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The magnetic field of a beam emerging from a filter facing a floodlight is given by B=12xx10^(-8)sin (1.20xx10^(7)z-3.60xx10^(15)t) T. What is the average intensity of the beam ? |
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Answer» Solution :Comparing `B=12xx10^(-8)sin(1.20xx10^(7)z-3.60xx10^(15)t)` with equation `B = B_(0)sin omega t` `B_(0)=12xx10^(-8)T` AVERAGE INTENSITY of beam. `I_("average")=(B_(0)^(2))/(2mu_(0))C` `=(1)/(2)xx((12xx10^(-8))^(2)xx3xx10^(8))/(4xx3.14xx10^(-7))` `therefore I_("average")=1.71 W//m^(2)` |
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