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Electromagnetic Waves question

2021 · 26 Aug · Shift 2 · Q51
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  5. /2021 · 26 Aug · Shift 2 · Q51

Electromagnetic Waves question

2021 · 26 Aug · Shift 2 · Q51

JEE MainPhysicsElectromagnetic WavesMCQ+4 / −1
A light beam is described by E=800sin⁡ω(t−xc)E = 800\sin \omega \left( {t - {x \over c}} \right)E=800sinω(t−cx​). An electron is allowed to move normal to the propagation of light beam with a speed of 3 ×\times× 107 ms −-− 1. What is the maximum magnetic force exerted on the electron?
  1. A
    1.28 ×\times× 10 −-− 18 N
  2. B
    1.28 ×\times× 10 −-− 21 N
  3. C
    12.8 ×\times× 10 −-− 17 N
  4. D
    12.8 ×\times× 10 −-− 18 N
View written solutionFree

Correct answer: D

  1. Given electromagnetic wave

    The electric field is E=800sin⁡ω(t−xc)E = 800\sin\omega\left(t-\frac{x}{c}\right)E=800sinω(t−cx​)

    So the amplitude of electric field is E0=800 V m−1E_0 = 800\ \text{V m}^{-1}E0​=800 V m−1

  2. Magnetic field amplitude in an electromagnetic wave

    For light in vacuum, E0=cB0E_0 = cB_0E0​=cB0​ hence B0=E0c=8003×108B_0 = \frac{E_0}{c} = \frac{800}{3\times 10^8}B0​=cE0​​=3×108800​

    B0=2.67×10−6 TB_0 = 2.67\times 10^{-6}\ \text{T}B0​=2.67×10−6 T

  3. Magnetic force on the electron

    Magnetic force magnitude is FB=∣q∣vBsin⁡θF_B = |q|vB\sin\thetaFB​=∣q∣vBsinθ

    The electron moves normal to the propagation direction of the light beam. In an electromagnetic wave, the magnetic field is also perpendicular to the propagation direction, so maximum force occurs when the electron velocity is perpendicular to BBB.

    Therefore, Fmax⁡=evB0F_{\max} = evB_0Fmax​=evB0​

    Given: e=1.6×10−19 Ce = 1.6\times 10^{-19}\ \text{C}e=1.6×10−19 C v=3×107 m s−1v = 3\times 10^7\ \text{m s}^{-1}v=3×107 m s−1 B0=2.67×10−6 TB_0 = 2.67\times 10^{-6}\ \text{T}B0​=2.67×10−6 T

    Substituting, Fmax⁡=(1.6×10−19)(3×107)(2.67×10−6)F_{\max} = (1.6\times 10^{-19})(3\times 10^7)(2.67\times 10^{-6})Fmax​=(1.6×10−19)(3×107)(2.67×10−6)

  4. Calculation

    First, 3×107×2.67×10−6=8.01×101≈803\times 10^7 \times 2.67\times 10^{-6} = 8.01\times 10^1 \approx 803×107×2.67×10−6=8.01×101≈80

    Then, Fmax⁡=1.6×10−19×80=128×10−19=1.28×10−17 NF_{\max} = 1.6\times 10^{-19}\times 80 = 128\times 10^{-19} = 1.28\times 10^{-17}\ \text{N}Fmax​=1.6×10−19×80=128×10−19=1.28×10−17 N

    This can also be written as 1.28×10−17 N=12.8×10−18 N1.28\times 10^{-17}\ \text{N} = 12.8\times 10^{-18}\ \text{N}1.28×10−17 N=12.8×10−18 N

  5. Match with options

    This corresponds to Option D.

Comparison with stored answer: Stored correct answer is D, which matches our result.

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