Answer:
N/l = 104
Explanation:
Energy stored in the inductor is given by the formula
[tex]U = \frac{1}{2}Li^2[/tex]
now we have
[tex]6\times 10^{-6} = \frac{1}{2}L(0.400)^2[/tex]
now we have
[tex]L = 7.5 \times 10^{-5}[/tex]
now we have
[tex]L = \frac{\mu_0 N^2 \pi r^2}{l}[/tex]
[tex]7.5 \times 10^{-5} = \frac{4\pi \times 10^{-7} N^2 \pi(0.05)^2}{0.7}[/tex]
[tex]N = 73 turns[/tex]
now winding density is turns per unit length
[tex]N/l = 104[/tex]
The winding density of the given solenoid is 104 turns per meter.
The formula for energy stored in the inductor can be used to determine the inductance of the solenoid as follows.
U = ¹/₂LI²
6 x 10⁻⁶ = ¹/₂ (L) x (0.4)²
6 x 10⁻⁶ = 0.0.8L
L = 7.5 x 10⁻⁵
The number of turns of the solenoid is calculated as follows;
[tex]L = \frac{\mu N^2\pi r^2}{l} \\\\7.5 \times 10^{-5} = \frac{(4\pi \times 10^{-7} ) \times N^2 \times \pi(0.05)^2}{0.7} \\\\N = 73 \ turns[/tex]
The winding density if the number of turns per length
N/l = 73/0.7
N/l = 104
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