Self Inductance of a Solenoid Problem

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SUMMARY

The inductance of a solenoid can be calculated using the formula L = (μ0 * μr * n² * A) / l. In this case, with 600 turns (n), a cross-sectional area of 450 mm² (A), a length of 500 mm (l), and a relative permeability (μr) of 25, the correct calculation yields an inductance of 10 mH. The user initially miscalculated the inductance as 2.55, indicating a misunderstanding of the formula or input values.

PREREQUISITES
  • Understanding of electromagnetic theory
  • Familiarity with inductance calculations
  • Knowledge of solenoid properties
  • Basic proficiency in unit conversions (mm² to m²)
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  • Study the derivation of the inductance formula for solenoids
  • Learn about the effects of core materials on inductance
  • Explore practical applications of inductors in circuits
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Students studying electromagnetism, electrical engineers, and anyone involved in designing or analyzing inductive components in circuits.

liamporter1702
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Homework Statement


Hi again guys, been stuck on this problem for a while now, thought it would be a simple bumping in some values in a given equation question but I can't get the correct answer :confused: Heres the problem:

What is the inductance of a 600 turn coil wound on a core of cross sectional area 450mm2 , length of 500mm and a relative permeability of 25?


Homework Equations


I assumed the equation I would use would be L=μ0μrn2Al
Where n = number of turns
A = cross sectional area
l = length
μ0 I took as 12.6x10^-7
and μr = 25

The Attempt at a Solution


When I put all these values into the equation I get 2.55, where as the given answer is 10mH. I'm stumped :confused:
 
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The correct equation is: L=\frac{\mu_0 \mu_r n^2 A}{l}
 

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