An air-filled cylindrical inductor with a inductance of 10 mH has 2600 turns. It

Question

An air-filled cylindrical inductor with a inductance of 10 mH has 2600 turns. It is 2.2 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 2.0 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

Hi,

In this case would be useful to remember the cancept of inductance as well as the method to calculate it.

The inductance is a property of the inductor and depends on its geometry and the magnetic permeability of the material used as core. This can be observed with the following equation:

L = u N2 A/l (1); where l is the length of the inductor, N is the number of turns, L is the inductance, A is the area of the cross section of the inductor and u is the absolute magnetic permeability which can be calculated as:

u = ur uo (2); where uo is the magnetic permeability of the free space and ur is the relative magnetic permeability. When you have an inductor with an empty core (for instance, an air-filled core) you can assume that ur = 1.

By the way, uo = 4*10-7 N/A2

(a) We find the length of the inductor using (1):

l = (4*10-7 N/A2)*(2600)2*(*(1.1*10-2)2) / (10*10-3 H) = 0.32 m = 32 cm

(b) For the second question, we use first (1) and then (2):

u = (10*10-3 H)*(0.32 m) / [ (1*10-2 m)2*(2600)2 ] = 1.507*10-6 N/A2

ur = u/ur = (1.507*10-6 N/A2) / (4*10-7 N/A2) = 1.2

I hope it helps.

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