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Question

A plane electromagnetic wave, has a frequency of 2.0×1010Hz and its energy density is 1.02×108Jm3 in vacuum. The amplitude of the magnetic field of the wave is close

(14πε0=9×109Nm2C2and speed of light =3x108ms-1)


A

160nT

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B

150nT

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C

180nT

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D

190nT

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Solution

The correct option is A

160nT


Step 1. Given Data:

Electromagnetic wave frequency f=2.0×1010Hz

Speed of light C=3x108ms

Energy density U=1.02×108Jm3

Assume,14πε0=9×109Nm2C2

Let B is Magnetic Field.

Step 2. Energy Density and Magnetic Field,

We know that, Energy density,

U=B22μ01 (where,μ0 is the permeability of free space)

And, Relation between the permeability of free space μ0 and permittivity of free space ϵ0 is,

C=1μ0ϵ0

μ0=1C2ε02

Step 3. To calculate B:

From equation 1 we can write,

U=B22μ0B=U×2μ03

Using equation 2 in equation 3 we have,

B=U×21C2ε0TB=1.02×108×2×19×1016×4π×9×109TB=1.02×108×2×4π×10-7TB=1.02×108×2×4×3.14×10-7TB=25.6×10-15TB=256×10-16TB=16×10-8TB=160×10-9TB=160nT

Therefore, B=160nT

Hence, the correct option is (A).


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