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Physics · Electromagnetic Waves

NEET (UG) 2023 — Question 6

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} and amplitude 48Vm−148 \mathrm{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} )

  1. Option A:

    1.6×10−8 T1.6 \times 10^{-8} \mathrm{~T}

  2. Option B:

    1.6×10−7 T1.6 \times 10^{-7} \mathrm{~T}

    Correct
  3. Option C:

    1.6×10−6 T1.6 \times 10^{-6} \mathrm{~T}

  4. Option D:

    1.6×10−9 T1.6 \times 10^{-9} \mathrm{~T}

Answer: B

Step-by-step solution

Sol. From the properties of electromagnetic wave we know that, C=E0B0C=\frac{E_0}{B_0} E0⇒E_0 \Rightarrow Amplitude of oscillating electric field B0⇒B_0 \Rightarrow Amplitude of oscillating magnetic field

⇒B0=483×108=1.6×10−7 T\Rightarrow B_0=\frac{48}{3 \times 10^8}=1.6 \times 10^{-7} \mathrm{~T}

Answer key and solution verified before publishing.

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Exam
NEET (UG) 2023
Paper
2023 paper
Subject
Physics
Chapter
Electromagnetic Waves
Topic
Properties of EM Waves and Electromagnetic Spectrum

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