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Suppose that the electric field amplitude of an electromagnetic wave is E_(0)=120 N/C and that its frequency is upsilon = 50.0 MHz.a.Determine, B_(0), omega, k, and lambda.b.Find the expressions for E and B. |
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Answer» SOLUTION :`E=120 NC^(-1), upsilon =50 MHZ = 50xx10^(6)HZ` a.`B_(0)=(E_(0))/(c )=(120)/(3XX10^(8))=4xx10^(-7)T` `omega = 2pi upsilon = 2xx3.14xx50xx10^(6)=3.14xx10^(8)RAD s^(-1)` `c=(omega)/(k)therefore k=(omega)/(c )=(3.14xx10^(8))/(3xx10^(8))=1.05 rad m^(-1)` `lambda = (2pi)/(k)=(2xx3.14)/(1.05)=5.98 m` b.`E=E_(0)sin (k x - omega t)=120 sin (1.05 x-3.14xx10^(8)t)Vm^(-1)` `B=B_(0)sin (k x-omega t)=4xx10^(-7)sin(1.05x -3.14xx10^(8)t)T`. |
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