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

An electromagnetic $(EM)$ wave is propagating in a medium with a velocity $\vec{v} = v\hat{i}$. The instantaneous oscillating electric field of this $EM$ wave is along the $+y$ axis. Then the direction of the oscillating magnetic field of the $EM$ wave will be along:

For electromagnetic waves,the electric field and the magnetic field are .......

$A$ radio transmitter transmits at $830 \, kHz$. At a certain distance from the transmitter,the magnetic field has an amplitude of $4.82 \times 10^{-11} \, T$. The electric field and the wavelength are respectively:

The magnetic field of a plane electromagnetic wave is $\overrightarrow{B} = 3 \times 10^{-8} \sin [200 \pi(y + ct)] \hat{i} \, T$,where $c = 3 \times 10^{8} \, m/s$ is the speed of light. The corresponding electric field is:

The radiation energy emitted per second by a point source is $100 \,W$. If the efficiency of the source is $4 \%$, then the rms value of the electric field at a distance of $2 \,m$ is [use $\frac{1}{4 \pi \varepsilon_0}=9 \times 10^9$ in $SI$ units].

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