The electric field in a plane electromagnetic | vedclass

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Question

(A) The wave equation is given by $E_z = E_0 \cos(kx + \omega t)$. Comparing this with the given equation,we have $E_0 = 60 	ext{ V/m}$,$k = 5 	ext{

Vedclass · Accepted Answer

$B_y = 2 	imes 10^{-7} \cos(5x + 1.5 	imes 10^9 t) 	ext{ T}$

Vedclass · Answer

(A) The wave equation is given by $E_z = E_0 \cos(kx + \omega t)$. Comparing this with the given equation,we have $E_0 = 60 	ext{ V/m}$,$k = 5 	ext{ rad/m}$,and $\omega = 1.5 	imes 10^9 	ext{ rad/s}$.The speed of the wave is $v = \frac{\omega}{k} = \frac{1.5 	imes 10^9}{5} = 3 	imes 10^8 	ext{ m/s}$.The amplitude of the magnetic field is $B_0 = \frac{E_0}{v} = \frac{60}{3 	imes 10^8} = 2 	imes 10^{-7} 	ext{ T}$.The direction of propagation is given by the vector $\vec{E} 	imes \vec{B}$. Since the wave propagates in the $-x$ direction and $\vec{E}$ is along the $z$-axis,we have $(-\hat{i}) = \hat{k} 	imes \vec{B}$. This implies $\vec{B}$ must be along the $y$-axis (since $\hat{k} 	imes \hat{j} = -\hat{i}$). Thus,the magnetic field is $B_y = B_0 \cos(kx + \omega t) = 2 	imes 10^{-7} \cos(5x + 1.5 	imes 10^9 t) 	ext{ T}$.

The electric field in a plane electromagnetic wave is given by $E_z = 60 \cos(5x + 1.5 \times 10^9 t) \text{ V/m}$. Then the expression for the corresponding magnetic field is (here subscripts denote the direction of the field):

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