How to Find the Inductance of a Long Solenoid?

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SUMMARY

The inductance L of a long solenoid can be calculated using the formula L = (μ₀N²A) / l, where μ₀ is the permeability of free space, N is the number of turns, A is the cross-sectional area, and l is the length of the solenoid. In this case, for a solenoid with a length of 39.0 cm, a radius of 2.30 cm, and 310 turns, the area A is calculated as πr². The user successfully derived the inductance formula after clarifying the relationship between magnetic flux and current.

PREREQUISITES
  • Understanding of electromagnetic theory, specifically inductance
  • Familiarity with the formula for magnetic flux (Φ = BA)
  • Knowledge of the properties of solenoids, including turns and dimensions
  • Basic algebra and geometry skills for calculating area and manipulating equations
NEXT STEPS
  • Study the concept of magnetic flux and its applications in inductance calculations
  • Learn about the permeability of free space (μ₀) and its significance in electromagnetism
  • Explore the derivation and applications of the inductance formula for different geometries
  • Investigate the effects of varying current on the inductance of solenoids
USEFUL FOR

Students studying electromagnetism, physics educators, and anyone interested in understanding the principles of inductance in solenoids.

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[SOLVED] Simple Self-Inductance Problem

Homework Statement



Find the inductance L of a long solenoid of length 39.0 cm, and radius 2.30 cm, which has 3.10E2 turns. (Express you answer in units of H for henrys).

Homework Equations



[tex]L=\frac{N\Phi_{B}}{i}[/tex]
[tex]\Phi_{B}=BA=\frac{\mu_{0}NiA}{2\pi*r}[/tex]

The Attempt at a Solution



Okay, so the way I am thinking to solve this problem is to first get the magnetic flux. But I have no idea how to find the current [tex]i[/tex] for a solenoid. Since I don't know how to find [tex]i[/tex], I simply plugged in the formula for flux into the formula for inductance to get:

[tex]L=\frac{N\Phi_{B}}{i}=\frac{\mu_{0}N^{2}A}{2\pi*r}[/tex]

That's where I get stuck. *If* I am in fact going about this the right way, do I just find area by using pi*r^2?
 
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Try

[tex] L=\frac{\mu_{0}N^{2}A}{l}[/tex]

Where A is cross sectional area and l is the length of the solenoid.

I think...
 
Thanks for the quick response. I got it.
 

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