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# During the propagation of electromagnetic waves in a medium:(A) Electric energy density is equal to magnetic energy density(B) Both electric and magnetic energy densities are zero(C) Electric energy density is double of the magnetic energy density(D) Electric energy density is half of the magnetic energy density

Last updated date: 17th Apr 2024
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Answer
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Hint: To solve this question, we need to use the formulae of the electric and the magnetic energy densities. Then from the properties of the electromagnetic waves, we can get a relation between the amplitudes of the electric and the magnetic fields. Substituting it in the formulae of the energy densities, we will get the required relation between the electric and the magnetic field densities.
Formula used: The formulae used for solving this question are given by
1. ${U_E} = \dfrac{1}{2}\varepsilon {E^2}$
2. ${U_B} = \dfrac{1}{{2\mu }}{B^2}$
3. $c = \dfrac{1}{{\sqrt {\mu \varepsilon } }}$
Here, ${U_E}$ and ${U_B}$ are respectively the electric and the magnetic field densities corresponding to an EM wave having the amplitude of the electric field as $E$ and the amplitude of magnetic field as $B$, $c$ is the speed of the electromagnetic field in a medium whose magnetic permeability is $\mu$ and the electric permittivity is $\varepsilon$.

Complete step-by-step solution:
We know that the electric energy density is given by the formula
${U_E} = \dfrac{1}{2}\varepsilon {E^2}$ (1)
Also, the magnetic field is given by the formula
${U_B} = \dfrac{1}{{2\mu }}{B^2}$ (2)
Dividing (1) by (2) we get
$\dfrac{{{U_E}}}{{{U_B}}} = \dfrac{{\dfrac{1}{2}\varepsilon {E^2}}}{{\dfrac{1}{{2\mu }}{B^2}}}$
$\Rightarrow \dfrac{{{U_E}}}{{{U_B}}} = \mu \varepsilon {\left( {\dfrac{E}{B}} \right)^2}$ (3)
Now, we know that the speed of an electromagnetic field is related to the amplitudes of the electric and the magnetic field as
$\dfrac{E}{B} = c$
Putting it in (3) we get
$\Rightarrow \dfrac{{{U_E}}}{{{U_B}}} = \mu \varepsilon {c^2}$ (4)
Now, the speed of the electromagnetic in a medium is given by
$c = \dfrac{1}{{\sqrt {\mu \varepsilon } }}$
On squaring both the sides, we get
${c^2} = \dfrac{1}{{\mu \varepsilon }}$ (5)
Putting (5) in (4) we finally get
$\dfrac{{{U_E}}}{{{U_B}}} = \mu \varepsilon \left( {\dfrac{1}{{\mu \varepsilon }}} \right)$
$\Rightarrow \dfrac{{{U_E}}}{{{U_B}}} = 1$
This can also be written as
${U_E} = {U_B}$
Thus, the electric energy density of an electromagnetic wave is equal to its magnetic energy density.

Hence, the correct answer is option A.

Note: The components of an electromagnetic wave are the electric and magnetic fields. They both are symmetric in terms of the phase, energy density, etc.