Ncert Exercises & Solutions

Show that the average value of radiant flux density S over a single period T is given by S = $\frac{1}{2 {cμ}_{0}} E_{0}^{2} .$

Hint: The radiant flux density is given by, S

= \frac{1}{μ_{0}} (\vec{E} \times \vec{B}) = c^{2} ε_{0} (\vec{E} \times \vec{B}) [∵ c = \frac{1}{\sqrt{μ_{0} ε_{0}}}]

Step 1: Find the equation of the electric field and the magnetic field.

Suppose the electromagnetic wave be propagating along the x-axis. The electric field vector of the electromagnetic wave be along the y-axis and the magnetic field vector is along the z-axis. Therefore,

and

\vec{E} = E_{0} \cos (kx - ωt) \hat{j}

\vec{B} = B_{0} \cos (kx - ωt) \hat{k}

\vec{E} \times \vec{B} = (E_{0} \times B_{0}) \cos^{2} (kx - ωt)

Step 2 : Find the radiant energy flux .

\vec{S} = c^{2} ε_{0} (\vec{E} \times \vec{B})

= c^{2} ε_{0} (E_{0} \times B_{0}) \cos^{2} (kx - ωt) \hat{k}

Step 3: Find the radiant energy flux for a cycle.

The average value of the magnitude of radiant flux density over a complete cycle is:

S_{av} = c^{2} ε_{0} |E_{0} \times B_{0}| \frac{1}{T} \int_{0}^{T} \cos^{2} (kx - ωt) dt

= c^{2} ε_{0} E_{0} B_{0} \times \frac{1}{T} \times \frac{T}{2} [∵ \int_{0}^{T} \cos^{2} (kx - ωt) dt = \frac{T}{2}]

S_{av} = \frac{c^{2}}{2} ε_{0} E_{0} (\frac{E_{0}}{c}) [As, c = \frac{E_{0}}{B_{0}}]

= \frac{c}{2} ε_{0} E_{0}^{2} = \frac{c}{2} \times \frac{1}{c^{2} μ_{0}} E_{0}^{2} [c = \frac{1}{\sqrt{μ_{0} ε_{0}}} or ε_{0} = \frac{1}{c^{2} μ_{0}}]

S_{av} = \frac{E_{0}^{2}}{2 μ_{0} c}, hence proved .

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