What is the energy density in the electromagnetic wave in free space?
The total energy stored in electromagnetic waves per unit volume due to the electric field and the magnetic field is called energy density in the electromagnetic wave in free space.
|
$U=\epsilon_{0} E^{2}=\frac{B^{2}}{\mu_{0}}$
|
Derivation of Energy density in electromagnetic waves in free space:
The energy per unit volume due to the electric field is
$U_{E}= \frac{1}{2} \overrightarrow{E}.\overrightarrow{D} \qquad(1)$
The energy per unit volume due to the magnetic field is
$U_{B}= \frac{1}{2} \overrightarrow{B}.\overrightarrow{H} \qquad(2)$
The total energy density of electromagnetic waves is
$U=U_{E}+U_{B} \qquad(3)$
Now substitute the value of $U_{E}$ and $U_{B}$ in equation$(3)$ then we get
$U=\frac{1}{2} \left( \overrightarrow{E}.\overrightarrow{D}+\overrightarrow{B}.\overrightarrow{H} \right)$
$U=\frac{1}{2} \left( \overrightarrow{E}.\epsilon_{0}\overrightarrow{E}+\overrightarrow{B}.\frac{1}{\mu_{0}}\overrightarrow{B} \right) \qquad ( \because \overrightarrow{B}= \mu_{0} \overrightarrow{H} \:OR \: \overrightarrow{D}= \epsilon_{0}\overrightarrow{E} )$
$U=\frac{1}{2} \left( \epsilon_{0} E^{2}+\frac{B^{2}}{\mu_{0}} \right) \qquad ( \because \overrightarrow{E}\overrightarrow{E}= E^{2} \:OR \: \overrightarrow{B}.\overrightarrow{B}=B^{2})$
$U=\frac{1}{2} \left( \epsilon_{0} E^{2}+\frac{E^{2}}{c^{2} \mu_{0}} \right) \qquad ( \because B=\frac{E}{c})$
$U=\frac{1}{2} \left( \epsilon_{0} E^{2}+\frac{E^{2}}{c^{2} \mu_{0}} \right) $
$U=\frac{1}{2} \left( \epsilon_{0} E^{2}+\epsilon_{0} E^{2} \right) \qquad ( \because c=\frac{1}{\sqrt{\mu_{0} \epsilon_{0}}})$
$U=\frac{1}{2} \left(2 \epsilon_{0} E^{2} \right) $
Similarly, the energy density of electromagnetic waves in free space in terms of the magnetic field $B$ can be written as:
|
$U= \frac{B^{2}}{\mu_{0}} $
|
The average value of energy density in the electromagnetic waves in free space:
Now we will find the average value of energy density in the electromagnetic wave in free space from the above equation $U= \epsilon_{0} E^{2} $. So we get
$\left< U \right> = \epsilon_{0} \left< E^{2} \right>$
$\left< U \right> = \epsilon_{0} \frac{E_{0}^{2}}{2} \qquad \left (\because \left< E^{2} \right>=\frac{E_{0}^{2}}{2} \right)$
$\left< U \right> = \epsilon_{0} E_{rms}^{2} \qquad \left (\because E_{rms}^{2}=\frac{E_{0}^{2}}{2} \right) \qquad (4)$
We know that
$\left< \overrightarrow{S} \right> = \frac{E_{rms}^{2}}{Z_{0}} .\hat{n} \qquad (5)$
Now divide the equation $(5)$ by equation$(4)$
$\frac{\left< \overrightarrow{S} \right>}{\left< U \right>}=\frac{\frac{E_{rms}^{2}}{Z_{0}} .\hat{n}}{\epsilon_{0} E_{rms}^{2}}$
$\frac{\left< \overrightarrow{S} \right>}{\left< U \right>}=\frac{\hat{n}}{\epsilon_{0} Z_{0}}$
$\frac{\left< \overrightarrow{S} \right>}{\left< U \right>}=\frac{\hat{n}}{\sqrt{\epsilon_{0} \mu_{0}}} \qquad(\because z_{0}= \sqrt{\frac{\mu_{0}}{\epsilon_{0}}})$
$\frac{\left< \overrightarrow{S} \right>}{\left< U \right>}=\hat{n} c \qquad(\because c= \frac{1}{\sqrt{\mu_{0} \epsilon_{0}}})$
|
$ \left< \overrightarrow{S} \right>=\hat{n} c \left< U \right> $
|
The energy flow per unit area per unit time in an electromagnetic wave is the product of energy density, speed of light, and the direction of propagation.
The ratio of the energy densities of the electric field and magnetic field:
So from above equation $U_{E}=\epsilon_{0} E^{2}$ and equation $U_{B}=\frac{B^{2}}{\mu_{0}}$, we can find the ratio between them i.e.
$\frac{U_{E}}{U_{B}}=\frac{\epsilon_{0} E^{2}}{\frac{B^{2}}{\mu_{0}}}$
$\frac{U_{E}}{U_{B}}=\frac{\epsilon_{0} \mu_{0} E^{2}}{B^{2}}$
$\frac{U_{E}}{U_{B}}=\frac{c^{2}}{c^{2}}$
$\frac{U_{E}}{U_{B}}=1$
So the energy density of the electric field is the same as the energy density of the magnetic field.
