The ratio of contributions made by the electr | vedclass

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

Question

Intensity of electromagnetic wave, $\mathrm{I}=\mathrm{U}_{\mathrm{av}} \mathrm{c}$

In terms of electric field, $\mathrm{U}_{\mathrm{av}}=\frac{1}{

Vedclass · Accepted Answer

$1:1$

Vedclass · Answer

Intensity of electromagnetic wave, $\mathrm{I}=\mathrm{U}_{\mathrm{av}} \mathrm{c}$

In terms of electric field, $\mathrm{U}_{\mathrm{av}}=\frac{1}{2} \varepsilon_{0} \mathrm{E}_{0}^{2}$

In terms of magnetic field $\mathrm{U}_{\mathrm{av}}=\frac{1}{2} \frac{\mathrm{B}_{0}^{2}}{\mu_{0}}$

Now, $U_{a v}$ (due to electric field) $=\frac{1}{2} \varepsilon_{0} E_{0}^{2}$

$=\frac{1}{2} \varepsilon_{0}\left(\mathrm{cB}_{0}ight)^{2}=\frac{1}{2} \varepsilon_{0} 	imes \frac{1}{\mu_{0} \varepsilon_{0}} \mathrm{B}_{0}^{2} \quad\left(\because \frac{\mathrm{E}_{0}}{\mathrm{B}_{0}}=\mathrm{c}ight)$

$=\frac{1 \mathrm{B}_{0}^{2}}{2 \mu_{0}}=\mathrm{U}_{\mathrm{av}}$ (due to magnetic field)

Therefore, the ratio of contributions by the electric field and magnetic field components to the intensity of electromagnetic wave is $1: 1 .$

The ratio of contributions made by the electric field and magnetic filed components to the intensity of an electromagnetic wave is

Similar Questions

Which of the following is $NOT$ true for electromagnetic waves ?

A flood light is covered with a filter that transmits red light. The electric field of the emerging beam is represented by a sinusoidal plane wave

$E_x=36\,sin\,(1.20 \times 10^7z -3.6 \times 10^{15}\,t)\,V/m$

The average intensity of the beam will be.....$W/m^2$

The kinetic energy possessed by a body of mass $m$ moving with a velocity $ v$ is equal to $\frac{1}{2}m{v^2}$, provided

The electric field component of a monochromatic radiation is given by

$\vec E = 2{E_0}\,\hat i\,\cos\, kz\,\cos\, \omega t$

Its magnetic field $\vec B$ is then given by

Light with an energy flux of $25 \times {10^4}$ $W/m^2$ falls on a perfectly reflecting surface at normal incidence. If the surface area is $15\,\, cm^2$ the average force exerted on the surface is