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Electromagnetic Waves question

2023 · 11 Apr · Shift 1 · Q59
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  5. /2023 · 11 Apr · Shift 1 · Q59

Electromagnetic Waves question

2023 · 11 Apr · Shift 1 · Q59

JEE MainPhysicsElectromagnetic WavesMCQ+4 / −1
The electric field in an electromagnetic wave is given as E→=20sin⁡ω(t−xc)j→NC−1\overrightarrow{\mathrm{E}}=20 \sin \omega\left(\mathrm{t}-\frac{x}{\mathrm{c}}\right) \overrightarrow{\mathrm{j}} \mathrm{NC}^{-1}E=20sinω(t−cx​)j​NC−1 where ω\omegaω and ccc are angular frequency and velocity of electromagnetic wave respectively. The energy contained in a volume of 5×10−4 m35 \times 10^{-4} \mathrm{~m}^{3}5×10−4 m3 will be (Given ε0=8.85×10−12C2/Nm2\varepsilon_{0}=8.85 \times 10^{-12} \mathrm{C}^{2} / \mathrm{Nm}^{2}ε0​=8.85×10−12C2/Nm2 )
  1. A
    17⋅7×10−13 J17 \cdot 7 \times 10^{-13} \mathrm{~J}17⋅7×10−13 J
  2. B
    28⋅5×10−13 J28 \cdot 5 \times 10^{-13} \mathrm{~J}28⋅5×10−13 J
  3. C
    8⋅85×10−13 J8 \cdot 85 \times 10^{-13} \mathrm{~J}8⋅85×10−13 J
  4. D
    88⋅5×10−13 J88 \cdot 5 \times 10^{-13} \mathrm{~J}88⋅5×10−13 J
View written solutionFree

Correct answer: C

  1. Given electric field

    The electromagnetic wave is 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}E=20sinω(t−cx​)j^​N C−1

    So the amplitude of electric field is E0=20  N C−1E_0 = 20\;\text{N C}^{-1}E0​=20N C−1

  2. Energy density of an electromagnetic wave

    The instantaneous total energy density in an electromagnetic wave is u=uE+uB=ε0E2u = u_E + u_B = \varepsilon_0 E^2u=uE​+uB​=ε0​E2 because electric and magnetic energy densities are equal, and uE=12ε0E2,uB=12ε0E2u_E = \frac{1}{2}\varepsilon_0 E^2, \qquad u_B = \frac{1}{2}\varepsilon_0 E^2uE​=21​ε0​E2,uB​=21​ε0​E2

    Since the field is sinusoidal, the average value of E2E^2E2 over a cycle is ⟨E2⟩=E022\langle E^2 \rangle = \frac{E_0^2}{2}⟨E2⟩=2E02​​

    Hence average energy density is ⟨u⟩=ε0⟨E2⟩=ε0E022\langle u \rangle = \varepsilon_0 \left\langle E^2 \right\rangle = \varepsilon_0 \frac{E_0^2}{2}⟨u⟩=ε0​⟨E2⟩=ε0​2E02​​

  3. Substitute values

    ⟨u⟩=8.85×10−12×(20)22\langle u \rangle = 8.85\times 10^{-12} \times \frac{(20)^2}{2}⟨u⟩=8.85×10−12×2(20)2​

    =8.85×10−12×4002= 8.85\times 10^{-12} \times \frac{400}{2}=8.85×10−12×2400​

    =8.85×10−12×200= 8.85\times 10^{-12} \times 200=8.85×10−12×200

    =1770×10−12= 1770\times 10^{-12}=1770×10−12

    =1.77×10−9  J m−3= 1.77\times 10^{-9}\;\text{J m}^{-3}=1.77×10−9J m−3

  4. Energy contained in given volume

    Given volume, V=5×10−4  m3V = 5\times 10^{-4}\;\text{m}^3V=5×10−4m3

    Therefore, U=⟨u⟩VU = \langle u \rangle VU=⟨u⟩V

    =1.77×10−9×5×10−4= 1.77\times 10^{-9} \times 5\times 10^{-4}=1.77×10−9×5×10−4

    =8.85×10−13  J= 8.85\times 10^{-13}\;\text{J}=8.85×10−13J

  5. Match with options

    8.85×10−13  J8.85\times 10^{-13}\;\text{J}8.85×10−13J corresponds to Option C.

Comparison with stored answer: Stored correct answer is C, which matches the derived answer.

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