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A plane EM wave travelling along z - direction is desctribed vec(E )=E_(0)sin(kz - omega t)hat(i)andveC(B)=B_(0)sin (kz - omega)hat(j). Show thatThe 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)). |
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Answer» Solution :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)` andenergy 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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