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if vecE=E(0)cos(kz)cos(omegat)hati then ...

if `vecE=E_(0)cos(kz)cos(omegat)hati` then `vec(B)` for electromagentic wave is

A

`barB = E_(0)/C hatj sin(kz) sin (omega g)`

B

`barB = E_(0)/C hatj sin(kz) cos(omega t)`

C

`barB= E_(0)/C hatj sin (kz) cos (omega t)`

D

`barB = E_(0)/C hatj cos (kz) sin (omega t)`

Text Solution

Verified by Experts

The correct Answer is:
C
`barE = Ehati cos(kz) cos (wt)`
It is made by superpossion of 2 waves.
`barE_(1) = E_(0)/2 hati cos(kz - wt)`, `barE_(2) =E_(0)/2hati cos(kz + wt)`
Corresponding `barB`
`barB_(1) = E_(0)/(2C) hatj cos(kz -wt)`, `barB_(2) =-E_(0)/(2C) hatj cos(kz+wt)`
So, `barB = barB_(1) + barB_(2) = hatj E_(0)/(2C) xx 2. sin(kz)sin(wt)`
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