If the magnetic field of a plane electromagnetic wave is given as,
\( \vec{B} = 30\cos(1600x + 48\times10^{10}t)\hat{\jmath}\,\mathrm{nT} \), where \( x \) is in meter and \( t \) is in second,
then, its associated electric field will be:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( 9\cos(1600x + 48\times10^{10}t)\hat{k}\,\mathrm{V/m} \)
In a plane electromagnetic wave, \( \vec{E} \) and \( \vec{B} \) oscillate in phase, their amplitudes are related by \( E_0 = cB_0 \), and \( \vec{E}\times\vec{B} \) points along the direction of propagation.
The phase \( (1600x + 48\times10^{10}t) \) has \( kx \) and \( \omega t \) with the same sign, so the wave travels along \( -x \). Its speed checks out as \( c \):
\( \displaystyle v = \frac{\omega}{k} = \frac{48\times10^{10}\,\mathrm{rad\,s^{-1}}}{1600\,\mathrm{rad\,m^{-1}}} = 3\times10^{8}\,\mathrm{m/s} \)
\( B_0 = 30\,\mathrm{nT} = 30\times10^{-9}\,\mathrm{T} \)
\( E_0 = cB_0 = 3\times10^{8}\,\mathrm{m/s}\times30\times10^{-9}\,\mathrm{T} = 9\,\mathrm{V/m} \)
Direction: \( \vec{B} \) is along \( \hat{\jmath} \) and \( \vec{E}\times\vec{B} \) must be along \( -\hat{\imath} \). Since \( \hat{k}\times\hat{\jmath} = -\hat{\imath} \), \( \vec{E} \) is along \( +\hat{k} \).
\( \vec{E} = 9\cos(1600x + 48\times10^{10}t)\hat{k}\,\mathrm{V/m} \)