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Electromagnetic Waves question

2023 · Q156
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Electromagnetic Waves question

2023 · Q156

NEETPhysicsElectromagnetic WavesMCQ+4 / −1

In a plane electromagnetic wave travelling in free space, the electric field component oscillates sinusoidally at a frequency of 2.0×1010 Hz2.0 \times 10^{10} \mathrm{~Hz}2.0×1010 Hz and amplitude 48 Vm−148 ~\mathrm{Vm}^{-1}48 Vm−1. Then the amplitude of oscillating magnetic field is : (Speed of light in free space =3×108 m s−1=3 \times 10^{8} \mathrm{~m} \mathrm{~s}^{-1}=3×108 m s−1 )

  1. A
    1.6×10−8 T1.6 \times 10^{-8} \mathrm{~T}1.6×10−8 T
  2. B
    1.6×10−7 T1.6 \times 10^{-7} \mathrm{~T}1.6×10−7 T
  3. C
    1.6×10−6 T1.6 \times 10^{-6} \mathrm{~T}1.6×10−6 T
  4. D
    1.6×10−9 T1.6 \times 10^{-9} \mathrm{~T}1.6×10−9 T
View written solutionFree

Correct answer: B

$$C=\frac{E_{0}}{B_{0}}$$

$$\mathrm{B}_{0}=\frac{\mathrm{E}_{0}}{\mathrm{C}}$$

$$=\frac{48}{3 \times 10^{8}}$$

$$=1.6 \times 10^{-7} \mathrm{~T}$$

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