The amplitude of the electric field (along \( x \)-axis) of an electromagnetic wave propagating in free space along the \( z \)-axis is \( 150\hat{\imath} \) N/C. If its frequency is \( 100 \) MHz, then the expression for magnetic field is:
(Given: \( c=3\times10^{8} \) m/s)
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( \vec{B}=500\sin\left[-\left(\displaystyle\frac{2\pi}{3}\right)z+(2\pi\times10^{8})t\right]\hat{\jmath} \) nT
In an electromagnetic wave \( \vec{E} \), \( \vec{B} \) and the direction of propagation are mutually perpendicular, with the direction of propagation along \( \vec{E}\times\vec{B} \), and the amplitudes are related by \( B_0 = E_0/c \).
Direction: the wave travels along +z and \( \vec{E} \) is along \( \hat{\imath} \). Since \( \hat{\imath}\times\hat{\jmath} = \hat{k} \), \( \vec{B} \) must be along \( \hat{\jmath} \), and the phase must contain z (not x).
Amplitude:
\( \displaystyle B_0 = \frac{E_0}{c} = \frac{150}{3\times10^{8}} = 5\times10^{-7}\,\mathrm{T} = 500\,\mathrm{nT} \)
Angular frequency and wave number:
\( \omega = 2\pi f = 2\pi\times10^{8}\,\mathrm{rad\,s^{-1}} \)
\( \displaystyle k = \frac{\omega}{c} = \frac{2\pi\times10^{8}}{3\times10^{8}} = \frac{2\pi}{3}\,\mathrm{m^{-1}} \)
\( \displaystyle \vec{B} = 500\sin\left[-\left(\frac{2\pi}{3}\right)z+(2\pi\times10^{8})t\right]\hat{\jmath}\ \mathrm{nT} \)
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 – …
