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An em wave going through vacuum is descr...

An em wave going through vacuum is described by `E=E_0sin(kx-omegat)`
`B=B_0sin(kx-omegat)`

A

`E_(0)k=B_(0)omega`

B

`E_(0)omega=B_(0)k`

C

`E_(0)B_(0)=omegak`

D

None of these

Text Solution

Verified by Experts

The correct Answer is:
A
`(E_(0))/(B_(0))=c`. Also `k=(2pi)/(lambda) and omega=2piv`
These relation gives `E_(0)k=B_(0)omega`
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