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JEE Main 2023 (11 Apr, Shift 1) · Energy Density & Intensity

Energy Density & Intensity question from JEE Main 2023 (11 Apr, Shift 1)

Q12JEE Main 2023 · 11 APR · SHIFT 1Easy

Energy Density & Intensity

The electric field in an electromagnetic wave is given as

E⃗=20sin⁡ω(t−xc)j^ N C−1\vec{E} = 20 \sin \omega \left(t - \frac{x}{c}\right) \hat{j}\ \text{N C}^{-1}

where ω\omega and cc are angular frequency and velocity of electromagnetic wave respectively. The energy contained in a volume of 5×10−4 m35 \times 10^{-4}\text{ m}^{3} will be

(Given ε0=8.85×10−12 C2/N m2\varepsilon_0 = 8.85 \times 10^{-12}\text{ C}^2/\text{N m}^2)

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Answer: (c)

  1. Average total energy density u=12ε0E02=12×8.85×10−12×202=1.77×10−9u = \tfrac{1}{2}\varepsilon_0 E_0^2 = \tfrac{1}{2} \times 8.85 \times 10^{-12} \times 20^2 = 1.77 \times 10^{-9} J/m3^3
  2. Energy =uV=1.77×10−9×5×10−4=8.85×10−13= uV = 1.77 \times 10^{-9} \times 5 \times 10^{-4} = 8.85 \times 10^{-13} J
  3. So the answer is (c) 8.85×10−138.85 \times 10^{-13} J.
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