An air-filled cylindrical inductor has 3100 turns. It is 2.6 cm in diameter and

Question

An air-filled cylindrical inductor has 3100 turns. It is 2.6 cm in diameter and 28.5 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 180 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? Incorrect: Your answer is incorrect. times

Explanation / Answer

(a)
Solenoid length L = 28.5 cm
N = 3100 turns
Coil radius r = 1.3 cm gives area A = 5.3 cm2
Relative permeability of the core, mu = 1 (air)
Then the inductance of the solenoid is
L = mu2NA/l
L = 1 * 3100 * 5.3 * 10^-4 / 0.285
L = 5.765 H

(b)
Again apply the above formula
L = mu2NA/l
5.765 = mu2 * 180 * 5.3 * 10^-4 / 0.285
mu2 = 17.222
mu = 4.15
mu/muo = 4.15 / 1 = 4.15

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