An air-filled cylindrical inductor with a inductance of 24 mH has 2700 turns. It

Question

An air-filled cylindrical inductor with a inductance of 24 mH has 2700 turns. It is 2.4 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 1.2 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

Solenoid length 28.5 cm with N = 2600 turns,
Coil dia=2.4 cm,

A = pi*2.4^2 cm2.

=50.24 cm^2
Relative permeability of the core k = 1 (air)
Indunctance =N^2*mu*A/l...(1)

Lenght of the inductor,

I=2700^2*4*pi*10^-7*50.24*10^-4/24*10^-3

=191.6 m

b.

b) as per equation 1
/0 =A0/A =d0^2/d^2= (2.4/1.2)^2= 4

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.