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Home » In a plane electromagnetic wave, the electric field oscillates sinusoidally at a frequency of and amplitude . Show that the average energy density of the field equals the average energy density of the B field.
In a plane electromagnetic wave, the electric field oscillates sinusoidally at a frequency of 2.0 \times 10^{10} \mathrm{~Hz} and amplitude 48 \mathrm{~V} \mathrm{~m}^{-1}. Show that the average energy density of the E field equals the average energy density of the B field. \left[\mathrm{c}=3 \times 10^{8} \mathrm{~m} \mathrm{~s}^{-1}\right]
CBSE | Class 12 | Electromagnetic Waves | NCERT | Physics
In a plane electromagnetic wave, the electric field oscillates sinusoidally at a frequency of 2.0 \times 10^{10} \mathrm{~Hz} and amplitude 48 \mathrm{~V} \mathrm{~m}^{-1}. Show that the average energy density of the E field equals the average energy density of the B field. \left[\mathrm{c}=3 \times 10^{8} \mathrm{~m} \mathrm{~s}^{-1}\right]

Frequency of the electromagnetic wave is given as \mathrm{v}=2 \times 10^{10} \mathrm{~Hz}

Electric field amplitude is given as E_{0}=48 \vee \mathrm{m}^{-1}

Speed of light is known as c=3 \times 10^{8} \mathrm{~m} / \mathrm{s}

Energy density of the electric field is represented as,

U_{E}=\frac{1}{2} \epsilon_{0} E^{2}

And, energy density of the magnetic field is given as:

U_{B}=\frac{1}{2 \mu_{0}} B^{2}

Where,

\epsilon_{0} is the permittivity of free space

\mu_{0}= is the permeability of free space

E=c B…(1)

Now,

c=\frac{1}{\sqrt{\epsilon_{0} \mu_{0}}} . \ldots(2)

On putting equation (2) in equation (1), we get

E=\frac{1}{\sqrt{\epsilon_{0} \mu_{0}}} B

On squaring on both sides, we get

E^{2}=\frac{1}{\epsilon_{0} \mu_{0}} B^{2}

\epsilon_{0} E^{2}=\frac{B^{2}}{\mu_{0}}

\frac{1}{2} \epsilon_{0} E^{2}=\frac{1}{2} \frac{B^{2}}{\mu_{0}}

\Rightarrow U_{E}=U_{B}

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