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  • Show that the average value of radiant flux density S over a single period T is given by S= frac{(frac{1}{2}c mu _{0})}{E{_{0}}^{2}}

Show that the average value of radiant flux density S over a single period T is given by

S= \frac{(\frac{1}{2}c\mu _{0})}{E{_{0}}^{2}}

Answers (2)

We know that radiant flux density is:

\overrightarrow{S}=\frac{1}{\mu _{0}}(\overrightarrow{E}\times \overrightarrow{B})=C^{2}\epsilon _{0} (\overrightarrow{E}\times \overrightarrow{B})              [as, c=\frac{1}{\sqrt{\mu _{0}\varepsilon _{0}}}]

Let, along the X-axis, electromagnetic waves are propagating. If along the Y-axis, the electric field vector of electromagnetic waves is there. In the same way, along the z-axis, magnetic field vectors will be there. Hence,

E=E_{0}\cos (kx-\omega t)

B=B_{0}\cos (kx-\omega t)

E\times B=\left ( E_{0}B_{0} \right )\cos^{2} (kx-\omega t)

S=C^{2}\varepsilon _{0}\left ( E\times B \right )=C^{2}\varepsilon _{0}\left ( E_{0}B_{0} \right )\cos ^{2}\left ( kx-\omega t \right )

The average value of the radiant flux density is

S_{av}=C^{2}\varepsilon _{0}\left | E_{0}\times B_{0} \right |\frac{1}{T}\int_{0}^{T} \cos ^{2}\left ( kx-\omega t \right )dt=c^{2}\varepsilon _{0}E_{0}B_{0}\left ( \frac{1}{T} \right )\left ( \frac{T}{2} \right )=\left ( c^{2}\frac{\varepsilon _{0}E_{0}}{2} \right ) \left ( \frac{E_{0}}{c} \right ) \; \; \; \; \;

S_{av}=\left ( \frac{\varepsilon _{0}c E{_{0}}^{2}}{2} \right )\; \; \; \; \; \; \; \left [ as, c=\frac{1}{\sqrt{\mu _{0}\varepsilon _{0}}} \right ]

Therefore,

S_{av}=\left ( \frac{E{_{0}}^{2}}{2c\mu _{0}} \right )

Posted by

infoexpert23

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 [ as,c=\frac{E_{0}B}{0} \right ]

Posted by

Safeer PP

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