Answer:
The self-inductance is [tex]L = 0.0053 \ H[/tex]
Explanation:
From the question we are told that
The number of loops is [tex]N = 900[/tex]
The length of the rod is [tex]l =6 \ cm = 0.06 \ m[/tex]
The radius of the rod is [tex]r = 1 \ cm = 0.01 \ m[/tex]
The self-inductance for the solenoid is mathematically represented as
[tex]L = \frac{\mu_o * A * N^2 }{l}[/tex]
Now the cross-sectional of the solenoid is mathematically evaluated as
[tex]A = \pi r^2[/tex]
substituting values
[tex]A =3.142 * 0.01 ^2[/tex]
[tex]A = 3.142 *10^{-4} \ m^2[/tex]
and [tex]\mu_o[/tex] is the permeability of free space with a value [tex]\mu_o = 4\pi * 10^{-7} N/A^2[/tex]
substituting values into above equation
[tex]L = \frac{ 4\pi * 10^{-7} ^2* 3.142*10^{-4} * 900^2 }{0.06}[/tex]
[tex]L = 0.0053 \ H[/tex]
