Verified Solution • 2025–26 Session moderate

QNEET-2023-Ph-239: NEET 2023 Physics - Class 12 NEET-2023-Ph-239 Solution: In a plane electromagnetic wave travelling in free space the…

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}\ \text{Hz} and amplitude 48 Vm−148\ \text{Vm}^{-1}. Then the amplitude of oscillating magnetic field is: (Speed of light in free space = 3×108 m s−13 \times 10^8\ \text{m s}^{-1}) 1.6×10−8 T1.6 \times 10^{-8}\ \text{T} 1.6×10−7 T1.6 \times 10^{-7}\ \text{T} 1.6×10−6 T1.6 \times 10^{-6}\ \text{T} 1.6×10−9 T1.6 \times 10^{-9}\ \text{T}

Multiple Choice Options

  • A 1.6×10−8 T1.6 \times 10^{-8} \mathrm{~T} 1.6 _ 1 0 _ 8   T
  • B 1.6×10−7 T1.6 \times 10^{-7} \mathrm{~T} 1.6 _ 1 0 _ 7   T
  • C 1.6×10−6 T1.6 \times 10^{-6} \mathrm{~T} 1.6 _ 1 0 _ 6   T
  • D 1.6×10−9 T1.6 \times 10^{-9} \mathrm{~T} 1.6 _ 1 0 _ 9   T

Step-by-Step Verified Solution

C = E 0 B 0 B 0 = E 0 C = 48 3 _ 10 8 = 1.6 _ 10 _ 7   T

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