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Suppose that the electric field amplitude of an electromagnetic wave is E_(0)=120N/C and that its frequency is v=50.0 MHz. Determine. B_(0), omega, k and lambda |
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Answer» Solution :`B_(0)=(E_(0))/(C)=(120)/(3XX10^(8))=4XX10^(-7)T` `omega=2piv=2xx3.14xx(50xx10^(6))=3.14xx10^(8)` rad/s K`=(omega)/(C)=(3.14)/(C)=(3.14xx10^(8))/(3xx10^(8))=1.05`rad/m `LAMBDA=(C)/(V)=(3xx10^(8))/(50xx10^(6))` =6.00 m |
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