Calculate the electric and magnetic fields produced by the radiation coming from a 100 W bulb at a distance of 3 m. Assume that the efficiency of the bulb is 2.5% and it is a point source.

Explanation

Here efficiency of given bulb is 2.5%. Hence power of radiation emitted from it
= 2.5 % of electrical power
`=100xx(2.5)/(100)`
`therefore P =2.5 W`
Now, intensity of radiation,
`I=(P)/(A)=(P)/(4pi R^(2))`
`therefore I=(2.5)/((4)(3.14)(3)^(2))=0.02212(W)/(m^(2))`
According to formula,
`I=in_(0)E_(rms)^(2)c`
`therefore E_(rms)^(2)=(I)/(in_(0)c)`
`therefore E_(rms)=sqrt((I)/(in_(0)c))`
`therefore E_(rms)=sqrt((0.02212)/(8.85xx10^(-12)xx3xx10^(8)))`
`=sqrt((221.2)/(8.85xx3))`
`=sqrt(8.33145)`
`therefore E_(rms)=2.8864 Vm^(-1)` (Ans 1)
According to formula,
`I=(B_(rms)^(2)c)/(mu_(0))`
`therefore B_(rms)^(2)=(I mu_(0))/(c )`
`therefore B_(rms)=sqrt((I mu_(0))/(c ))`
`therefore B_(rms)=sqrt((0.02212xx4xx3.14xx10^(-7))/(3xx10^(8)))`
`=sqrt((2212xx12.56xx10^(-20))/(3))`
`=96.23xx10^(-10)T`
`therefore B_(rms)=9.623xx10^(-9)T`
Note : Here amplitudes of oscillations of electric field and magnetic field can be found out as follows :
`E_(rms)=(E_(0))/(sqrt(2))rArr E_(0)=sqrt(2)E_(rms)`
`therefore E_(0)=(1.414)(2.8864)`
`therefore E_(0)=4.0814 Vm^(-1)`
`B_(rms)=(B_(0))/(sqrt(2))rArr B_(0)=sqrt(2)B_(rms)`
`therefore B_(0)=(1.414)(9.623xx10^(-9))`
`therefore B_(0)=13.61xx10^(-9)T=1.361xx10^(-8)T`.