Answer:
the number of turns in the winding is 387 turns.
Explanation:
Given;
radius of the solenoid, r = 0.12 m
cross sectional area of the solenoid, A = 20 x 10⁻⁴ m²
current flowing through the solenoid, I = 20 A
the energy of the flowing current, E = 0.1 J
The energy stored in solenoid is given as;
[tex]U = \frac{1}{2} LI^2\\\\L = \frac{2U}{I^2} \\\\L = \frac{2 \times 0.1}{20^2} \\\\L = 5 \times 10^{-4} \ H[/tex]
The number of turns in the winding is calculated as follows;
[tex]L = \frac{\mu_o N^2A}{2\pi r} \\\\N^2 = \frac{2\pi r L}{\mu_o A} \\\\N = \sqrt{\frac{2\pi r L}{\mu_o A}} \\\\N = \sqrt{\frac{2\pi \times 0.12 \times 5\times 10^{-4} }{4\pi \times 10^{-7} \times 20\times 10^{-4} }}\\\\N = 387 \ turns[/tex]
Therefore, the number of turns in the winding is 387 turns.
