The electric field of an electromagnetic wave in vacuum is given by the equation $E_x = 30 \sin(ky - \omega t) \text{ V/m}$ . The average intensity of the electromagnetic wave is nearly:
(Take $c = 3 \times 10^8 \text{ m/s}$ and $\varepsilon_0 = 8.85 \times 10^{-12} \text{ C}^2/\text{N m}^2$ )

Question

The electric field of an electromagnetic wave in vacuum is given by the equation

E_x = 30 \sin(ky - \omega t) \text{ V/m}

. The average intensity of the electromagnetic wave is nearly:
(Take

c = 3 \times 10^8 \text{ m/s}

and

\varepsilon_0 = 8.85 \times 10^{-12} \text{ C}^2/\text{N m}^2

)

Quantrex Academy · Accepted Answer

Correct answer: B. 1.2 W/m^2

From the given equation, the amplitude of the electric field is

E_0 = 30 \text{ V/m}

.

The average intensity of an electromagnetic wave is given by the formula:

I = \dfrac{1}{2} c \varepsilon_0 E_0^2

Substituting the standard values:

I = \dfrac{1}{2} \times (3 \times 10^8) \times (8.85 \times 10^{-12}) \times (30)^2

I = 1.5 \times 10^8 \times 8.85 \times 10^{-12} \times 900

I = 1350 \times 8.85 \times 10^{-4}

I = 1.19475 \text{ W/m}^2 \approx 1.2 \text{ W/m}^2

Note: Forgetting the factor of

\dfrac{1}{2}

leads to the peak intensity (

2.4 \text{ W/m}^2

), which is a common error.

Answer:

1.2 \text{ W/m}^2

The electric field of an electromagnetic wave in vacuum is given by...