Answer:
L = 0.0319 H
Explanation:
Given that,
Number of loops in the solenoid, N = 900
Radius of the wire, r = 3 cm = 0.03 m
Length of the rod, l = 9 cm = 0.09 m
To find,
Self inductance in the solenoid
Solution,
The expression for the self inductance of the solenoid is given by :
[tex]L=\dfrac{\mu_o N^2 A}{l}[/tex]
[tex]L=\dfrac{4\pi\times10^{-7}\times(900)^{2}\times\pi(0.03)^{2}}{0.09}[/tex]
L = 0.0319 H
So, the self inductance of the solenoid is 0.0319 henries.
