Question

If a beam of electromagnetic radiation has an intensity of 120 W/m2, what is the maximum value of the electric field?

Answer 1

Answer:

Electric field, E = 300.65 N/C

Explanation:

Given that,

Intensity of a beam of electromagnetic radiation, $I=120\ W/m^2$

We need to find the maximum value of the electric field. The intensity of electromagnetic wave in terms of electric field is given by :

$I=\dfrac{1}{2}\epsilon_oE^2c$

c is the speed of light

$E=\sqrt{\dfrac{2I}{\epsilon_o c}}$

$E=\sqrt{\dfrac{2\times 120}{8.85\times 10^{-12}\times 3\times 10^8}}$

E = 300.65 N/C

So, the maximum value of the electric field is 300.65 N/C. Hence, this is the required solution.

Answer 2

The maximum value of the electric field for the given beam of electromagnetic radiation.

the electric field of an electromagnetic radiation can be calculated by,

$\bold {I = \dfrac 12 \epsilon_0 E^2C}$

$\bold {E = \sqrt {\dfrac {2I}{\epsilon_0 C}}}$

Where,

I - intensity of the beam = 120 W/m2

C- speed of light = $\bold { 3x10^8\ m/s}}$

E - electric field

Put the value of the formula

$\bold {E = \sqrt {\dfrac {2\times 20 }{8.85x10^{-12} \times 3x10^8}}}\\\\\bold {E = 300.65\ N/C}$

Therefore, the maximum value of the electric field for the given beam of electromagnetic radiation.

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