NEET2023Physics-Electromagnetism

NEET 2023 Physics Electromagnetic Waves MCQ Question

Type: MCQ-numerical-Medium-Class 12

In a plane electromagnetic wave travelling in free space, the electric field component oscillates sinusoidally at a frequency of 2.0×10¹⁰ Hz and amplitude 48 Vm⁻¹. Then the amplitude of oscillating magnetic field is: (Speed of light in free space = 3 × 10⁸ m s⁻¹)

A

1.6×10⁻⁸ T

B

1.6×10⁻⁷ T

C

1.6×10⁻⁶ T

D

1.6×10⁻⁹ T

Correct Answer

Option B

Detailed Explanation

The amplitude of the magnetic field B0B_0 is given by B_0 = rac{E_0}{c}. Substituting E0=48Vm1E_0 = 48 \, Vm^{-1} and c=3imes108ms1c = 3 imes 10^8 \, ms^{-1}, we get B_0 = rac{48}{3 imes 10^8} = 1.6 imes 10^{-7} \, T.

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