An air-filled cylindrical inductor has 3500 turns. It is 2.6 cm in diameter and

Question

An air-filled cylindrical inductor has 3500 turns. It is 2.6 cm in diameter and 28.1 cm long.

(a) What is its inductance?
H
(b) Assume a second cylindrical inductor with the same length and diameter as the first, but with only 120 turns and a core filled with a material instead of air. If the second inductor has the same inductance as the first, what is the magnetic permeability of the material filling the second inductor relative to that of free space? That is, what is?/?0 for this material?
times

Explanation / Answer

apply the formula for inductance L =    uo N^2 A/ l

uo = 4pi e-7

N   = No. of turns = 3500

A is area    = pi r^2

r = d/2 = 2.6/2 = 1.3 cm or 0.013 m

l = 0.281 m

so Inductance L = 4*3.14e-7 * 3500*3500 * 3.14*0.013* 0.013/(0.281)

L =    29 mH
--------------------------------------------

inductance L2 =   4*3.14e-7 * 120*120 * 3.14*0.013* 0.013/(0.281)

L2 = 34.14 uH

so

as L is directly proportional to uo

u/uo = L2/Lo

u/uo = 34.14e-6/29e-3

u/uo = 1.1*10^-3

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