InterviewSolution
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InterviewSolution
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In a plane electromagnetic wave, the electric field oscillates sinusoidally at a frequency of `2.0xx10^10 H_z` and amplitude `48V_m^-1` (a) What is the wavelength o f the wave? (b) What is the amplitude of the oscillating magnetic field. (c) Show that the average energy density of the field `E` equals the average energy density of the field `B.[c=3xx10^8 ms^-1]`. |
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Answer» Here, `v=2.0xx10^10Hz,E_0=48Vm^-1, c=3xx10^8ms^-1` (a) Wavelength of the wave, `lambda=c/v =(3xx10^8)/(2.0xx10^10)=1.5xx10^-2m` (b) Ampllitude of the oscillating magnetic field, `B_0=(E_0)/c=48/(3xx10^8)=1.6xx10^-7T` |
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