A solenoid having 500 turns and length 2 m has radius of 2 cm then self inductance of solenoid
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Given
Total number of turns of the solenoid N = 500
Length of the solenoid l = 2 m
Radius of the coil R = 2 cm = 0.02 m
Now from self inductance formula we have
L = (μ₀ A N²)/ l -----(A)
Reducing the above mentioned given values in (A) we get,
L = {4π ×10⁻⁷ ×2π × 0.02²× 500²}/ 2
L = 0.0394
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