Inductance Of A Solenoid Problem

AI Thread Summary
The discussion centers on calculating the inductance of a solenoid wound with insulated copper wire. The solenoid has 904 turns, a diameter of 3.000 cm, and a length of 1.900 m. The inductance formula used is based on the magnetic constant, number of turns, cross-sectional area, and length. There is confusion regarding the calculation of inductance per meter, with clarification that the total inductance should be divided by the solenoid's length to find the per meter value. Additionally, there is a suggestion to verify the cross-sectional area calculation using the correct radius.
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Homework Statement



A solenoid is wound with a single layer of insulated copper wire of diameter 2.100 mm and has a diameter of 3.000 cm and is 1.900 m long. Assume that adjacent wires touch and that insulation thickness is negligible.

a) How many turns are on the solenoid?
Solved: determined to be 904

b)What is the inductance per meter (H/m) for the solenoid near its center?

Homework Equations



Inductance = (constant mu0)* (N^2)*A / l

mu0 = 4*PI*10^-7
N = 904 turns
A = PI*R^2 (R = 2.100 mm/2)
l = length of solenoid = 1.900 m

The Attempt at a Solution



Plugged all those variables above (converted R to meters) and got 0.00019 for the inductance.

I don't understand what it means to find the Inductance per Meter at the center of the solenoid? Do I have to divide by 1.900 m again? Confused as to what I'm supposed to do.
 
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hi pierretong! :smile:
pierretong said:
I don't understand what it means to find the Inductance per Meter at the center of the solenoid? Do I have to divide by 1.900 m again?

yes, your formula µN2A/l gives the inductance of the whole solenoid

so divide by l again for the inductance per metre :wink:

(if you cut the solenoid into x pieces, then for one piece N2 would be 1/x2, so N2/l would be 1/x)

see also http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/indcur.html#c2 and http://hyperphysics.phy-astr.gsu.edu/hbase/electric/indsol.html
 
pierretong said:

Homework Statement



A solenoid is wound with a single layer of insulated copper wire of diameter 2.100 mm and has a diameter of 3.000 cm and is 1.900 m long. Assume that adjacent wires touch and that insulation thickness is negligible.

a) How many turns are on the solenoid?
Solved: determined to be 904

b)What is the inductance per meter (H/m) for the solenoid near its center?

Homework Equations



Inductance = (constant mu0)* (N^2)*A / l

mu0 = 4*PI*10^-7
N = 904 turns
A = PI*R^2 (R = 2.100 mm/2)
l = length of solenoid = 1.900 m

The Attempt at a Solution



Plugged all those variables above (converted R to meters) and got 0.00019 for the inductance.

I don't understand what it means to find the Inductance per Meter at the center of the solenoid? Do I have to divide by 1.900 m again? Confused as to what I'm supposed to do.

Check the area calculation. You want the cross sectional area of the solenoid, so which radius should you use?
 
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