In a plane electromagnetic wave travelling in free space, the electric field component oscillates sinusoidally

Question

In a plane electromagnetic wave travelling in free space, the electric field component oscillates sinusoidally at a frequency of $2.0 	imes 10^{10} 	ext{ Hz}$ and amplitude $48 	ext{ V m}^{-1}$. Then the amplitude of the oscillating magnetic field is: (Speed of light in free space $= 3 	imes 10^8 	ext{ m s}^{-1}$)

Accepted Answer

1. **Formula:** For an electromagnetic wave propagating in free space, the amplitude of the magnetic field ($B_0$) is related to the amplitude of the electric field ($E_0$) and the speed of light ($c$) by the relationship:
   $$B_0 = \frac{E_0}{c}$$
   (Reference: NCERT Physics Class 12, Chapter 8, Electromagnetic Waves).
2. **Given Data:**
   - Electric Field Amplitude, $E_0 = 48 	ext{ V m}^{-1}$.
   - Speed of light, $c = 3 	imes 10^8 	ext{ m s}^{-1}$.
   - Frequency, $
u = 2.0 	imes 10^{10} 	ext{ Hz}$ (Note: Frequency is not required to find the amplitude ratio).
3. **Calculation:**
   Substituting the values into the formula:
   $$B_0 = \frac{48}{3 	imes 10^8} 	ext{ T}$$
   $$B_0 = 16 	imes 10^{-8} 	ext{ T}$$
   Expressing in standard scientific notation:
   $$B_0 = 1.6 	imes 10^{-7} 	ext{ T}$$

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