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NEET 2023
Exam
NEET
Year
2023
Subject
Physics
Chapter
Electromagnetic Waves

NEET 2023 · Physics · Electromagnetic Waves

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 2.0×1010Hz2 . 0 \times {10}^{10}Hz and amplitude 48Vm−148V{m}^{- 1}. Then the amplitude of oscillating magnetic field is: (Speed of light in free space =3×108ms−13 \times {10}^{8}m{s}^{- 1})

Options

  1. A1.6×10−6T1 . 6 \times {10}^{- 6}T
  2. B1.6×10−9T1 . 6 \times {10}^{- 9}T
  3. C1.6×10−8T1 . 6 \times {10}^{- 8}T
  4. D1.6×10−7T1 . 6 \times {10}^{- 7}T

Correct answer

D. 1.6×10−7T1 . 6 \times {10}^{- 7}T

Step-by-step solution

Frequency of oscillation =2×1010Hz= 2 \times {10}^{10}Hz
Given c=3×108ms−1c = 3 \times {10}^{8}m{s}^{- 1},

Electric field amplitude E0=48Vm−1{E}_{0} = 48V{m}^{- 1}

Using the formula,c=E0B0\text{Using the formula,}c = \dfrac{{E}_{0}}{{B}_{0}}

The amplitude of the oscillating magnetic field

B0=E0c=483×108=1.6×10−7TB{}_{0} = \dfrac{E{}_{0}}{c} = \dfrac{48}{3 \times 1{0}^{8}} = 1 . 6 \times 1{0}^{- 7}T.

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