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Find E_(0), B_(0), intensity I and force exerted on surface for a spherical surface 20 m away from the point source bulb of 2000 W. Efficiency of bulb is 2% and consider the bulb as point source. (epsilon_(0)=8.85xx10^(-12)SI and c=3xx10^(8)ms^(-1)) |
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Answer» SOLUTION :Here, `P=2000 W, R=20 m, eta = 2%` `epsilon_(0)=8.85xx10^(-12)C^(2)N^(-1)m^(-2), c=3xx10^(8)ms^(-1)` Energy per second by the bulb, U = 2% of `P=2000xx0.02=40 W` AREA of circular surface `A=4pi R^(2)=4xx3.14xx(20)^(2)=5024 m^(2)` Intensity of radiation, `I=("incidient radiated energy in one second (U)")/("area (A)")` `therefore I=(40)/(5024)` `therefore I=7.96xx10^(-3)(W)/(m^(2))` `I=[(1)/(epsilon_(0)c)]^((1)/(2))` `E_(rms)=[(7.96xx10^(-3))/(8.85xx10^(-12)xx3xx10^(8))]^((1)/(2))=[(7.96)/(8.85xx3)]^((1)/(2))` `therefore E_(rms)=1.73 NC^(-1)` `E_(0)=sqrt(2)XX E_(rms)` `=1.414xx1.73` `therefore E_(0)=2.45 NC^(-1)` Now,`c=(E_(0))/(B_(0))` `therefore B_(0)=(E_(0))/(c )=(2.45)/(3xx10^(8))` `therefore B_(0)=8.167xx10^(-7)T` Momentum per second gained by the surface means force, `therefore F=(U)/(c )=(40)/(3xx10^(8)) ""[because P=(U)/(c )]` `therefore F=1.333xx10^(-7)N` Energy density on surface `rho=(I)/(c )=(7.96xx10^(-3))/(3xx10^(8))` `therefore rho=2.653xx10^(-11)Jm^(-3)` |
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