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 $\mu_0 = 4\pi \times 10^{-7} \text{ SI units}$)

  • A
    $5 \times 10^{-5} \text{ T}$
  • B
    $4 \times 10^{-6} \text{ T}$
  • C
    $6 \times 10^{-6} \text{ T}$
  • D
    $3 \times 10^{-5} \text{ T}$

Explore More

Similar Questions

If the electric field intensity of a uniform plane electromagnetic wave is given as $E = -301.6 \sin (kz - \omega t) \hat{a}_{x} + 452.4 \sin (kz - \omega t) \hat{a}_{y} \text{ V/m}$. Then,the magnetic intensity $H$ of this wave in $\text{A/m}$ will be (Given: Speed of light in vacuum $c = 3 \times 10^{8} \text{ m/s}$,permeability of vacuum $\mu_{0} = 4\pi \times 10^{-7} \text{ N/A}^{2}$)

The relationship between the amplitude of the electric field $(E_0)$ and the magnetic field $(B_0)$ in an electromagnetic wave is given by:

Select the correct statement from the following:

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:

Light is an electromagnetic wave. Its speed in vacuum is given by the expression

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo