Em Waves Question 147
Question: In a plane electromagnetic wave, the electric field oscillates sinusoidally at a frequency of $ 2.0\times 10^{10} $ Hz and amplitude $ 48,V,{{m}^{-1}} $ . Then
Options:
A) the wavelength of the wave is $ 1.5\times {{10}^{-5}}m $
B) the amplitude of the oscillating magnetic field is $ 16\times {{10}^{-3}}T $
C) the average energy density of the E is equal to the average energy density of the B. $ [c=3\times 10^{8},m,{{s}^{-1}}.] $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
[c] Here c $ =3\times 10^{8}m/s $ , $ v=2\times 10^{10},Hz $ , $ E _{0}=48V/m $
$ \lambda =? $ , $ B _{0}=? $ , $ \overrightarrow{u _{E}}=? $ , $ \overrightarrow{u _{B}}=? $
$ \because c=v\lambda $
$ \Rightarrow ,\lambda =\frac{c}{v}=\frac{3\times 10^{8}}{2\times 10^{10}}=1.5\times {{10}^{-2}}m $
$ \frac{E _{0}}{B _{0}}=c\Rightarrow B _{0}=\frac{48}{3\times 10^{8}}=16\times {{10}^{-8}}T $ . Energy density due to electric field and magnetic field $ u _{E}=\frac{1}{2}{\varepsilon _{o}}E^{2},\And ,u _{B}=\frac{1}{2}{\mu _{o}}B^{2} $ We have $ E=cB $ and $ C^{2}=\frac{1}{{\mu _{o}}{\varepsilon _{o}}} $ so, $ U _{E}=\frac{1}{2}{\varepsilon _{o}}{{(cB)}^{2}}=\frac{1}{2}{\varepsilon _{o}}( \frac{1}{{\mu _{o}}{\varepsilon _{o}}} )B^{2}=\frac{1}{2{\mu _{o}}}B^{2}={U _{B.}} $
BETA
