Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

Electromagnetic Waves question

2006 · Shift 0 · Q83
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Main
  3. /Physics
  4. /Electromagnetic Waves
  5. /2006 · Shift 0 · Q83

Electromagnetic Waves question

2006 · Shift 0 · Q83

JEE MainPhysicsElectromagnetic WavesMCQ+4 / −1
The rmsrmsrms value of the electric field of the light coming from the Sun is 720N/C.720N/C.720N/C. The average total energy density of the electromagnetic wave is
  1. A
    4.58×10−6 J/m34.58 \times {10^{ - 6}}\,J/{m^3}4.58×10−6J/m3
  2. B
    6.37×10−9 J/m36.37 \times {10^{ - 9}}\,J/{m^3}6.37×10−9J/m3
  3. C
    81.35×10−12 J/m381.35 \times {10^{ - 12}}\,J/{m^3}81.35×10−12J/m3
  4. D
    3.3×10−3 J/m33.3 \times {10^{ - 3}}\,J/{m^3}3.3×10−3J/m3
View written solutionFree

Correct answer: A

  1. For an electromagnetic wave, the average total energy density is

uavg=ϵ0Erms2u_{avg} = \epsilon_0 E_{rms}^2uavg​=ϵ0​Erms2​

This is because the average electric and magnetic energy densities are equal, and using RMS value directly gives the average total energy density.

  1. Given:

Erms=720 N/CE_{rms} = 720\,\text{N/C}Erms​=720N/C ϵ0=8.85×10−12 F/m\epsilon_0 = 8.85 \times 10^{-12}\,\text{F/m}ϵ0​=8.85×10−12F/m

So,

uavg=8.85×10−12×(720)2u_{avg} = 8.85 \times 10^{-12} \times (720)^2uavg​=8.85×10−12×(720)2

  1. Compute 7202720^27202:

7202=518400=5.184×105720^2 = 518400 = 5.184 \times 10^57202=518400=5.184×105

  1. Now multiply:

uavg=8.85×10−12×5.184×105u_{avg} = 8.85 \times 10^{-12} \times 5.184 \times 10^5uavg​=8.85×10−12×5.184×105

uavg=45.8784×10−7u_{avg} = 45.8784 \times 10^{-7}uavg​=45.8784×10−7

uavg=4.58784×10−6 J/m3u_{avg} = 4.58784 \times 10^{-6}\,\text{J/m}^3uavg​=4.58784×10−6J/m3

  1. Therefore,

uavg≈4.58×10−6 J/m3u_{avg} \approx 4.58 \times 10^{-6}\,\text{J/m}^3uavg​≈4.58×10−6J/m3

  1. Checking options:
  • A: 4.58×10−6 J/m34.58 \times 10^{-6}\,J/m^34.58×10−6J/m3 ✅
  • B: incorrect
  • C: incorrect
  • D: incorrect

Hence, the correct answer is Option A.

PreviousNext

More from Electromagnetic Waves

  • An electromagnetic wave of frequency v=3.0MHz passes from vacuum into a dielectric medium with permittivity ∈=4.0. Then2004 · MCQ
  • A radiation of energy E falls normally on a perfectly reflecting surface. The momentum transferred to the surface is2004 · MCQ
  • Which of the following are not electromagnetic waves?2002 · MCQ
  • Electromagnetic waves are transverse in nature is evident by2002 · MCQ
  • The unit of ε0​c2I​​ is : (I = intensity of an electromagnetic wave, c = speed of light)2025 · MCQ
  • A parallel plate capacitor of area A=16 cm2 and separation between the plates 10 cm , is charged by a DC current. Consider a hypothetical plane surface of area A0​=3.2 cm2 inside the capacitor and…2025 · Numerical
  • The electric field of an electromagnetic wave in free space is E=57cos[7.5×106t−5×10−3(3x+4y)](4i^−3j^​)N/C. The associated magnetic field in Tesla is2025 · MCQ
  • A plane electromagnetic wave of frequency 20 MHz travels in free space along the +x direction. At a particular point in space and time, the electric field vector of the wave is Ey​=9.3Vm−1. Then, the magnetic…2025 · MCQ