Question

An air-filled cylindrical inductor with a inductance of 28 mH has 3300 turns. It is 2.8 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 2.8 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 μ/μ₀for this material?
_____ times

Answer 1

Solenoid length L = ?

NO. of turns N = 3300 turns,

Coil diameter D = 2.8 cm

Radius R = D/2 = 2.6/2 = 1.4 cm

A = pi r^2 = 3.14 * 0.014* 0.014 = 0.000615 m^2

Relative permeability of the core k = 1 (air)

Indunctance =N^2*mu*A/L

L = 3300*3300 * 4*3.14 e-7 * 0.000615/(28e -3)

L = 30 cm
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b.

μ/μ0 =A0/A

u/uo =d0^2/d^2

u/uo = 2.8^2/2.8^2

u/uo = 1