Inductance Of A Solenoid Problem

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SUMMARY

The discussion centers on calculating the inductance of a solenoid wound with insulated copper wire. The solenoid has a diameter of 3.000 cm, a length of 1.900 m, and contains 904 turns of wire. The inductance formula used is Inductance = (μ₀) * (N²) * A / l, where μ₀ = 4π × 10^-7 H/m, N = 904 turns, A = π * (R²) with R = 1.050 mm, and l = 1.900 m. The inductance per meter is determined by dividing the total inductance by the length of the solenoid.

PREREQUISITES
  • Understanding of solenoid geometry and dimensions
  • Familiarity with the inductance formula and its components
  • Knowledge of the physical constant μ₀ (permeability of free space)
  • Ability to perform unit conversions (e.g., mm to meters)
NEXT STEPS
  • Learn about the calculation of inductance for different solenoid configurations
  • Study the effects of wire insulation on inductance
  • Explore the relationship between solenoid dimensions and inductance
  • Investigate the significance of inductance per meter in practical applications
USEFUL FOR

Students studying electromagnetism, electrical engineers, and anyone involved in designing or analyzing inductive components in circuits.

pierretong
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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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