Question

What is intensity of electromagnetic wave? Give its relation in terms of electric field E and magnetic field B.

Answer 1

The intensity of an electromagnetic wave is the average power per unit area carried by the wave.

Its relation in terms of electric field ($E_0$) and magnetic field ($B_0$) amplitudes is:

1. $I = \frac{E_0 B_0}{2\mu_0}$

2. $I = \frac{1}{2} \epsilon_0 E_0^2 c$

3. $I = \frac{1}{2\mu_0} B_0^2 c$

where $\epsilon_0$ is the permittivity of free space, $\mu_0$ is the permeability of free space, and $c$ is the speed of light in vacuum.

Answer 2

The intensity of an electromagnetic wave is a measure of the average power it carries per unit area. It represents the average amount of energy transported by the wave per unit time across a unit area perpendicular to the direction of wave propagation. Its SI unit is Watts per square meter ($W/m^2$).

The intensity ($I$) of an electromagnetic wave in vacuum can be expressed in terms of the amplitudes of the electric field ($E_0$) and magnetic field ($B_0$) as follows:

The instantaneous Poynting vector, which represents the instantaneous power per unit area, is given by: $\vec{S} = \frac{1}{\mu_0} (\vec{E} \times \vec{B})$ For a plane electromagnetic wave propagating in the x-direction, the electric field $\vec{E}$ and magnetic field $\vec{B}$ are perpendicular to each other and to the direction of propagation. Their magnitudes vary sinusoidally: $E = E_0 \sin(kx - \omega t)$ $B = B_0 \sin(kx - \omega t)$ The magnitude of the instantaneous Poynting vector is $S = \frac{EB}{\mu_0}$. The intensity $I$ is the time average of the magnitude of the Poynting vector over one complete cycle: $I = \langle S \rangle = \left\langle \frac{E_0 B_0}{\mu_0} \sin^2(kx - \omega t) \right\rangle$ Since the average value of $\sin^2(\theta)$ over a full cycle is $1/2$: $I = \frac{E_0 B_0}{2\mu_0}$ This is the fundamental relation in terms of both $E_0$ and $B_0$.

Using the relationship between $E_0$ and $B_0$ for an electromagnetic wave in vacuum, $E_0 = c B_0$ (where $c$ is the speed of light in vacuum), we can express the intensity solely in terms of $E_0$ or $B_0$:

1. In terms of $E_0$ (peak electric field amplitude): Substitute $B_0 = E_0/c$ into the expression for $I$: $I = \frac{E_0 (E_0/c)}{2\mu_0} = \frac{E_0^2}{2\mu_0 c}$ Alternatively, using $c = \frac{1}{\sqrt{\mu_0 \epsilon_0}}$, which implies $\frac{1}{\mu_0 c} = c \epsilon_0$: $I = \frac{1}{2} E_0^2 (c \epsilon_0) = \frac{1}{2} \epsilon_0 E_0^2 c$

2. In terms of $B_0$ (peak magnetic field amplitude): Substitute $E_0 = c B_0$ into the expression for $I$: $I = \frac{(c B_0) B_0}{2\mu_0} = \frac{c B_0^2}{2\mu_0}$

Explanation of the solution: Intensity of an electromagnetic wave is the average power per unit area. It is obtained by averaging the magnitude of the Poynting vector $S = \frac{EB}{\mu_0}$ over one cycle. For sinusoidal fields $E=E_0\sin(\omega t)$ and $B=B_0\sin(\omega t)$, the average of $E_0B_0\sin^2(\omega t)$ is $\frac{E_0B_0}{2}$. Thus, $I = \frac{E_0B_0}{2\mu_0}$. Using $E_0=cB_0$ and $c=\frac{1}{\sqrt{\mu_0\epsilon_0}}$, this can be written as $I = \frac{1}{2}\epsilon_0 E_0^2 c$ or $I = \frac{1}{2\mu_0} B_0^2 c$.