An air-filled cylindrical inductor has 2900 turns. It is 3.0 cm in diameter and

Question

An air-filled cylindrical inductor has 2900 turns. It is 3.0 cm in diameter and 26.5 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 100 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

Here is what I solved before, please modify the figures as per your question. Please let me know if you have further questions. If this helps then kindly rate 5-stars

An air-filled cylindrical inductor has 2700 turns. It is 2.2 cm in diameter and 26.2 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 110 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?

solution:

Part A)

Inductance = L = uN^2A/l

L = (4pi X 10^-7)(2700)^2(pi)(.011)^2/.262

L = .0133 H

Part B)

Set up the ratio uN^2 = uN^2

u/u = N^2/N^2

u/u = 2700^2/110^2

u/u = 602.5

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