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))`

Explanation

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)`