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Suppose that the electric field part of ...

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.

Text Solution

Verified by Experts

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 field part 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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