Question

The energy of an electromagnetic wave in vacuum is given by the relation:
(A) \[\dfrac{{{E}^{2}}}{2{{\varepsilon }_{0}}}+\dfrac{{{B}^{2}}}{2{{\mu }_{0}}}\]
(B) \[\dfrac{1}{2}{{\varepsilon }_{0}}{{E}^{2}}+\dfrac{1}{2}{{\mu }_{0}}{{B}^{2}}\]
(C) \[\dfrac{1}{2}{{\varepsilon }_{0}}{{E}^{2}}+\dfrac{1}{2}{{\mu }_{0}}{{B}^{2}}\]
(D) \[\dfrac{1}{2}{{\varepsilon }_{0}}{{E}^{2}}+\dfrac{{{B}^{2}}}{2{{\mu }_{0}}}\]

Answer 1

Hint: In the given question, we have been asked to establish a relation or an expression for the energy of an electromagnetic wave in a vacuum. Since the electromagnetic waves can be said to be composed of electric and magnetic fields, we will use their properties to find and establish the required relation. Let us proceed to the detailed solution to the problem.

Complete step by step solution:
The electromagnetic wave is a synchronous oscillation or vibration of the magnetic field and the electric field. The direction of oscillations of the electric field is perpendicular to the direction of oscillations of the magnetic field, and both the fields are mutually perpendicular to the direction of propagation of the electromagnetic wave.
To find the energy of an electromagnetic wave, we will have to analyse the constituent electric and magnetic fields.
The energy density of the electric field can be given as \[\dfrac{1}{2}{{\varepsilon }_{0}}{{E}^{2}}\] where \[{{\varepsilon }_{0}}\] is the permittivity of free space and \[E\] is the strength of the electric field.
Similarly, the energy density of the magnetic field is given as \[\dfrac{{{B}^{2}}}{2{{\mu }_{0}}}\] where \[{{\mu }_{0}}\] is the permeability constant and \[B\] is the strength of the magnetic field
Since the oscillations of the electric and the magnetic field together form up an electromagnetic wave, the energy of an electromagnetic wave is the combined energy of the underlying electric and magnetic fields, that is
Energy of an electromagnetic wave \[\left( E \right)=\dfrac{1}{2}{{\varepsilon }_{0}}{{E}^{2}}+\dfrac{{{B}^{2}}}{2{{\mu }_{0}}}\]

Now we can compare the given options against the answer obtained and say that option (D) is the correct answer to the given question.

Note:
While finding the electric field density, the permittivity constant is in the numerator. But in the expression for magnetic field density, the permeability constant lies in the denominator. Often, students get confused with this placement of the constants, and they either place both constants in the numerator or both in the denominator. This mistake is very small yet very frequent, so you should be well aware of which constant to place where.