Question

. A solenoid coil consists of a single layer of 250 circular turns of wire with each turn having a 0.02m radius. The axial length of the coil is 0.3m. The coil is self-supporting, containing only air. (a) Determine the inductance of the coil, assuming that the magnetic field intensity is uniform inside the coil and zero elsewhere. (b) Find the stored energy in the magnetic field of the coil when the coil current is 18A.

Answer 1

a) L = 3.29 10⁻⁴ H, b)U = 5.33 10⁻² J

Step-by-step explanation:

a) The inductance is a solenoid this given carrier

L =
`(N \ \phi_B )/(I)`

The magnetic field inside the solenoid is

B = μ₀
`(N)/(l) i`

hence the magnetic flux

Ф_B = B. A = μ₀
`(N \ A)/(l \ i)`

we substitute in the expression of inductance

L = N² μ₀ A /l

let's find the area of ​​each turn

A = π r²

A = π 0.02²

A = 1.2566 10⁻³ m²

let's calculate

L = 250² 4π 10⁻⁷ 1.2566 10⁻² / 0.3

L = 3.29 10⁻⁴ H

b) The stored energy is

U = ½ L i²

let's calculate

U = ½ 3.29 10⁻⁴ 18²

U = 5.33 10⁻² J