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An observer is at 2 m from an isotropic ...

An observer is at 2 m from an isotropic point source of light emitting 40 W power. What are the rms values of the electric and magnetic fields due to the source at the position of the observer ? `[c=3xx10^(8)ms^(-1), epsilon_(0)=8.854xx10^(-12)C^(2)N^(-1)m^(-2)]`

Text Solution

Verified by Experts

Intensity `I=(P)/(A)`
`therefore epsilon_(0)c E_(rms)^(2)=(P)/(4pi R^(2))`
`therefore E_(rms)=sqrt((P)/(4pi R^(2)epsilon_(0)c)) " " (P=40 W, R=2m)`
`therefore E_(rms)=sqrt((40)/(12.56xx4xx8.85xx3xx10^(-12)xx10^(8)))`
`=sqrt((10xx10^(4))/(331.875))`
`=17.3 Vm^(-1)`
Now, `B_(rms)=(E_(rms))/(c )=(17.3)/(3)xx10^(-8)=5.77xx10^(-8)T`.
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