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Physics · PYQ NEET UG 2023

हिन्दी

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

  1. A
    \( 1.6 \times 10^{-9}\text{ T} \)
  2. B
    \( 1.6 \times 10^{-8}\text{ T} \)
  3. C
    \( 1.6 \times 10^{-6}\text{ T} \)
  4. D
    \( 1.6 \times 10^{-7}\text{ T} \)
View Answer & Solution

Correct answer · Option D

\( 1.6 \times 10^{-7}\text{ T} \)

Explanation

Amplitude of magnetic field \( B_0 = \frac{E_0}{c} = \frac{48}{3 \times 10^8} = 1.6 \times 10^{-7}\text{ T} \).
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