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A plane EM wave travelling in vacuum alo...

A plane EM wave travelling in vacuum along z - direction is given by `vec(E )=E_(0)sin (kz - omega t)hat(i)` and `vec(B)=B_(0)sin(kz - omega t)hat(j)`.

Use equation `int vec(E ).vec(d)l =-(d phi_(B))/(dt)` to prove `(E_(0))/(B_(0))=c`.

Text Solution

Verified by Experts

Now `oint vec(E ).vec(d)l = -(d)/(dt)oint vec(B).vec(d )s`
`thereofre` From equation (1) and (2),
`E_(0)h[sin (kz_(2)-omega t)-sin (kz_(1)-omega t)] =-(d)/(dt)`
`[(B_(0)h)/(k){cos (kz_(2)-omega t)-cos(kz_(1)-wt)}]`
`therefore E_(0)h[sin(kz_(2)-omega t)-sin (kz_(1)-omega t)]`
`=(B_(0)h)/(k)(omega)[sin (kz_(2)-omega t)-sin (kz_(1)-omega t)]`
`therefore E_(0)h=R_(0)h((omega)/(k))`
`therefore E_(0)=B_(0)c " " [because (omega)/(k)=c]`
`therefore (E_(0))/(B_(0))=c`
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