In an electromagnetic wave propagating along the \( +x \) direction in vacuum, the magnetic field oscillates at a frequency of \( 3\times10^{14}\ \mathrm{Hz} \) and has an amplitude of \( 2\times10^{-6}\ \mathrm{T} \), acting along \( +y \) direction. Expression representing the corresponding oscillating electric field is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( \vec{E} = -600\sin(6\pi\times10^{14}t - 2\pi\times10^{6}x)\hat{k}\ \mathrm{V/m} \)
In an electromagnetic wave in vacuum, \( \vec{E} \) and \( \vec{B} \) are perpendicular to each other and to the direction of travel, the direction of propagation is that of \( \vec{E}\times\vec{B} \), and the amplitudes are related by \( E_{0} = cB_{0} \).
Amplitude:
\( E_{0} = cB_{0} = (3\times10^{8}\ \mathrm{m\,s^{-1}})(2\times10^{-6}\ \mathrm{T}) = 600\ \mathrm{V/m} \)
Angular frequency and propagation constant:
\( \omega = 2\pi\nu = 2\pi \times 3\times10^{14} = 6\pi\times10^{14}\ \mathrm{rad\,s^{-1}} \)
\( \displaystyle k = \frac{\omega}{c} = \frac{6\pi\times10^{14}}{3\times10^{8}} = 2\pi\times10^{6}\ \mathrm{m^{-1}} \)
Direction: with \( \vec{B} = B_{0}\sin(\omega t - kx)\,\hat{\jmath} \), the product \( \vec{E}\times\vec{B} \) must point along \( +\hat{\imath} \). Since \( \hat{k}\times\hat{\jmath} = -\hat{\imath} \), the electric field must point along \( -\hat{k} \) when \( \vec{B} \) points along \( +\hat{\jmath} \):
\( (-\hat{k})\times\hat{\jmath} = +\hat{\imath} \)
\( \vec{E} = -600\sin(6\pi\times10^{14}t - 2\pi\times10^{6}x)\,\hat{k}\ \mathrm{V/m} \)
Note: It seems that there is a mistake in the original NTA paper, which prints Options (3) and (4) identically. The question can still be solved, because both describe an electric field along \( \hat{\jmath} \), parallel to the magnetic field, and so neither can be correct.