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  •  An electromagnetic wave travelling in the x-direction has frequency of and electric

 An electromagnetic wave travelling in the x-direction has frequency of 2\times 10^{14}Hz and electric field amplitude of 27 Vm-1. From the options given below, which one describes the magnetic field for this wave ?

Option: 1

\vec{B} (x,t)=(3*10^{-8}T)\vec{j}sin[2\pi(1.5*10^{-8}x-2*10^1^4t)]


Option: 2

\vec{B} (x,t)=(9*10^{-8}T)\vec{k}sin[2\pi(1.5*10^{-6}x-2*10^1^4t)]


Option: 3

\vec{B} (x,t)=(9*10^{-8}T)\vec{i}sin[2\pi(1.5*10^{-8}x-2*10^1^4t)]


Option: 4

\vec{B} (x,t)=(9*10^{-8}T)\vec{j}sin[1.5*10^{-6}x-2*10^1^4t]


Answers (1)

best_answer

Using Maxwell's equation 
\frac{E_{0}}{B_{0}}=c

where c is velocity of light, E?? and B?? are amplitudes of electric and magnetic field.

given, \ \ E_{\circ}=27 N / C$\\ \\ The amplitude of magnetic field is $B_{0}=\frac{27}{3 \times 10^{8}}=9 \times 10^{-8}$ \\ \\ frequency of wave is given as $f=2 X 10^{14} Hz ,$ \\ \\ it's angular frequency is given as $\omega=2 \pi f=2 \pi \times$ $2 X 10^{14}$

The\ equation\ of \ magnetic\ field \ can \ be \ given \ by \ \\ \\ \left.B_{o} \sin (k x-\omega t)\right)=B_{o} \sin \left(2 \pi\left(\frac{x}{\lambda}-f t\right)\right)\\ \\ \\ =9 \times 10^{-8} \sin \left(2 \pi\left(\frac{x}{\lambda}-2 X 10^{14} t\right)\right)(\hat{j} \text { or } \hat{k})$

 

The direction of electromagnetic wave is given as x axis, the direction of magnetic field will be perpendicular to it, hence option B and C can only be the answer, however, in option B the angular frequency is not as calculated above.

Posted by

Suraj Bhandari

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