In an electromagnetic wave of frequency 4 x 1010 Hz, propagating in vacuum, the amplitude of the electric field is 65 V/m. The amplitude of the magnetic field and the wavelength of the wave, respectively, will be:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 2.166 x 10−7 T, 7.5 x 10−3 m
In an electromagnetic wave in vacuum, the field amplitudes are related by \( E_0 = cB_0 \), and the wavelength follows from \( c = \nu\lambda \).
\( \displaystyle B_0 = \frac{E_0}{c} = \frac{65}{3\times10^{8}} \)
\( \displaystyle B_0 \approx 2.17\times10^{-7}\,\mathrm{T} \)
\( \displaystyle \lambda = \frac{c}{\nu} = \frac{3\times10^{8}}{4\times10^{10}} \)
\( \displaystyle \lambda = 7.5\times10^{-3}\,\mathrm{m} \)
Shortcut: The wavelength alone decides it: \( \lambda = c/\nu = 7.5\times10^{-3} \) m appears only in Option (1). Note that Option (2) has the same magnetic field written differently (21.66 × 10−8 T = 2.166 × 10−7 T) but the wrong wavelength.
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 – …
