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Suppose that the electric field part of an electromagnetic wave in vacuum is vecE={(3.1N//C)cos[(1.8"rad/m")y+(5.4xx10^(6)"rad/s")t]}hati. (a) What is the direction of propagation? (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» Solution :Here electric field part is of the form `vecE=E_(0)cos(ky+omegat)hati` where `E_(0)=3.1N//, K=1.8" rad/m and "omega=5.4xx10^(6)"rad/s"` (a) The equation for `vecE` clearly shows that dreiction of PROPAGATION is along negative DIRECTION of y - axis. (a) The equation for `vecE` clearly shows that direction of propagation is along negative direction of y - axis. (b) `lambda=(2pi)/(k)=(2xx3.14)/(1.8)=3.5m` (c ) Frequency `v=(omega)/(2pi)=(5.4xx10^(6))/(2xx3.14)=8.6xx10^(5)Hz or 0.86MHz` (d) Amplitude of the MAGNETIC fieldpart of the wave `B_(0)=(E_(0))/(c )=(3.1)/(3xx10^(8))=1.033xx10^(-8)T=10nT.` (e) As `vecE` is along - ve y - axis, hence `vecB` must be along - ve - z axis and, therefore, expression for the magnetic field part of the wave should be `vecB=B_(0)cos(ky+omegat)hatk` `rArr""vecB=(10nT)cos[(1.8"rad/m")y+(5.4xx10^(6)"rad/s")t]hatk`. |
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