Answer:
a) L = 3.29 10⁻⁴ H, b)U = 5.33 10⁻² J
Explanation:
a) The inductance is a solenoid this given carrier
L = [tex]\frac{N \ \phi_B }{I}[/tex]
The magnetic field inside the solenoid is
B = μ₀ [tex]\frac{N}{l} i[/tex]
hence the magnetic flux
Ф_B = B. A = μ₀ [tex]\frac{N \ A}{l \ i}[/tex]
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
