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Electromagnetic Waves question

2023 · 10 Apr · Shift 2 · Q48
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Electromagnetic Waves question

2023 · 10 Apr · Shift 2 · Q48

JEE MainPhysicsElectromagnetic WavesMCQ+4 / −1
The amplitude of magnetic field in an electromagnetic wave propagating along y-axis is 6.0×10−7 T6.0 \times 10^{-7} \mathrm{~T}6.0×10−7 T. The maximum value of electric field in the electromagnetic wave is
  1. A
    6.0×10−7 Vm−16.0 \times 10^{-7} ~\mathrm{Vm}^{-1}6.0×10−7 Vm−1
  2. B
    5×1014 Vm−15 \times 10^{14} ~\mathrm{Vm}^{-1}5×1014 Vm−1
  3. C
    180 Vm−1180 ~\mathrm{Vm}^{-1}180 Vm−1
  4. D
    2×1015 Vm−12 \times 10^{15} ~\mathrm{Vm}^{-1}2×1015 Vm−1
View written solutionFree

Correct answer: C

  1. For an electromagnetic wave in vacuum (or air), the amplitudes of electric and magnetic fields are related by

E0=cB0E_0 = c B_0E0​=cB0​

where:

  • E0E_0E0​ = maximum electric field
  • B0B_0B0​ = maximum magnetic field
  • c=3×108 m/sc = 3 \times 10^8 \, \text{m/s}c=3×108m/s
  1. Given:

B0=6.0×10−7 TB_0 = 6.0 \times 10^{-7} \, \text{T}B0​=6.0×10−7T

So,

E0=(3×108)(6.0×10−7)E_0 = (3 \times 10^8)(6.0 \times 10^{-7})E0​=(3×108)(6.0×10−7)

  1. Calculate:

E0=18×101=180 V/mE_0 = 18 \times 10^1 = 180 \, \text{V/m}E0​=18×101=180V/m

  1. Therefore, the maximum electric field is

180 V m−1\boxed{180 \, \text{V m}^{-1}}180V m−1​

  1. Checking options:
  • A: 6.0×10−7 V m−16.0 \times 10^{-7} \, \text{V m}^{-1}6.0×10−7V m−1 ✗
  • B: 5×1014 V m−15 \times 10^{14} \, \text{V m}^{-1}5×1014V m−1 ✗
  • C: 180 V m−1180 \, \text{V m}^{-1}180V m−1 ✓
  • D: 2×1015 V m−12 \times 10^{15} \, \text{V m}^{-1}2×1015V m−1 ✗

Hence, the correct option is C.

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