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The electric field associted with an electromagnetic wave in vacuum is given by `vecE=hati 40 cos(kz=6xx10^(8)t)`, when `E`, `z` and `t` are in volt/m metre and second respectivelyfind the wave vector.A. `2m^-1`B. `0.5m^-1`C. `6m^-1`D. `3m^-1` |
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Answer» Correct Answer - a Here, `E=40cos (kz-6xx10^8t)` Comparing it with the relation `E=E_0 cos .(2pi)/lambda(z-vt)=E_0cos ((2pi)/lambdaz-(2pi)/lambdavt)` we have `(2pi)/lambda=k` and `(2piv)/lambda=6xx10^8s^-1 or kv=6xx10^8` or `k=(6xx10^8)/v=(6xx10^8 s^-1)/(3xx10^8ms^-1)=2m^-1` |
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