Question

Show that the average value of radiant flux density $S$ over a single period $T$ is given by $S = \dfrac{1}{{2c{\mu _0}}}{E_0}^2$.

Answer 1

In electromagnetism, the radiant flux density simple means the net power of an electromagnetic wave passing through a unit area of surface and its SI unit is $W{m^{ - 2}}$ . We will use the general formula of flux density $\vec S = \dfrac{1}{{{\mu _0}}}(\vec E \times \vec B)$ where, $\vec E$ is the associated electric field component of the electromagnetic wave and $\vec B$ is the associated magnetic field component of electromagnetic wave, ${\mu _0}$ is called the permeability of the free space.

Complete step by step answer:
As we know that, the speed of light is related with ${\mu _0}$ permeability of free space and ${ \in _0}$ permittivity of free space as $c = \dfrac{1}{{\sqrt {{\mu _0}{ \in _0}} }}$ or we can write it as
${c^2} = \dfrac{1}{{{\mu _0}{ \in _0}}}$
$\Rightarrow \dfrac{1}{{{\mu _0}}} = {c^2}{ \in _0}$ .
Now, from the flux density formula $S = \dfrac{1}{{{\mu _0}}}(\vec E \times \vec B)$ we can write it’s as
$\vec S = {c^2}{ \in _0}(\vec E \times \vec B)$
Let the electromagnetic wave is travelling across the X axes the, electric field and magnetic field component will travel in Y and Z direction and can be written as:
$\vec E = {E_0}\cos (kx - \omega t)\hat j$
$\Rightarrow \vec B = {B_0}\cos (kx - \omega t)\hat k$

Taking the cross product of $\vec E = {E_0}\cos (kx - \omega t)\hat j$ and $\vec B = {B_0}\cos (kx - \omega t)\hat k$ we have,
$\vec E \times \vec B = {E_0}{B_0}{\cos ^2}(kx - \omega t)\hat i$
Now, we have to take the average value of this flux density $\vec E \times \vec B = {E_0}{B_0}{\cos ^2}(kx - \omega t)\hat i$ over the time period of $T$ ,
${S_{avg}} = {c^2}{ \in _0}{E_0}{B_0} \times \dfrac{1}{T}\int\limits_0^T {{{\cos }^2}(kx - \omega t)dt}$
Now, we need to find the average value of $\int\limits_0^T {{{\cos }^2}(kx - \omega t)dt}$ over the time period of $T$ ,
Using $\cos (2\theta ) = 2{\cos ^2}(\theta ) - 1$
So, $\int\limits_0^T {{{\cos }^2}(kx - \omega t)dt} = \dfrac{1}{2}\int\limits_0^T {(1 + \cos 2(kx - \omega t))dt}$
Since, $\int {dt} = t$ and $\int {\cos (kx - \omega t)dt} = \dfrac{{\sin (kx - \omega t)}}{{( - \omega )}}$
$\int\limits_0^T {{{\cos }^2}(kx - \omega t)dt} = \dfrac{1}{2}[t - \dfrac{{\sin (kx - \omega t)}}{{( - \omega )}}]$

Putting the limits of $t = 0$ to $t = T$ we have,
Since $\sin (0) = 0$ and $\sin (\pi ) = 0$ so we have,
$\int\limits_0^T {{{\cos }^2}(kx - \omega t)dt} = \dfrac{1}{2}[T - 0]$
$\Rightarrow \int\limits_0^T {{{\cos }^2}(kx - \omega t)dt} = \dfrac{T}{2}$
So put the value of $\int\limits_0^T {{{\cos }^2}(kx - \omega t)dt} = \dfrac{T}{2}$ in the equation ${S_{avg}} = {c^2}{ \in _0}{E_0}{B_0} \times \dfrac{1}{T}\int\limits_0^T {{{\cos }^2}(kx - \omega t)dt}$ we get,
${S_{avg}} = {c^2}{ \in _0}{E_0}{B_0} \times \dfrac{1}{T}[\dfrac{T}{2}]$
Also we know that, ${B_0} = \dfrac{{{E_0}}}{c}$ and put from equation $\dfrac{1}{{{\mu _0}}} = {c^2}{ \in _0} \to (i)$ in above equation we get,
${S_{avg}} = \dfrac{1}{{{\mu _0}}}{E_0}^2 \times \dfrac{1}{{2c}}$
$\therefore {S_{avg}} = \dfrac{1}{{2c{\mu _0}}}{E_0}^2$

Hence, the average value of radiance flux density is proved which is ${S_{avg}} = \dfrac{1}{{2c{\mu _0}}}{E_0}^2$.

Note: It should be remembered that while solving integrations of type $\int {f(ax)dx = \dfrac{{F(x)}}{a}}$ the constant which is in multiple of integral function will be divided by its integration and the speed of light is a ratio of magnitude of electric field component to the magnetic field component of the electromagnetic wave travelling in free space which can be written as $c = \dfrac{{{E_0}}}{{{B_0}}}$.