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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=50.0MHz. (a) Determine B_(0),omega,k and lambda (b) Find expressions for vecE and vecB. |
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Answer» Solution :Here `E_(0)=120NC^(-1) and v=50.0MHz=50xx10^(6)HZ` (a) `B_(0)=(E_(0))/(c )=(120)/(3xx10^(8))=4xx10^(-7)T or 400nT` Angular FREQUENCY `omega=2piv=2xx3.14xx50xx10^(6)=3.14xx10^(8)"rad s"^(-1)` Wavelength `lambda=(c )/(v)=(3xx10^(8))/(50xx10^(6))=6m` Propagation constant `k=(2pi)/(lambda)=(2xx3.14)/(6)="1.05 rad m"^(-1)` (b) If electromagnetic wave is being PROPAGATED along x - axis, then `vecE=E_(0)sin(kx-omegat)hatj=120sin(1.05x-3.14xx10^(8)t)hatj` `and""vecB=B_(0)sin(kx-omegat)hatk=4xx10^(-7)sin(1.05x-3.14xx10^(8)t)hatj` and`""vecB=B_(0)sin(kx-omegat)hatk=4xx10^(-7)sin(1.05x3.14xx10^(8)t)hatk` |
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