Energy Density And Intensity Of EM Waves MCQ - Practice Questions with Answers

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 $u=\frac{1}{2} \varepsilon_0 E^2+\frac{1}{2 \mu Z_0} B^2$

$$
u=\frac{1}{2} \varepsilon_0 E_0^2 \sin ^2 \omega(t-x / c)+\frac{1}{2 \mu 0} B_0^2 \sin ^2 \omega(t-x / c)
$$

For taking 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, $E_0=c B_0$ and $\mu_0 \varepsilon_0=\frac{1}{c^2}$ so that, $\mu_0=\frac{1}{\varepsilon_0 c^2}$ and,

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

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 that that the linear momentum is associated with energy and speed, similarly if a portion of EM wave of energy u propagating with speed c, then linear momentum

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 totally reflected surface, then the momentum will be equals 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,

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

Importance of poynting vector is that the direction of the poynting vector $\vec{S}$ at any point gives the wave's direction of travel and direction of energy transport the point.

Average value of 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 vector product so, The direction of $\vec{S}$ does not oscillate but it's magnitude varies between zero and a maximum each quarter of period.

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

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

And for perfectly reflecting surface -

$$
P_r=\frac{2 S}{c}
$$
 

 

Wave impedance (Z) : As the word impedance tell that obstruction in flowing of something, similarly here ,the medium offers hindrance to the propagation of wave. Such hindrance is called wave impedance and it is given by - 

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

For vacuum or free space -

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