Answer:
102.8 μH
Explanation:
The (self) inductance of a coil based on its own geometry is given as
L = (μ₀N²A)/l
where
μ₀ = magnetic constant = (4π × 10⁻⁷) H/m
N = number of turns = 189
A = Cross sectional Area = (πD²/4) = (π×0.00799²/4) = 0.00005014 m²
l = length of the solenoid = 2.19 cm = 0.0219 m
L = (4π × 10⁻⁷ × 189² × 0.00005014)/0.0219
L = 0.0001028131 H = (1.028 × 10⁻⁴) H = (102.8 × 10⁻⁶) H = 102.8 μH
Hope this Helps!!!
