Answer:
L = 0.109 H
Step-by-step explanation:
Given that,
Number of loops in the solenoid, N = 1500
Radius of the wire, r = 4 cm = 0.04 m
Length of the rod, l = 13 cm = 0.13 m
To find,
Self inductance in the solenoid
Solution,
The expression for the self inductance of the solenoid is given by :
![L=(\mu_o N^2 A)/(l)](https://img.qammunity.org/2020/formulas/physics/college/kl448qioha665jwy8d6zrnt3lugwaaho7m.png)
![L=(4\pi * 10^(-7)* (1500)^2* \pi (0.04)^2)/(0.13)](https://img.qammunity.org/2020/formulas/physics/college/fextfeyvwo8lukpm5jecpmuedi4ueih4c4.png)
L = 0.109 H
So, the self inductance of the solenoid is 0.109 henries.