Asolenoid that is 94.9 cm long has a cross-sectional area of 23.0 cm2. there are 809 turns of wire carrying a current of 9.70 a. (a) calculate the energy density of the magnetic field inside the solenoid. (b) find the total energy in joules stored in the magnetic field there (neglect end effects).

a) 43.2 Jm⁻³

b) 0.0941 J

Explanation:

A = area of cross-section = 23 cm² = 23 x 10⁻⁴ m²

l = length of solenoid = 94.9 cm = 0.949 m

L = inductance of solenoid

N = Number of turns = 809

i = current = 9.70 A

Inductance is given as

L = μ₀ N² A l⁻¹

L = (12.56 x 10⁻⁷) (809)² (23 x 10⁻⁴) (0.949)⁻¹

L = 0.002

V = Volume = Al = (23 x 10⁻⁴) (0.949) = 21.8 x 10⁻⁴ m³

Energy density is given as

U = (0.5) L i² V⁻¹

U =  (0.5) (0.002) (9.70)² (21.8 x 10⁻⁴)⁻¹

U = 43.2 Jm⁻³

b)

Total energy is given as

E = (0.5) L i²

E = (0.5) (0.002) (9.70)²

E = 0.0941 J

found in the nucleus

has mass of one amu