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NEET PHYSICSEasy

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 V m−148 \text{ V m}^{-1}. Then the amplitude of the oscillating magnetic field is: (Speed of light in free space =3×108 m s−1= 3 \times 10^8 \text{ m s}^{-1})

A

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

B

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

C

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

D

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

Step-by-Step Solution

  1. Formula: For an electromagnetic wave propagating in free space, the amplitude of the magnetic field (B0B_0) is related to the amplitude of the electric field (E0E_0) and the speed of light (cc) by the relationship: B0=E0cB_0 = \frac{E_0}{c} (Reference: NCERT Physics Class 12, Chapter 8, Electromagnetic Waves).
  2. Given Data:
  • Electric Field Amplitude, E0=48 V m−1E_0 = 48 \text{ V m}^{-1}.
  • Speed of light, c=3×108 m s−1c = 3 \times 10^8 \text{ m s}^{-1}.
  • Frequency, ν=2.0×1010 Hz\nu = 2.0 \times 10^{10} \text{ Hz} (Note: Frequency is not required to find the amplitude ratio).
  1. Calculation: Substituting the values into the formula: B0=483×108 TB_0 = \frac{48}{3 \times 10^8} \text{ T} B0=16×10−8 TB_0 = 16 \times 10^{-8} \text{ T} Expressing in standard scientific notation: B0=1.6×10−7 TB_0 = 1.6 \times 10^{-7} \text{ T}
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