Inductance/unit length of solenoid

In summary: Expert summarizerIn summary, the question asks for the inductance of a solenoid wound with a single layer of copper wire and 1 cm in diameter and 2.6 cm long. In order to find the inductance per unit length, the number of turns per unit length must be calculated using the diameter of the solenoid. Then, the inductance equation can be used with the radius of the solenoid to find the inductance per unit length. There were some minor errors in the calculation of the number of turns and the use of the solenoid's diameter instead of its length.
  • #1
Luca13
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Homework Statement



A solenoid is wound with a single layer of copper wire of radius 0.24 cm. The solenoid is 1 cm in diameter and 2.6 cm long. What is the inductance of the solenoid per unit length? Assume that adjacent wires touch each other and that insulation thickness is negligible.

Homework Equations



L = μon^2Ad



The Attempt at a Solution



So I am thinking that in order to find n (the number of turns of the copper wire per unit length) I need to be using the radius given for the wire. Then I can put that n in the inductance equation. The length, l, of the solenoid will cancel because the question asks for the inductance/unit length. The problem is, I keep getting this wrong and I would like to find out what I do wrong!

if r(Cu wire) = 0.0024 m --> d(Cu wire) = 0.0048
so, the number of turns per unit length = l/d(Cu wire) = 0.01/0.0048 = 2.0833

Now, L/l = (μon^2∏r(solenoid)^2l)/l = (4∏E-7*2.0833^2*∏*0.01^2) = 1.71E-9 H/m

Where is my mistake? Thank you!
 
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  • #2


Thank you for your question. Your approach is correct, however, there are a few minor errors in your calculations.

Firstly, for the calculation of the number of turns per unit length, the diameter of the solenoid should be used instead of the diameter of the copper wire. This is because the question specifies that the solenoid is 1 cm in diameter, not the copper wire.

So, the number of turns per unit length, n = l/d(solenoid) = 0.01/0.01 = 1 turn/meter.

Secondly, in the inductance equation, the radius of the solenoid should be used, not the length. So, the inductance per unit length, L/l = μon^2∏r(solenoid)^2 = (4∏E-7*1^2*∏*0.005^2) = 3.14E-9 H/m.

I hope this helps. Let me know if you have any further questions.
 

1. What is the formula for calculating the inductance per unit length of a solenoid?

The formula for calculating the inductance per unit length of a solenoid is L = μN²A/l, where L is the inductance per unit length, μ is the permeability of the material, N is the number of turns, A is the cross-sectional area, and l is the length of the solenoid.

2. How does the number of turns affect the inductance per unit length of a solenoid?

The inductance per unit length of a solenoid is directly proportional to the square of the number of turns. This means that as the number of turns increases, the inductance per unit length will also increase.

3. What is the significance of the cross-sectional area in calculating the inductance per unit length of a solenoid?

The cross-sectional area is an important factor in calculating the inductance per unit length of a solenoid because it determines the amount of magnetic flux that can pass through the solenoid. A larger cross-sectional area allows for more magnetic flux and thus, a higher inductance per unit length.

4. How does the length of a solenoid affect its inductance per unit length?

The length of a solenoid is inversely proportional to its inductance per unit length. This means that as the length of the solenoid increases, the inductance per unit length will decrease. Therefore, longer solenoids have a lower inductance per unit length than shorter solenoids.

5. What is the unit of measurement for inductance per unit length of a solenoid?

The unit of measurement for inductance per unit length of a solenoid is Henry per meter (H/m). This unit is used to measure the amount of inductance produced per unit length of the solenoid.

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