Answer:
The inductance of the solenoid is 71.98 mH
Explanation:
Given;
number of loops per length of the solenoid, n = 100 loops/cm
radius of the solenoid, r = 4.53 cm
length of the solenoid, L = 8.88 cm
current carried by the solenoid, I = 500 mA
The inductance of the solenoid is calculated as;
[tex]L = \frac{N^2\times \mu_0 \times A}{l}[/tex]
where;
A is the area of the coil
[tex]A =\pi r^2 = \pi (0.0453)2 = 0.00645 \ m^2[/tex]
N is the number of turns, = 100 x 8.88 = 888 loops
l is the length = 0.0888 m
[tex]L = \frac{N^2\times \mu_0 \times A}{l}\\\\L = \frac{(888)^2\times \(4\pi \times 10^{-7}) \times (0.00645)}{0.0888}\\\\L = 0.07198 \ H\\\\L = 71.98 \ mH[/tex]
Therefore, the inductance of the solenoid is 71.98 mH
