridge

... hmmm, it's usually the first thing they do right after the Biot-Savart law. As in these utex lecture notes eq.909.
"Self Inductance" doesn't mean much without it.

Oh well...
You got the formula in the end.

Code (Text):

> 7.643/((4*pi*10^-7)*(pi*(0.0385/2)^2) *(10000^2))
ans =  52.245

... which is rather long so probably impractical.