Solve Length of Solenoid Homework: Find N & L

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The discussion focuses on calculating the self-inductance of a solenoid that stores 9.3 J of energy with a 1.56-A current. The first part was solved, yielding an inductance of 7.643 H, but the second part posed challenges in determining the solenoid's length based on coil density. Participants clarified that the number of coils (N) is related to the turns per unit length (n) and the length (L) of the solenoid. The final calculation for length, using the formula L = μ0*A*l*n², resulted in a length of approximately 52.24 m, which is considered impractical. The discussion highlights the importance of understanding the relationships between the solenoid's dimensions and its inductance.
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Homework Statement



(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?


Homework Equations



l=D*N
D=π*d


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
 
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N = (10000 coils/m) * (0.0385 m) = 385 coils
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.
 
Simon Bridge said:
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.

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?
 
N = (10/(1*10^-3)) * 7.643 = 76430 turns
... um... no: not what I wrote.
me said:
If n=#turns per unit length, and N is the number of turns[/color] in the solenoid, and L is the length of the solenoid, then n=N/L
Hmmm... L was not a good choice for a variable-name for length was it?
You pick one.
 
Simon Bridge said:
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.

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?
 
Actually I don't. I couldn't find one in my book, but searching online I believe I found it:
... 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:
> 7.643/((4*pi*10^-7)*(pi*(0.0385/2)^2) *(10000^2))
ans =  52.245

... which is rather long so probably impractical.
 
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