JEE Main · 2021 · 31 Aug 2021 · Shift 1 · EM Waves

The electric field in an electromagnetic wave is given by

E = (50 NC^{- 1}) sin ω (t - x / c)

The energy contained in a cylinder of volume V is

5.5 \times 1 0^{- 12} J

. The value of V is _____

cm^{3}

.
(given

ϵ_{0} = 8.8 \times 1 0^{- 12} C^{2} N^{- 1} m^{- 2}

)

Answer: 500

Solution

The time-averaged energy density of the wave is

⟨ u ⟩ = \frac{1}{2} ϵ_{0} E_{0}^{2}

, with amplitude

E_{0} = 50 N C^{- 1}

⟨ u ⟩ = \frac{1}{2} \times 8.8 \times 1 0^{- 12} \times 5 0^{2} = 1.1 \times 1 0^{- 8} J m^{- 3} .

The energy stored in volume

V

U = ⟨ u ⟩ V

, so

V = \frac{U}{⟨ u ⟩} = \frac{5.5 \times 1 0 ^{- 12}}{1.1 \times 1 0 ^{- 8}} = 5 \times 1 0^{- 4} m^{3} .

Converting,

V = 500 c m^{3}

JEE Main · 2021 · 31 Aug 2021 · Shift 1 · Integer answer

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