Question

How many turns should a solenoid of cross-sectional area 3.3×10−2 m2 and length 0.30 m have if its inductance is to be 47 mh ?

Answer 1

The inductance of a solenoid is given by

`L=\mu (N^2)/(l) A`
where

`\mu = 12.56 \cdot 10^(-7)NA^(-2)` is the magnetic permittivity
N is the number of turns of the solenoid
l is its lenght
A is its cross-sectional area

By re-arranging the formula and replacing the numbers, we get the number of turns of the solenoid:

`N= \sqrt{ (lL)/(\mu A) }= \sqrt{ ((4.7\cdot 10^(-3)H)(0.3 m))/((12.56\cdot 10^(-7)NA^(-2))(3.3\cdot10^(-2)m^2)) } =583`