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  • A plane EM wave travelling along z-direction is described by E=E_{0} sin (kz- omega t) hat{i} and B=B_{0}sin (kz-omega t) hat{j} show that, i) the average energy density of the wave is given by

A plane EM wave travelling along z-direction is described by E=E_{0}\sin (kz-\omega t)\hat{i}\; and\; B=B_{0}\sin (kz-\omega t)\hat{j} show that,

i) the average energy density of the wave is given by

u_{av}=\frac{(u_{E}+u_{B})}{2}=\frac{1}{4}.\varepsilon _{0}E^{2}+\frac{1}{4}.\frac{B^{2}}{\mu _{0}}

ii) the time - averaged intensity of the wave is given by

I_{av}=u_{av}c=\frac{1}{2} (\varepsilon _{0}E^{2})c

Answers (1)

i) The energy density due to electric field E is

u_{\varepsilon }=\frac{1}{2}(\varepsilon _{0}E^{2})

The energy density due to magnetic field B is

u_{B}=\frac{1}{2}\left ( \frac{B^{2}}{\mu _{0}} \right )

The average energy density of the wave is given by :

u_{av}=\frac{\left ( u_{E}+u_{B} \right )}{2}=\frac{1}{4}.\varepsilon _{0}E^{2}+\frac{1}{4}.\frac{B^{2}}{\mu _{0}}

ii) u_{av}=\frac{\left ( u_{E}+u_{B} \right )}{2}=\frac{1}{4}.\varepsilon _{0}E^{2}+\frac{1}{4}.\frac{B^{2}}{\mu _{0}}

Assuming u_{E}=u_{B},

u_{av}=\frac{1}{2}\left ( \varepsilon _{0}E^{2} \right )=\frac{1}{2}\left ( \frac{B^{2}}{\mu _{0}} \right )

We know that c=\frac{1}{\sqrt{\mu _{0}\varepsilon _{0}}}=\left ( \frac{E_{0}}{B_{0}} \right )

The time -averaged intensity of the wave is given as

I_{av}=u_{av}c=\frac{1}{2}\left ( \varepsilon _{0}E^{2} \right )c

Posted by

infoexpert23

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