An air-filled cylindrical inductor with a inductance of 8 mH has 3500 turns. It

Question

An air-filled cylindrical inductor with a inductance of 8 mH has 3500 turns. It is 3.0 cm in diameter.

(a) What is its length?
cm
(b) Assume a second cylindrical inductor with the same length and turns as the first, and a core filled with a material instead of air, but its diameter is 2.4 cm. 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

a) Inductance = N^2 u0 A / L

8 x 10^-3 H = 3500^2 x (4pi x 10^-7 ) x(pi x 0.015^2) / L

L = 1.36 m

b) for material filled :

Inductance = N^2 u0 ur A / L

8 x 10^-3   = 3500^2 x (4pi x 10^-7) x ur x (pi x0.012^2) / 1.36

ur = 1.56

ur/u0 = 1.56 / (4pi x 10^-7) = 1.24 x 10^6

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