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
$L=\mu \frac{N^2}{l} A$
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:
$N= \sqrt{ \frac{lL}{\mu A} }= \sqrt{ \frac{(4.7\cdot 10^{-3}H)(0.3 m)}{(12.56\cdot 10^{-7}NA^{-2})(3.3\cdot10^{-2}m^2)} } =583$