Question

What is the intensity of a traveling plane electromagnetic wave if ${\mathit{B}}_{\mathbf{m}}$is$\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{4}}\mathbf{}\mathbf{T}\mathbf{}$?

Short answer

The intensity of a traveling plane electromagnetic wave is $1.2\times {10}^6{\text{W/m}}^{\text{2}}$.

Step-by-step solution

Listing the given quantities

The magnetic field in the wave, $B_m=1.0\times {10}^{-4}\mathrm{T}$ .

Permeability constant,${\mu }_0=1.26\times {10}^{-6}{\text{H/m}}^{\text{2}}$.

The velocity of light, $c=3\times {10}^8\text{m/s}$.

Understanding the concepts of intensity

The rate of energy transport per unit area in such a wave is described by a vector $\overrightarrow{S}$ called the Poynting vector. The average energy transported over time S, i.e., $S_{average}$is called the intensity$I$ of the wave. The mathematical expression for intensity is $I=S_{avg}=c{B^2}_m/2{\mu }_0,$, where c is the velocity of light and $B_m$ is the magnetic field in the wave.

Calculations of the intensity of the traveling plane

The intensity of the plane can be calculated as,

$I=\frac{c{B}_m^2}{2{\mu }_0}$

Substitute the values in the above expression, and we get,

$\begin{array}{rcl}I& =& \frac{\left(3\times {10}^8\text{m/s}\right){\left(1.0\times {10}^{-4}\text{T}\right)}^2}{2\left(1.26\times {10}^{-6}{\text{H/m}}^{\text{2}}\right)}\\ & =& 1.2\times {10}^6·\left(\frac{1\text{m/s}}{1{\text{H/m}}^{\text{2}}}\times 1{\text{T}}^2\times \frac{1{\text{W/m}}^{\text{2}}}{1\text{T}}\times \frac{1{\text{H/m}}^{\text{2}}}{1\text{m/s}}\times \frac{1}{1\text{T}}\right)\\ & =& 1.2\times {10}^6{\text{W/m}}^{\text{2}}\end{array}$

Thus, the intensity of a traveling plane electromagnetic wave is $1.2\times {10}^6{\text{W/m}}^{\text{2}}$.