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A plane EM wave travelling along z - dir...

A plane EM wave travelling along z - direction is desctribed `vec(E )=E_(0)sin(kz - omega t)hat(i)` and `veC(B)=B_(0)sin (kz - omega)hat(j)`. Show that
The average energy density of the wave is given by `U_(av)=(1)/(4)in_(0)E_(0)^(2)+(1)/(4).(B_(0)^(2))/(mu_(0))`.

Text Solution

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In electromagnetic waves due to electric field vector magnetic field vector waves carry energy `vec(E )` and `vec(B)` varies with time nad position.
Let E and B is average value at given time,
`U_(E )=(1)/(2) in_(0)E^(2)` and energy density due to magnetic field B,
`U_(B)=(1)/(2)(B^(2))/(mu_(0))`
Total average energy density of EM waves,
`U_("average")=U_(E )+U_(B)=(1)/(2)E_(0)E^(2)+(1)/(2)(B^(2))/(mu_(0))`
Consider EM wave propagating in z - direction hence electric field and magnetic field will be represented as
`vec(E )=E_(0)sin(kz - omega t)` and
`vec(B)=B_(0)sin (kz - omega t)`
Average value of `E^(2)` over one cycle (periodic time)
`lt E^(2)gt = (E_(0)^(2))/(2)`
Hence `U_("average")=(1)/(2)E_(0)(E_(0)^(2))/(2)+(1)/(2)mu_(0)((B_(0)^(2))/(2))`
So, `U_("average")=(1)/(4)[in_(0)E_(0)^(2)+(E_(0)^(2))/(4 mu_(0))] " "` ...(1)
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