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.

Question

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Respuesta :

DeniceSandidge DeniceSandidge · Answer 1 · 2019-08-19T12:06:32+02:00

Answer:

electric ﬁeld amplitude is 0.1133 V/m

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×[tex]10^{-7}[/tex]×5.69 × [tex]10^{-14}[/tex])

B = 3.780645 × [tex]10^{-10}[/tex]

so from equation 1

E = 3.780645 × [tex]10^{-10}[/tex] (2.99792 × 10^8)

E = 0.1133

electric ﬁeld amplitude is 0.1133 V/m