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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 time averaged intensity of the wave is given by `I_(av)=(1)/(2)c in_(0)E_(0)^(2)`.

Text Solution

Verified by Experts

Now `E_(0) = c B_(0)` and `c=(1)/(sqrt(mu_(0)in_(0)))`
In equation (1) if `U_("average")=0`, then
`(1)/(4)(B_(0)^(2))/(mu_(0))=-(1)/(4)in_(0)E_(0)^(2)`
`=-(1)/(4).(E_(0)^(2))/(mu_(0)c^(2))[because c=(1)/(sqrt(mu_(0)in_(0)))]`
Neglecting negative sign,
`(1)/(4)(B_(0)^(2))/(mu_(0))=(1)/(4)(E_(0)^(2)//c^(2))/(mu_(0))=(E_(0)^(2))/(4mu_(0)).mu_(0)in_(0) " " [c^(2)=(1)/(mu_(0)E_(0))]`
`therefore (1)/(4)(B_(0)^(2))/(mu_(0))=(1)/(4)in_(0)E_(0)^(2)`
or `(1)/(2)(B_(0)^(2))/(mu_(0))=(1)/(2)in_(0)E_(0)^(2)`
`therefore U_(B)=U_(E )`
`therefore U_("average")=(1)/(4)in_(0)E_(0)^(2)+(1)/(4)(B_(0)^(2))/(mu_(0))[because c=(E_(0))/(B_(0))]`
`=(1)/(4)in_(0)E_(0)^(2)+(1)/(4)(E_(0)^(2))/(mu_(0)c^(2))`
`= (1)/(4)in_(0)E_(0)^(2)+(1)/(4)in_(0)E_(0)^(2)[because c=(1)/(sqrt(mu_(0)in_(0)))]`
`=(1)/(2)in_(0)E_(0)^(2)`
`=(1)/(2)in_(0)xx(B_(0)^(2))/(mu_(0)in_(0))[because E_(0)=(B_(0))/(mu_(0)in_(0))]`
`=(1)/(2)(B_(0)^(2))/(mu_(0))`
`therefore U_(E )= U_(B)`
Now average intensity of wave,
`I_("average")=U_("average")`
`=(1)/(2)in_(0)B_(0)^(2)c`
`=(1)/(2)in_(0)E_(0)^(2)c`
`therefore I_("average")=(1)/(2)in_(0)E_(0)^(2)c`
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