In a plane EM wave, the electric field oscillates sinusoidally at a frequency of $5 \times 10^{10} \mathrm{~Hz}$ and an amplitude of $50 \mathrm{Vm}^{-1}$. The total average energy density of the electromagnetic field of the wave is : [Use $\varepsilon_{0}=8.85 \times 10^{-12} \mathrm{C}^{2} / \mathrm{Nm}^{2}$ ]

(A) $1.106 \times 10^{-8} \mathrm{Jm}^{-3}$

(B) $4.425 \times 10^{-8} \mathrm{Jm}^{-3}$

(C) $2.212 \times 10^{-8} \mathrm{Jm}^{-3}$

(D) $2.212 \times 10^{-10} \mathrm{Jm}^{-3}$

JEE Main · 2024 · 31 Jan 2024 · Shift 1 · EM Waves

In a plane EM wave, the electric field oscillates sinusoidally at a frequency of

5 \times 1 0^{10} Hz

and an amplitude of

50 Vm^{- 1}

. The total average energy density of the electromagnetic field of the wave is :
[Use

ε_{0} = 8.85 \times 1 0^{- 12} C^{2} / Nm^{2}

]

Answer: (A)

For a plane EM wave, writing the instantaneous energy density in terms of the peak electric field

E_{0}

(using

B = E / c

and

μ_{0} ε_{0} c^{2} = 1

) gives

u (t) = ε_{0} E_{0}^{2} sin^{2} (ω t)

, whose time average is the total average energy density of the wave:

u_{a v g} = \frac{1}{2} ε_{0} E_{0}^{2}

With

E_{0} = 50 V m^{- 1}

and

ε_{0} = 8.85 \times 1 0^{- 12} C^{2} N^{- 1} m^{- 2}

u_{a v g} = \frac{1}{2} \times 8.85 \times 1 0^{- 12} \times (50)^{2} = 1.106 \times 1 0^{- 8} J m^{- 3}

JEE Main · 2024 · 31 Jan 2024 · Shift 1 · Single correct option