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In a plane electromagnetic wave, the electric field oscillates sinusoidally at a frequency of 2.0 xx 10^(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 = 3 xx 10^(8) m s^(-1).] |
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Answer» SOLUTION :(a) `lamda =(c//v) =1.5 xx10^(-2) m` (b) `B_(0)=(E_(0)//c)=1.6 xx10^(-7) T` (c ) Energy density in E field: `u_(E) =(1//2)epsi_(0) E^(2)` Using `E= cB, and c=(1)/(SQRT(mu_(0) epsi_(0)), u_(E)=u_(B)` |
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