Suppose the electric field amplitude of an electromagnetic wave in vacuum is \( 120\ \mathrm{N/C} \) and its frequency is \( 50\ \mathrm{MHz} \). Then, for this electromagnetic wave:
A. magnetic field amplitude is \( 400\ \mathrm{nT} \).
B. angular frequency is \( 3.14\times10^{8}\ \mathrm{rad/s} \).
C. wavelength is \( 6\ \mathrm{cm} \).
D. the angular wave number is \( \sim 1.05\ \mathrm{rad/m} \).
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → A, B and D only
For an electromagnetic wave in vacuum, the amplitudes are linked by \( E_{0} = cB_{0} \), and the wave obeys \( c = \nu\lambda \), \( \omega = 2\pi\nu \) and \( k = 2\pi/\lambda \).
\( E_{0} = 120\,\mathrm{N\,C^{-1}},\ \nu = 50\,\mathrm{MHz} = 5\times10^{7}\,\mathrm{Hz},\ c = 3\times10^{8}\,\mathrm{m\,s^{-1}} \)
(A) magnetic field amplitude is \( 400\ \mathrm{nT} \). — Correct.
\( \displaystyle B_{0} = \frac{E_{0}}{c} = \frac{120\,\mathrm{N\,C^{-1}}}{3\times10^{8}\,\mathrm{m\,s^{-1}}} = 4\times10^{-7}\,\mathrm{T} = 400\,\mathrm{nT} \)
(B) angular frequency is \( 3.14\times10^{8}\ \mathrm{rad/s} \). — Correct.
\( \omega = 2\pi\nu = 2\pi\times(5\times10^{7}\,\mathrm{Hz}) = 3.14\times10^{8}\,\mathrm{rad\,s^{-1}} \)
(C) wavelength is \( 6\ \mathrm{cm} \). — Incorrect. The wavelength is 6 m, not 6 cm.
\( \displaystyle \lambda = \frac{c}{\nu} = \frac{3\times10^{8}\,\mathrm{m\,s^{-1}}}{5\times10^{7}\,\mathrm{Hz}} = 6\,\mathrm{m} \)
(D) the angular wave number is \( \sim 1.05\ \mathrm{rad/m} \). — Correct.
\( \displaystyle k = \frac{2\pi}{\lambda} = \frac{2\pi}{6\,\mathrm{m}} \approx 1.05\,\mathrm{rad\,m^{-1}} \)
Note: This question is from NCERT (2026-27). NTA sets numerical questions from NCERT too, so do not ignore NCERT while practising numericals.
NCERT (2026-27):
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