Respuesta :
Answer:
Following are the solution to the given points:
Explanation:
For point a:
Following are the expression to the self- inductance and the solenoid:
[tex]L=\frac{\mu_0 N^2 A}{l}[/tex]
[tex]=\frac{\mu_0 (\frac{l}{d})^2 (\pi r^2)}{l}\\\\=\frac{\mu_0 ( \pi(\frac{D}{2})^2)}{d^2}\\\\=\frac{Ld^2}{\mu_0( \pi (\frac{D}{2})^2)}\\\\=\frac{(1.0 \ H)(0.77 \times 10^{-3} \ m)^2}{ 4 \pi \times 10^{-7} \frac{H}{m} (\pi (\frac{13 \times 10^{-2} \ m }{2})^2)}\\\\=35.5 \ m[/tex]
For point b:
Calculting the length of the copper wire:
[tex]L'=N(\pi D)[/tex]
[tex]=\frac{l}{d}(\pi D)\\\\=\frac{35.5 m}{0.77 \times 10^{-3} m}\pi (13\times 10^{-2} m)\\\\=18829.2 \ m(\frac{1 km}{10^3 \ m})\\\\=18.8 \ km[/tex]
For point c:
The resistance of the solenoid is given by,
[tex]R=\frac{\rho L'}{A}[/tex]
[tex]=\frac{\rho L'}{\pi r'^2}\\\\=\frac{\rho L'}{\pi (\frac{d}{2})^2}\\\\=\frac{(1.68\times 10^{-8} \Omega . m) (18829.2 \ m)}{\pi (\frac{0.77 \times 10^{-3} \ m}{2})^2}\\\\=6793 \ \Omega[/tex]
