Question

In 1965 Penzias and Wilson discovered the cosmic microwave radiation left over from the Big Bang expansion of the universe. The energy density of this radiation is 5.69 × 10−14 J/m3. The speed of light 2.99792 × 108 m/s and the permeability of free space is 4π × 10−7 N/A2. Determine the corresponding electric field amplitude. Answer in units of V/m.

Answer 1

electric field amplitude is 0.1133 V/m

Step-by-step explanation:

given data

energy density = 5.69 × 10^−14 J/m3

speed of light = 2.99792 × 10^8 m/s

permeability of free space = 4π × 10^−7 N/A2

to find out

corresponding electric field amplitude

solution

we know electric filed amplitude E is

E = BC ..............1

so first we find magnetic filed B from energy density

that is energy density

u = B²/ 2µ

so B = √2µu

put value

B = √2(4π×
`10^(-7)`×5.69 ×
`10^(-14)`)

B = 3.780645 ×
`10^(-10)`

so from equation 1

E = 3.780645 ×
`10^(-10)` (2.99792 × 10^8)

E = 0.1133

electric field amplitude is 0.1133 V/m