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Intensity: Radiation of a single frequency reaches the upper atmosphere of the earth with an intensity of 1350 W/m2. What is the maximum value of the electric field associated with this radiation? (c = 3.00 × 108 m/s, μ0 = 4π × 10-7 T ∙ m/A, ε0 = 8.85 × 10-12 C2/N ∙ m2)

Respuesta :

Cricetus

Answer:

The right answer is "1010 V/m".

Explanation:

The given values are:

Intensity,

[tex]I=1350 \ W/m^2[/tex]

[tex]c = 3.00\times 108 \ m/s[/tex]

[tex]\mu_0= 4\pi\times 10^{-7} \ T.m/A[/tex]

Now,

The electric field's maximum value will be:

=  [tex]\sqrt{2\times u\times c\times I}[/tex]

On substituting the values in the above formula, we get

=  [tex]\sqrt{2\times 4\times \pi\times 10^{-7}\times 3\times 10^8\times 1350}[/tex]

=  [tex]\sqrt{32400\times 3.14\times 10^{-7}\times 10^8}[/tex]

=  [tex]1010 \ V/m[/tex]

batolisis

The maximum value of the electric field associated with this radiation is :

  • 1010 V/m

Given data :

Intensity ( I ) = 1350 W/m²

C = 3.00 * 10⁸ m/s

The maximum value of the electric field can be calculated using the equation below

[tex]E_{max} = \sqrt{2*u*c*I}[/tex]  ----- ( 1 )

where :  I = 1350 W/m²,  μ =  4π * 10⁻⁷,  c = 3.00 * 10⁸

Insert values into equation ( 1 )

∴ [tex]E_{max}[/tex] = 1010 V/m.

Hence we can conclude that the maximum value of the electric field is 1010 V/m .

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