- Home
- Standard 12
- Physics
8.Electromagnetic waves
medium
A radio transmitter transmits at $830\, kHz$. At a certain distance from the transmitter magnetic field has amplitude $4.82\times10^{-11}\,T$. The electric field and the wavelength are respectively
A
$0.014\,N/C, 36\,m$
B
$0.14\,N/C, 36\,m$
C
$0.14\,N/C, 360\,m$
D
$0.014\,N/C, 360\,m$
(AIEEE-2012)
Solution
Frequency of $EM$ wave $v=830 \,\mathrm{KHz}$
$=830 \times 10^{3} \,\mathrm{Hz}$
Magnetic field. $B=4.82 \times 10^{-11}\, \mathrm{T}$
As we know. frequency, $v=\frac{c}{\lambda}$ or
$\lambda=\frac{c}{r}=\frac{3 \times 10^{8}}{830 \times 10^{2}}$
$\lambda =360\, \mathrm{m}$
$\text { And. } E =B C=4.82 \times 10^{-11} \times 3 \times 10^{8} $
$=0.014 \,\mathrm{N} / \mathrm{C} $
Standard 12
Physics
