Intensity of sunlight is observed as 0.092 , | vedclass

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

Question

(D) The intensity $I$ of an electromagnetic wave is related to the peak magnetic field $B_0$ by the formula: $I = \frac{B_0^2 c}{2 \mu_0}$.Since $c =

Vedclass · Accepted Answer

$2.77 	imes 10^{-8} \, T$

Vedclass · Answer

(D) The intensity $I$ of an electromagnetic wave is related to the peak magnetic field $B_0$ by the formula: $I = \frac{B_0^2 c}{2 \mu_0}$.Since $c = \frac{1}{\sqrt{\mu_0 \epsilon_0}}$,we have $\frac{1}{\mu_0} = \epsilon_0 c^2$.Substituting this into the intensity formula: $I = \frac{B_0^2 c}{2} (\epsilon_0 c^2) = \frac{1}{2} \epsilon_0 c^3 B_0^2$.Rearranging for $B_0$: $B_0 = \sqrt{\frac{2I}{\epsilon_0 c^3}}$.Given $I = 0.092 \, W/m^2$,$\epsilon_0 = 8.85 	imes 10^{-12} \, F/m$,and $c = 3 	imes 10^8 \, m/s$:$B_0 = \sqrt{\frac{2 	imes 0.092}{8.85 	imes 10^{-12} 	imes (3 	imes 10^8)^3}}$.$B_0 = \sqrt{\frac{0.184}{8.85 	imes 10^{-12} 	imes 27 	imes 10^{24}}} = \sqrt{\frac{0.184}{238.95 	imes 10^{12}}} = \sqrt{0.00077 	imes 10^{-12}} = \sqrt{7.7 	imes 10^{-16}} \approx 2.77 	imes 10^{-8} \, T$.

Intensity of sunlight is observed as $0.092 \, W/m^2$ at a point in free space. What will be the peak value of the magnetic field at that point? (Given: $\epsilon_0 = 8.85 \times 10^{-12} \, C^2 N^{-1} m^{-2}$,$c = 3 \times 10^8 \, m/s$)

Similar Questions

During the propagation of electromagnetic waves in a medium,

The electric field in a plane electromagnetic wave is given by $E_z = 60 \cos(5x + 1.5 \times 10^9 t) \text{ V/m}$. Then the expression for the corresponding magnetic field is (here subscripts denote the direction of the field):

For plane electromagnetic waves propagating in the positive $Z$-direction,the combination which gives the correct possible direction for $\vec{E}$ and $\vec{B}$ fields respectively is:

If the electric field of a plane electromagnetic wave is $E_z = 60 \sin(0.5 \times 10^3 x + 1.5 \times 10^{11} t) \ Vm^{-1}$,then the magnetic field of the wave is

$A$ plane electromagnetic wave of frequency $25 \text{ MHz}$ travels in free space along the $X$-direction. At a particular point in space and time,the magnetic field is $\overrightarrow{B} = 2.1 \times 10^{-8} \hat{k} \text{ T}$. Find the electric field $\overrightarrow{E}$ at this point.

Vedclass Test Series

Exam Paper Generator

Online Exam Module