Solve Length of Solenoid Homework: Find N & L

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Homework Help Overview

The discussion revolves around calculating the self-inductance of a solenoid and determining its length based on given parameters, including energy storage and coil density. The subject area includes concepts from electromagnetism and circuit theory.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relationship between the number of coils, solenoid length, and inductance. There is confusion regarding the application of coil density and whether it pertains to the diameter or length of the solenoid. Some participants attempt to derive formulas and express variables in terms of others.

Discussion Status

Several participants have provided calculations and attempted to clarify the relationships between the variables involved. There is ongoing exploration of the correct application of formulas, with some participants questioning the assumptions made about coil density and the dimensions of the solenoid.

Contextual Notes

There is a lack of consensus on the correct interpretation of coil density and its application to the problem. Some participants express uncertainty about the formulas available in their resources, which may affect their calculations.

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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 through 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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