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In a plane electromagnetic wave, the electric field oscillates sin usoidally at a frequency of 2.0xx10^(10) Hz and amplitude 48 V m^(-1).a.What is the wavelength of the wave ?b.What is the amplitude of the oscillating magnetic field ?c.Show that the average energy density of the E field equals the average energy density of the B field. [c=3xx10^(8)ms^(-1)]. |
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Answer» SOLUTION :a.`lambda = (c )/(upsilon)=(3xx10^(8))/(2xx10^(10))=1.5xx10^(-2)m` B.`c=(E_(0))/(B_(0)), B_(0)=(E_(0))/(c )=(48)/(3xx10^(8))=16xx10^(-8)T` c.`U_(E )=(1)/(2)epsilon_(0)E^(2), U_(B)=(1)/(2mu_(0))B^(2)` But `E=cB, c=(1)/(sqrt(mu_(0)epsilon_(0))) therefore U_(E )=U_(B)` |
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