Energy Density And Intensity Of EM Waves - Practice Questions & MCQ

Energy Density and Intensity of EM waves-

1. Energy Density-

The electric and magnetic fields in a plane electromagnetic wave are given by

$
\begin{aligned}
& E=E_0 \sin \omega(t-x / c) \\
& \text { and, } B=B_0 \sin \omega(t-x / c)
\end{aligned}
$

In any small volume $d V$, the energy of the electric field is

$
U_E=\frac{1}{2} \varepsilon_0 E^2 d V
$

And the energy of the magnetic field is $U_B=\frac{1}{2 \mu_0} B^2 d V$

Thus, the total energy is $U=\frac{1}{2} \varepsilon_0 E^2 d V+\frac{1}{2 \mu_0} B^2 d V$

The energy density is $u=\frac{1}{2} \varepsilon_0 E^2+\frac{1}{2 \mu_o} B^2$

$
u=\frac{1}{2} \varepsilon_0 E_0^2 \sin ^2 \omega(t-x / c)+\frac{1}{2 \mu} B_0^2 \sin ^2 \omega(t-x / c)
$
 

$
U_B = \frac{1}{2\mu_0} \int B^2 \, dV
$

$
U = \frac{1}{2} \int \varepsilon_0 E^2 \, dV + \frac{1}{2\mu_0} \int B^2 \, dV
$

$
u = \frac{1}{2} \varepsilon_0 E^2 + \frac{1}{2\mu_0} B^2
$

$
u = \frac{1}{2} \varepsilon_0 E_0^2 \sin^2\left(\omega \left(t - \frac{x}{c}\right)\right) + \frac{1}{2\mu_0} B_0^2 \sin^2\left(\omega \left(t - \frac{x}{c}\right)\right)
$
 

 

For taking the average over a long time, the $\sin ^2$ terms have an average value of $1 / 2$
So, $u_{\mathrm{av}}=\frac{1}{4} \varepsilon_0 E_0^2+\frac{1}{4 \mu_0} B_0^2$
Now, as we know, $\quad E_0=c B_0$ and $\mu_0 \varepsilon_0=\frac{1}{c^2}$ so that, $\mu_0=\frac{1}{\varepsilon_0 c^2}$ and,

$
B_0=\frac{E_0}{c} \frac{1}{4 \mu_0} B_0^2=\frac{\varepsilon_0 c^2}{4}\left(\frac{E_0}{c}\right)^2=\frac{1}{4} \varepsilon_0 E_0^2
$

Thus, the electric energy density is equal to the magnetic density in average,
or, $\quad u_{\mathrm{av}}=\frac{1}{4} \varepsilon_0 E_0^2+\frac{1}{4} \varepsilon_0 E_0^2=\frac{1}{2} \varepsilon_0 E_0^2$
Also $u_{\mathrm{av}}=\frac{1}{4 \mu_0} B_0^2+\frac{1}{4 \mu_0} B_0^2=\frac{1}{2 \mu_0} B_0^2$

Intensity (I): The energy crossing per unit area per unit time, perpendicular to the direction of propagation of EM wave is called intensity.

So,

     

$
\begin{gathered}
\text { i.e. } I=\frac{\text { Total EM energy }}{\text { Surface area } \times \text { Time }}=\frac{\text { Total energy density } \times \text { Volume }}{\text { Surface area } \times \text { Time }} \\
\Rightarrow I=u_{a v} \times c=\frac{1}{2} \varepsilon_0 E_0^2 c=\frac{1}{2} \frac{B_0^2}{\mu_0} \cdot c \quad \frac{\text { Watt }}{m^2}
\end{gathered}
$

Momentum: Electromagnetic waves also carries momentum. As we know linear momentum is associated with energy and speed.
So we can write that if wave incident on a completely absorbing surface then momentum delivered will be equal to -

$
p=\frac{u}{c}
$

But if the wave is incident on a totally reflected surface, then the momentum will be equal to -

$
-p=\frac{2 u}{c}
$

Poynting vector $(\vec{S})$ : It is defined as the rate of flow of energy crossing a unit area in electromagnetic waves. So,

$
\vec{S}=\frac{1}{\mu_o}(\vec{E} \times \vec{B})=c^2 \varepsilon_0(\vec{E} \times \vec{B})
$

Unit of Poynting vector is Watt $/ \mathrm{m}^2$. Now, as we know that in electromagnetic waves, $\vec{E}$ and $\vec{B}$ are perpendicular to each other. So,

$
|\vec{S}|=\frac{1}{\mu_0} E B \sin 90^{\circ}=\frac{E B}{\mu_0}=\frac{E^2}{\mu C}
$
 

The importance of the Poynting vector is that the direction of the Poynting vector $S$ at any point gives the wave's direction of travel and direction of energy transport to the point.
The average value of the Poynting vector -

$
\vec{S}=\frac{1}{2 \mu_0} E_0 B_0=\frac{1}{2} \varepsilon_0 E_0^2 c=\frac{c B_0^2}{2 \mu_0}
$

As we can notice that direction of Poynting vector can be given by the vector product so, The direction of $\vec{S}$ does not oscillate but its magnitude varies between zero and a maximum each quarter of the period.

Radiation pressure: It is defined as the momentum imparted per second per unit area on which the light falls
So, for the perfectly absorbing body, we can write in terms of the Poynting vector -

$
P_a=\frac{S}{c}
$

And for perfectly reflecting surface -

$
P_r=\frac{2 S}{c}
$

$
Z=\sqrt{\frac{\mu}{\varepsilon}}=\sqrt{\frac{\mu_r}{\varepsilon_r}} \sqrt{\frac{\mu_0}{\varepsilon_0}}
$

For vacuum or free space -

$
Z=\sqrt{\frac{\mu_0}{\varepsilon_0}}=376.6 \Omega
$