The average electric field of electromagnetic waves in certain region of free space is \( 9\times10^{-4}\,\mathrm{N\,C^{-1}} \). Then the average magnetic field in the same region is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 3\times10^{-12}\,\mathrm{T} \)
In an electromagnetic wave in free space the electric and magnetic fields are in phase, and at every instant their magnitudes are in the ratio \( E/B = c \). The same ratio therefore holds for their average values.
\( \displaystyle B = \frac{E}{c} = \frac{9\times10^{-4}\,\mathrm{N\,C^{-1}}}{3\times10^{8}\,\mathrm{m\,s^{-1}}} = 3\times10^{-12}\,\mathrm{T} \)
Shortcut: The magnetic field of a light wave is always very small: dividing by \( c = 3\times10^{8}\,\mathrm{m\,s^{-1}} \) lowers the power of ten by 8, from \( 10^{-4} \) to about \( 10^{-12} \). Only Options (2) and (3) have that size, and \( 9/3 = 3 \) picks Option (2).