InterviewSolution
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InterviewSolution
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Suppose that the electric field part of an electromagnetic wave in vacuume is `E=3.1N//C cos [(1.8rad//m)y+(5.4xx10^8rad//s)t]hati` (a) Wavelength is the direction of motion? (b) What is the wavelength `lambda`? (c) What is the frequency v? (d) What is the amplitude of the magnetic field part of the wave? (e) Write an expression for the magnetic field part of the wave. |
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Answer» (a) From the given equation, it is clear that the direction of motion of e.m. wave is along negative y direction i.e., along `-hatj`. (b) Comparing the given equation with the equation `E=E_0cos (ky+omegat)`, we have `k=1.8rad//s, omega=5.4xx10^8rad//s, E_0=3.1 N//C` `lambda=(2pi)/k=(2xx(22//7))/1.8=3.492m~~3.5m` (c) `v=omega/(2pi) =(5.4xx10^8)/(2xx(22//7))=8.5xx10^6~~86MHz` (d) `B_0=(E_0)/c=3.1/(3xx10^8)=1.03xx10^-8T~~10.3nT` (e) `B=B_0cos (ky+omegat)hatk=(10.3nT)cos [(1.8rad//m)y+(5.4xx10^8rad//s)t]hatk` |
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