InterviewSolution
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InterviewSolution
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A plane EM wave travelling along z direction is described by `E=E_0sin (kz-omegat) hati and B=B_0 sin (kz-omegat)hatj`. show that (i) The average energy density of the wave is given by `u_(av)=1/4 epsilon_0 E_0^2+1/4 (B_0^2)/(mu_0)`. (ii) The time averaged intensity of the wave is given by `I_(av)=1/2 cepsilon_0E_0^2`. |
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Answer» (i)The e.m. wave carry energy which is due to electric field vector and magnetic field vector. In e.m. wave E and B vary from point to point and from moment to moment. Let E and B be their time average. The energy density due to electric field E is `u_E=1/2in_0E^2` The energy density due to magnetic field B is `u_B=1/2 (B^2)/(mu_0)` Total average energy density of em wave `u_(av)=u_E+u_B=1/2in_0E^2+1/2 (B^2)/(mu_0).....(i)` Let the em wave be propagation along z-direction. The electric filed vector and magnetic field vectors be represented by `E=E_0sin (kz-omegat)` `B=B_0sin (kz-omegat)` The time average value of `E^2` over complete cycle `=E_0^2//2`. and time average value of `B^2` over complete cycle `=B_0^2//2`. `:. u_(av)=1/2 in_0 (E_0^2)/2+1/2 mu_0 ((B_0^2)/2)=1/4 in_0E_0^2+(B_0^2)/(4mu_0)` (ii) we know that `E_0=cB_0 and c=1/sqrt(mu_0in_0)`. `:. 1/4 (B_0^2)/(mu_0)=1/4 (E_0^2//c^2)/(mu_0)=(E_0^2)/(4mu_0)xxmu_0in_0=1/4in_0E_0^2` `:. u_B=u_E` Hence, `u_(av)=1/4 in_0E_0^2+1/4 (B_0^2)/(mu_0)=1/4in_0E_0^2+1/4in_0E_0^2=1/4in_0E_0^2=1/2 (B_0^2)/(mu_0)` Time average intensity of the wave `I_(av)=u_(av)c=1/2 in E_0^2c=1/2in_0cE_0^2` |
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