Answer:
The number of turns of the wire needed is 243 turns
Explanation:
Given;
self inductance of the solenoid, L = 1.8 mH
cross sectional area of the inductor, A = 1.6 x 10⁻³ m²
length of the inductor, l = 0.066 m
The self inductance of long solenoid is given by;
L = μ₀n²Al
where;
μ₀ is permeability of free space = 4π x 10⁻⁷ H/m
n is number of turns per length
A is the area of the solenoid
l is length of the solenoid
[tex]n = \sqrt{\frac{L}{\mu_o Al} } \\\\n = \sqrt{\frac{1.8*10^{-3}}{(4\pi*10^{-7}) (1.6*10^{-3})(0.066)} } \\\\n = \sqrt{13562583.184} \\\\n = 3682.74 \ turns/m[/tex]
The number of turns is given by;
N = nL
N = (3682.74)(0.066)
N = 243 turns
Therefore, the number of turns of the wire needed is 243 turns
