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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)`.

By using similar process and the equation `int vec(B).vec(d)l = mu_(0)I+ in_(0)(d phi_(E ))/(dt)`, prove that `c=(1)/(sqrt(mu_(0)in_(0)))`

Text Solution

Verified by Experts

As shown in figure square loop 1234 is in yz - plane.

`oint vec(B).vec(d)l=int_(1)^(2)vec(B).vec(d)l+int_(2)^(3)vec(B).vec(d)l+ int_(3)^(4)vec(B).vec(d)l+int_(4)^(1)vec(B).vec(d)l`
`= int_(1)^(2) B dl cos 0^(@)+int_(2)^(3) B dl cos 90^(@)+int_(3)^(4)B dl cos 180^(@)+ int_(4)^(1)B dl cos 90^(@)`
`= B_(0)h[sin(kz_(1)-omega t)-sin (kz_(2)-omega t)] " " [because " taking " dl = h] " "` ...(3)
Now to find `phi_(0)=int vec(E ).vec(d)s` consider square loop 1234 which is made up of no. of small section of area ds = hdz.

`phi_(E )=int vec(E ).vec(d)s = int Eds cos 0^(@) = int Eds " " [because cos 0^(@)=1]`
`= int_(z_(1))^(z_(2))E_(0)sin (kz_(1)-omega t)hdz`
`=-(E_(0)h)/(k)[cos (kz_(2)-omega t)-cos (kz_(1)-omega t)]`
`therefore (d phi_(E ))/(dt)=(E_(0)h omega)/(k)[sin (kz_(2)-omega t)-sin (kz_(1)-omega t)] " "` ...(4)
From Ampere law,
`oint B dl = mu_(0)(I_(c )+in_(0)(d phi_(E ))/(dt))`
where `I_(c )` is conduction current for vacuum `I_(c )= 0`
Hence, `oint B dl=(mu_(0)in (d phi_(E ))/(dt))`
From equation (3) and (4),
`B_(0)h[sin (kz_(1)-omega t)-sin (kz_(2)-omega t)]`
`= mu_(0)mu_(0)(E_(0)h omega)/(k)[sin (kz_(1)-omega t)-sin (kz_(2)-omega)]`
`therefore B_(0) = (mu_(0)in_(0)E_(0)omega)/(k)`
`therefore (1)/(mu_(0)in_(0))=(E_(0)omega)/(B_(0)k)`
`therefore (1)/(mu_(0)in_(0))= c xx (ck)/(k)[because (E_(0))/(B_(0))=c " and " omega = ck]`
`therefore (1)/(mu_(0)in_(0))=c^(2)`
`therefore c=(1)/(sqrt(mu_(0)in_(0)))`
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