Question

A solenoid with 3,000.0 turns is 70.0 cm long. If its self-inductance is 25.0 mH, what is 5) its radius? (The value of o is 4 × 10-7 N/A2.)

Answer 1

The self-inductance of a solenoid is given by:
$L= \frac{\mu_0 N^2 A}{l}$
where
$\mu_0$ is the vacuum permeability
N is the number of turns
A is the cross-sectional area of the solenoid
l is the length of the solenoid

For the solenoid in our problem, N=3000, l=70.0 cm=0.70 m and the self-inductance is L=25.0 mH=0.025 H, therefore the cross-sectional area is
$A= \frac{Ll}{\mu N^2}= \frac{(0.025 H)(0.70 m)}{(4\pi \cdot 10^{-7}N/A^2)(3000)^2}= 1.55 \cdot 10^{-3}m^2$
And since the area is related to the radius by
$A=\pi r^2$
The radius of the solenoid is
$r= \sqrt{ \frac{A}{\pi} } = \sqrt{ \frac{1.55 \cdot 10^{-3} m^2}{\pi} } =0.022 m=2.2 cm$