1.

An electromagnetic wave going through vacuum is described by` E= E_0 sin(kx- omega t), B=B_0sin(kx-omega t)`. ThenA. `E_(0)k=B_(0)omega`B. `E_(0)omega=B_(0)k`C. `E_(0)B_(0)=omega k`D. `None of these

Answer» Correct Answer - a
`E_(0)/(B_(0)=C`. Also `k=(2pi)/(lambda)` and `omega=2piv`.
These relation gives `E_(0)K=B_(0)omega`


Discussion

No Comment Found

Related InterviewSolutions