A plane electromagnetic wave is moving in free space with velocity $c=3 imes 10^8 \mathrm{~m} / \m

Question

A plane electromagnetic wave is moving in free space with velocity $c=3 	imes 10^8 \mathrm{~m} / \mathrm{s}$ and its electric field is given as $\vec{E}=54 \sin (k z-\omega t) \hat{j} \mathrm{~V} / \mathrm{m}$, where $\hat{j}$ is the unit vector along $y$-axis. The magnetic field vector $\vec{B}$ of the wave is:

Professor PC Thomas Classes · Answer

The amplitude of the magnetic field is $B_0 = \frac{E_0}{c} = \frac{54}{3 	imes 10^8} = 18 	imes 10^{-8} = 1.8 	imes 10^{-7}$ T. The wave propagates in the $+z$ direction (deduced from $kz - \omega t$), which corresponds to $\hat{k}$. The electric field oscillates along the $+y$ direction ($\hat{j}$). The direction of wave propagation is given by the cross product of the electric and magnetic fields, $\hat{E} 	imes \hat{B} = \hat{c}$. Substituting the unit vectors: $\hat{j} 	imes \hat{B} = \hat{k}$. Using the cyclic property of cross products, $\hat{j} 	imes (-\hat{i}) = \hat{k}$. Thus, the magnetic field must oscillate along the $-\hat{i}$ direction. The complete equation is $\vec{B} = -1.8 	imes 10^{-7} \sin(kz - \omega t) \hat{i} 	ext{ T}$.