Answer: b) 3.47 nj
Step-by-step explanation:
Given that;
length l = 5m
radius of inner conductor r = 10cm = 0.1m
radius of outer conductor D = 20cm = 0.2m
current I = 100A = 100×10⁻³ = 0.1
medium between conductor in air u₀ = 4π × 10⁻⁷
Energy in a coaxial cable transmission line is
w = u₀ /2π I² en(b/a)
we substitute
L = 4π × 10⁻⁷ /4π ×10⁻²× 5 en (20/10)
L =3.4657 × 10⁻⁹ J
L = 3.4657 nJ ≈ 3.47 nJ