On a particular day,the sun delivers an avera | vedclass

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

Question

(B) The average intensity $I$ of an electromagnetic wave is related to the amplitude of the magnetic field $B_0$ by the formula: $I = \frac{B_0^2 C}{2

Vedclass · Accepted Answer

$4 	imes 10^{-6} 	ext{ T}$

Vedclass · Answer

(B) The average intensity $I$ of an electromagnetic wave is related to the amplitude of the magnetic field $B_0$ by the formula: $I = \frac{B_0^2 C}{2 \mu_0}$.Given,$I = \frac{6}{\pi} 	imes 10^3 	ext{ W/m}^2$,$C = 3 	imes 10^8 	ext{ m/s}$,and $\mu_0 = 4\pi 	imes 10^{-7} 	ext{ T m/A}$.Rearranging the formula for $B_0^2$: $B_0^2 = \frac{2 \mu_0 I}{C}$.Substituting the values: $B_0^2 = \frac{2 	imes (4\pi 	imes 10^{-7}) 	imes (\frac{6}{\pi} 	imes 10^3)}{3 	imes 10^8}$.Simplifying the expression: $B_0^2 = \frac{8\pi 	imes 10^{-7} 	imes 6 	imes 10^3}{\pi 	imes 3 	imes 10^8} = \frac{48 	imes 10^{-4}}{3 	imes 10^8} = 16 	imes 10^{-12}$.Taking the square root: $B_0 = \sqrt{16 	imes 10^{-12}} = 4 	imes 10^{-6} 	ext{ T}$.

On a particular day,the sun delivers an average power of $\left(\frac{6}{\pi} \times 10^3\right) \frac{W}{m^2}$ to the top of Earth's atmosphere. Find the amplitude of the magnetic field for the electromagnetic waves above the atmosphere. (Take $\mu_0 = 4\pi \times 10^{-7} \text{ SI units}$)

Similar Questions

$A$ plane electromagnetic wave of frequency $50\, MHz$ travels in free space along the positive $x-$ direction. At a particular point in space and time, $\vec E = 6.3\,\hat j\,V/m$. The corresponding magnetic field $\vec B$ at that point will be:

$A$ charged particle oscillates about its mean equilibrium position with a frequency of $10^{9} \,Hz$. The frequency of electromagnetic waves produced by the oscillator is . . . . . . .

In an electromagnetic wave,the electric field vector and magnetic field vector are given as $\vec{E} = E_{0} \hat{i}$ and $\vec{B} = B_{0} \hat{k}$ respectively. The direction of propagation of the electromagnetic wave is along:

An electromagnetic wave with angular frequency $\omega$ and wavelength $\lambda$ propagates in the $+y$ direction. Its magnetic field is directed along the $-x$ direction. The vector expression for the associated electric field (with amplitude $E_0$) is ...........

In the given electromagnetic wave $E_y = 600 \sin (\omega t - kx) \ Vm^{-1}$,the intensity of the associated light beam is (in $W/m^2$); (Given $\epsilon_0 = 9 \times 10^{-12} \ C^2 N^{-1} m^{-2}$ and $c = 3 \times 10^8 \ m/s$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module