Q 12-08-007NEETNEET 2023Top questionEasy
In a plane electromagnetic wave travelling in free space, the electric field component oscillates sinusoidally at a frequency of $2.0 \times 10^{10}$ Hz and amplitude $48\ \text{V m}^{-1}$. Then the amplitude of oscillating magnetic field is : (Speed of light in free space $= 3 \times 10^8\ \text{m s}^{-1}$)
Answer: (C) $1.6 \times 10^{-7}$ T
$$B_0 = \frac{E_0}{c} = \frac{48}{3 \times 10^8} = 1.6 \times 10^{-7}\ \text{T}$$
(The frequency is not needed.)
Solution by Sreeraj P, M.Sc Physics