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An electromagnetic wave of electric field E=10 sin (omega t-Kx)N//C is incident normal to the cross - sectional area of a cylinder of 10 cm^(2) and having length 100 cm, lying along X - axis. Find (a) the energy density,(b) energy contained in the cylinder, (c ) the intensity of the wave, (d) momentum transferred to the cross - sectional area of the cylinder in 1 s, considering total absorption, (e ) radiation pressure.[epsilon_(0)=8.854xx10^(12)C^(2)N^(-1)m^(-2), c=3xx10^(8)ms^(-1)] |
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Answer» Solution :For relative cylinder, area `A=10cm^(2)=10xx10^(-4)m^(2)` length l = 100 cm = 1 meter (unit length) For the electromagnetic WAVE incident perpendicularly, `E=10 sin(omega t-kx)N//C` (a)Energy density `rho = epsilon_(0)E_(rms)^(2)` `=epsilon_(0)(E_(0)^(2))/(2) ""(E_(0)=10 N//C)` `therefore rho = (8.85xx10^(-12)xx100)/(2)=4.425xx10^(-10)J//m^(3)` (b)Energy contained in the cylinder U = energy density `xx` volume `therefore U = 4.425xx10^(-10)xx10xx10^(-4)XX1` `=4.425xx10^(-13)J` (c ) Intensity `I = rho` `therefore I=4.425xx10^(-10)xx3xx10^(8)` `=1.3275xx10^(-1)Wm^(-2)` (d)MOMENTUM transferred to the cross - section area of the cylinder in 1 sec considering total absorption (means forces) `P=(rho Al)/(c )=(4.425xx10^(-10)xx10^(-3)xx1)/(3xx10^(8))` `=1.475xx10^(-21)N` (e ) Radiation pressure `=("force")/("area")=(1.475xx10^(-21))/(10^(-3))` `=1.475xx10^(-18)Nm^(-2)`. |
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