A lamp emits monochromatic green light unifor | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The power of the electromagnetic radiation emitted by the lamp is $P = 100\,W 	imes 0.03 = 3\,W$.The intensity $I$ at a distance $r = 5\,m$ is gi

Vedclass · Accepted Answer

$2.68$

Vedclass · Answer

(B) The power of the electromagnetic radiation emitted by the lamp is $P = 100\,W 	imes 0.03 = 3\,W$.The intensity $I$ at a distance $r = 5\,m$ is given by $I = \frac{P}{4\pi r^2} = \frac{3}{4\pi (5)^2} = \frac{3}{100\pi}\,W/m^2$.The relationship between intensity and the amplitude of the electric field $E_0$ is $I = \frac{1}{2} c \varepsilon_0 E_0^2$.Rearranging for $E_0$,we get $E_0 = \sqrt{\frac{2I}{c \varepsilon_0}}$.Substituting the values $c = 3 	imes 10^8\,m/s$ and $\varepsilon_0 = \frac{1}{4\pi 	imes 9 	imes 10^9}$,we have $E_0 = \sqrt{\frac{2 	imes (3 / 100\pi)}{(3 	imes 10^8) 	imes (1 / (4\pi 	imes 9 	imes 10^9))}}$.Simplifying the expression: $E_0 = \sqrt{\frac{6}{100\pi} 	imes (4\pi 	imes 9 	imes 10^9) / (3 	imes 10^8)} = \sqrt{\frac{6 	imes 36 	imes 10^9}{100 	imes 3 	imes 10^8}} = \sqrt{\frac{216 	imes 10}{300}} = \sqrt{7.2} \approx 2.68\,V/m$.

Similar Questions

Suppose that the electric field amplitude of an electromagnetic wave is $E_{0}=120 \text{ NC}^{-1}$ and its frequency $f=50 \text{ MHz}$. Then,which of the following values is incorrectly computed?

The electric field of an electromagnetic wave in free space is $\overrightarrow{E}=57 \cos \left[7.5 \times 10^6 t-5 \times 10^{-3}(3 x+4 y)\right]\ (4 \hat{i}-3 \hat{j})\ N/C$. The associated magnetic field in Tesla is

What is the dimension of $\frac{1}{\mu_0 \varepsilon_0}$? ($\mu_0 =$ magnetic permeability and $\varepsilon_0 =$ permittivity of free space)

An electromagnetic wave is propagating in vacuum along $-\hat{j}$ direction. The magnetic field of the wave is given by $\vec{B} = (2 \times 10^{-8}) \cos [\pi \times 10^{15}(t + \frac{y}{c})] \hat{k} \text{ T}$. The electric field $\vec{E}$ of this wave is $(c = \text{speed of light})$

The electric field in an electromagnetic wave is given by $E = 50 \sin \frac{2 \pi}{\lambda}(ct - x) \ N/C$. Find the energy contained in a cylindrical region of cross-section $10 \ cm^2$ and length $50 \ cm$ along the $x$-axis.

Vedclass Test Series

Exam Paper Generator

Online Exam Module