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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)] |
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Answer» Solution :`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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