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The maximum electric field at a distance...

The maximum electric field at a distance of 10m from an isotropic point source of light is `3.0 Vm^(-1)`. Calculate (a) the maximum value of magnetic field, (b) average intensity of light at that place and (c ) the power of the source. `[c=3xx10^(8)ms^(-1), epsilon_(0)=8.854xx10^(-12)C^(2)N^(-1)m^(-2)]`

Text Solution

Verified by Experts

`E_(0)=3 V//m, c=3xx10^(8)m//c`
`c=(E_(0))/(B_(0))rArr B_(0)=(E_(0))/(c )`
`therefore B_(0)=(3)/(3xx10^(8))=1xx10^(-8)T`
Average intensity of radiation `I=epsilon_(0)c^(2)E_(rms)^(2)`
`E_(rms)^(2)=(E_(0)^(2))/(2)`
`therefore I=(epsilon_(0)cE_(0)^(2))/(2)`
`(epsilon_(0)=8.85xx10^(-12)c^(2)//Nm, E_(0)=3 V//m)`
`therefore I=(8.85xx10^(-2)xx3xx10^(8))xx(2)/(2)`
`=1.195xx10^(-2)W//m^(2)`
Now power P = `("energy")/("time")`
`=("intensity"xx"area"xx"time")/("time")`
`therefore` Power P `=I xx 4pi R^(2) " " [R=10m]`
`=1.195xx4xx3.14xx100xx10^(-2)`
= 15 Watt.
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