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Electromagnetic Waves question

2022 · 25 Jul · Shift 2 · Q53
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  5. /2022 · 25 Jul · Shift 2 · Q53

Electromagnetic Waves question

2022 · 25 Jul · Shift 2 · Q53

JEE MainPhysicsElectromagnetic WavesMCQ+4 / −1
Light wave travelling in air along x-direction is given by Ey=540sin⁡π×104(x−ct) Vm−1{E_y} = 540\sin \pi \times {10^4}(x - ct)\,V{m^{ - 1}}Ey​=540sinπ×104(x−ct)Vm−1. Then, the peak value of magnetic field of wave will be (Given c = 3 ×\times× 108 ms −-− 1)
  1. A
    18 ×\times× 10 −-− 7 T
  2. B
    54 ×\times× 10 −-− 7 T
  3. C
    54 ×\times× 10 −-− 8 T
  4. D
    18 ×\times× 10 −-− 8 T
View written solutionFree

Correct answer: A

  1. Identify the electric field amplitude

    The wave is given by Ey=540sin⁡(π×104(x−ct)) V m−1E_y = 540\sin\big(\pi\times 10^4(x-ct)\big)\,\text{V m}^{-1}Ey​=540sin(π×104(x−ct))V m−1

    So the peak value of electric field is E0=540 V m−1E_0 = 540\,\text{V m}^{-1}E0​=540V m−1

  2. Use the relation between electric and magnetic fields in an electromagnetic wave

    For a light wave in air (approximately vacuum), E0=cB0E_0 = cB_0E0​=cB0​ where B0B_0B0​ is the peak magnetic field.

    Hence, B0=E0cB_0 = \frac{E_0}{c}B0​=cE0​​

  3. Substitute the given values

    B0=5403×108B_0 = \frac{540}{3\times 10^8}B0​=3×108540​

    B0=180×10−8B_0 = 180\times 10^{-8}B0​=180×10−8

    B0=1.8×10−6 TB_0 = 1.8\times 10^{-6}\,\text{T}B0​=1.8×10−6T

  4. Write in the form of the options

    1.8×10−6=18×10−7 T1.8\times 10^{-6} = 18\times 10^{-7}\,\text{T}1.8×10−6=18×10−7T

  5. Match with the options

    Therefore the correct option is: A: 18×10−7 T\boxed{\text{A: }18\times 10^{-7}\,\text{T}}A: 18×10−7T​

  6. Comparison with stored answer

    Stored correct answer: A

    Derived answer: A

    They match.

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