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In a plane electromagnetic wave, the electric field oscillates sinusoidally at a frequency of 2.0xx10^(10) Hz and amplitude 48 V m^(-1). |
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Answer» SOLUTION :Here, v=`2.0xx10^(10) Hz, E_(0)=48 VM^(-1),C=3xx10^(8) m//s` For average energy density `U_(E)=1/2 epsi_(0)E_(0)^(2)………………. (1)` We KNOWN that `(E_(0))/(B_(0))=C` Putting in Eq (1) `U_(E)=(1)/(4)epsi_(0).C^(2)B_(0)^(2)..............(2)` Speed of ELECTRO magnetic waves, `C=(1)/(sqrt(mu_(0)E_(0)))` Putting in Eq (2) We get . `U_(E)=(1)/(4) epsi_(0) B_(0)^(2)* (1)/(mu_(0)epsi_(0))` `U_(E)=1/4 * (B_(o)^(2))/(mu_(0))=(Bo^(2))/(2mu_(0))=mu_(B)` Thus, the average energy density of E field equals the average energy density of B field. |
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