The amplitude of the magnetic field of a harmonic electromagnetic wave in vacuum is \( B_o = 610\,\mathrm{nT} \). The amplitude of the electric field of the wave is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( 183\,\mathrm{NC^{-1}} \)
In an electromagnetic wave in vacuum, the field amplitudes are related by the speed of light:
\( \displaystyle E_0 = c\,B_0 \)
\( B_0 = 610\,\mathrm{nT} = 610\times10^{-9}\,\mathrm{T} \)
\( c = 3\times10^{8}\,\mathrm{m\,s^{-1}} \)
\( \displaystyle E_0 = 3\times10^{8} \times 610\times10^{-9} = 183\,\mathrm{N\,C^{-1}} \)
Note: NTA picked this question from an NCERT (2026-27) exercise and changed the figures. Do not ignore NCERT while practising numericals.
NCERT (2026-27):
Exercises – …
