Question 48: EM waves

Question

In the given electromagnetic wave $$\mathrm{E}_{\mathrm{y}}=600 \sin (\omega t-\mathrm{kx}) \mathrm{Vm}^{-1}$$, intensity of the associated light beam is (in $$\mathrm{W} / \mathrm{m}^2$$ : (Given $$\epsilon_0=9 	imes 10^{-12} \mathrm{C}^2 \mathrm{~N}^{-1} \mathrm{~m}^{-2}$$ )

Accepted Answer

To find the intensity of the given electromagnetic wave, we need to use the formula for the intensity of an electromagnetic wave:
$$ I = \frac{1}{2} \epsilon_0 c E_0^2 $$
where:
$$I$$ is the intensity in $$\mathrm{W} / \mathrm{m}^2$$
$$\epsilon_0$$ is the permittivity of free space, given as $$9 	imes 10^{-12} \mathrm{C}^2 \mathrm{~N}^{-1} \mathrm{~m}^{-2}$$
$$c$$ is the speed of light in vacuum, approximately $$3 	imes 10^8 \mathrm{~m/s}$$
$$E_0$$ is the peak electric field, given as $$600 \mathrm{Vm}^{-1}$$
Now, substitute these values into the formula:
$$ I = \frac{1}{2} 	imes 9 	imes 10^{-12} 	imes 3 	imes 10^8 	imes (600)^2 $$
Simplify the expression step-by-step:
$$ I = \frac{1}{2} 	imes 9 	imes 10^{-12} 	imes 3 	imes 10^8 	imes 360000 $$
First, calculate $$9 	imes 3 	imes 360000$$:
$$ I = \frac{1}{2} 	imes 9.72 	imes 10^{-4} 	imes 360000 $$
Combine 9 and 3 into 27, giving you:
$$ I = 13.5 	imes 10^{-4} 	imes 360000 $$
Then, calculate the multiplication:
$$ I = 13.5 	imes 36 $$
Finally, multiply the remaining values:
$$ I = 486 	imes 10^{-4} $$
The final value is:
The intensity, $$I$$, is 486 $$\mathrm{W} / \mathrm{m}^2$$. Therefore, the correct option is:
Option A: 486

Answer

486

Answer

729

Answer

243

Answer

972

EM waves

In the given electromagnetic wave $\mathrm{E}_{\mathrm{y}}=600 \sin (\omega t-\mathrm{kx}) \mathrm{Vm}^{-1}, intensity of the associated light beam is (in \mathrm{W} / \mathrm{m}^2 : (Given \epsilon_0=9 \times 10^{-12} \mathrm{C}^2 \mathrm{~N}^{-1} \mathrm{~m}^{-2}$ )

Choose the correct answer: