Question

A solenoid having 500 turns and length 2 m has radius of 2 cm then self inductance of solenoid

Answer 1

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

Answer 2

Answer:

See Above the answer it attached