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Electromagnetic Waves question

2019 · 10 Jan · Shift 1 · Q48
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Electromagnetic Waves question

2019 · 10 Jan · Shift 1 · Q48

JEE MainPhysicsElectromagnetic WavesMCQ+4 / −1
If the magnetic field of a plane electromagnetic wave is given by (the speed of light = 3 × 108 B = 100 × 10–6 sin [2π×2×1015(t−xc)]\left[ {2\pi \times 2 \times {{10}^{15}}\left( {t - {x \over c}} \right)} \right][2π×2×1015(t−cx​)] then the maximum electric field associated with it is -
  1. A
    4.5 ×\times× 104 N/C
  2. B
    4 ×\times× 104 N/C
  3. C
    6 ×\times× 104 N/C
  4. D
    3 ×\times× 104 N/C
View written solutionFree

Correct answer: D

  1. Given magnetic field of the electromagnetic wave

    The magnetic field is of the form B=100×10−6sin⁡[2π×2×1015(t−xc)]B = 100\times 10^{-6} \sin\left[2\pi\times 2\times 10^{15}\left(t-\frac{x}{c}\right)\right]B=100×10−6sin[2π×2×1015(t−cx​)]

    Hence, the amplitude (maximum value) of magnetic field is B0=100×10−6 T=10−4 TB_0 = 100\times 10^{-6}\,\text{T} = 10^{-4}\,\text{T}B0​=100×10−6T=10−4T

  2. Relation between electric and magnetic fields in an electromagnetic wave

    For a plane electromagnetic wave, E0=cB0E_0 = cB_0E0​=cB0​

    where c=3×108 m/sc = 3\times 10^8\,\text{m/s}c=3×108m/s

  3. Calculate the maximum electric field

    E0=(3×108)(10−4)E_0 = (3\times 10^8)(10^{-4})E0​=(3×108)(10−4) E0=3×104 N/CE_0 = 3\times 10^4\,\text{N/C}E0​=3×104N/C

  4. Match with the options

    3×104 N/C3\times 10^4\,\text{N/C}3×104N/C corresponds to Option D.

  5. Comparison with stored answer

    Stored correct answer: D

    Derived answer: D

    So, the derived answer agrees with the stored answer.

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