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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 time averaged intensity of the wave is given by I_(av)=(1)/(2)c in_(0)E_(0)^(2). |
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Answer» Solution :Radiant flux density, `S=(1)/(mu_(0))(vec(E )xx vec(B))` `therefore S=c^(2)in_(0)(vec(E )xx vec(B)) "" …(1) [because c=(1)/(sqrt(mu_(0)in_(0)))]` Let electromagnetic WAVES propagate in X - DIRECTION. Electric field vector in y - direction and magnetic field vector in z - direction. Hence, `vec(E )=vec(E )_(0)cos(K x-omega t)` `vec(B)=vec(B)_(0)cos(kx - omega t)` `vec(E )xx vec(B)=(vec(E )_(0)xx vec(B)_(0))cos^(2)(kx-omega t)` `S=c^(2)in_(0)(vec(E )xx vec(B))` [`because` From equation (1)] `c^(2)in_(0)(vec(E )_(0)xx vec(B)_(0))cos^(2)(kx-omega t)` Average value of magnitude of radiant flux density over complete cycle is, `S_("average")=c^(2)in_(0)|vec(E )_(0)xx vec(B)_(0)|(1)/(T)int_(0)^(T)cos^(2)(kx-omega t)dt` `c^(2)in_(0)E_(0)B_(0)xx(1)/(T)xx(T)/(2) "" [because |vec(E )_(0)xx vec(B)_(0)|=E_(0)B_(0)sin 90^(@)=E_(0)B_(0)]` and`int_(0)^(T)cos^(2)(kx-omega t)dt=(T)/(2)` `therefore S_("average")=(c^(2))/(2).in_(0)E_(0)((E_(0))/(c )) "" [because c=(E_(0))/(B_(0))rArr B_(0)=(E_(0)^(2))/(c )]` `=(c )/(2)in_(0)E_(0)^(2)` `=(c )/(2)xx(1)/(c^(2)mu_(0))xx E_(0)^(2) "" [because c=(1)/(sqrt(mu_(0)in_(0)))rArr in_(0)=(1)/(c^(2)mu_(0))]` `=(E_(0)^(2))/(2mu_(0)c)` |
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