The electric field of a plane electromagnetic | vedclass

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Question

(A) The given electric field is $E_y = E_0 \sin(kx - \omega t)$, where $E_0 = 69 	ext{ V/m}$, $k = 0.6 	imes 10^3 	ext{ rad/m}$, and $\omega = 1.8

Vedclass · Accepted Answer

$B_z = 2.3 	imes 10^{-7} \sin[0.6 	imes 10^3 x - 1.8 	imes 10^{11} t]$

Vedclass · Answer

(A) The given electric field is $E_y = E_0 \sin(kx - \omega t)$, where $E_0 = 69 	ext{ V/m}$, $k = 0.6 	imes 10^3 	ext{ rad/m}$, and $\omega = 1.8 	imes 10^{11} 	ext{ rad/s}$.Since the wave propagates in the $+x$ direction $(\hat{i})$ and the electric field is in the $y$ direction $(\hat{j})$, the magnetic field must be in the $z$ direction $(\hat{k})$ because $\vec{B} = \frac{1}{c} (\hat{c} 	imes \vec{E}) = \frac{1}{c} (\hat{i} 	imes E_y \hat{j}) = \frac{E_y}{c} \hat{k}$.The amplitude of the magnetic field is $B_0 = \frac{E_0}{c} = \frac{69}{3 	imes 10^8} = 23 	imes 10^{-8} = 2.3 	imes 10^{-7} 	ext{ T}$.The phase of the magnetic field is the same as the electric field, so $B_z = B_0 \sin(kx - \omega t) = 2.3 	imes 10^{-7} \sin[0.6 	imes 10^3 x - 1.8 	imes 10^{11} t] 	ext{ T}$.

The electric field of a plane electromagnetic wave is given by: $E_y = 69 \sin[0.6 \times 10^3 x - 1.8 \times 10^{11} t] \text{ V/m}$. The expression for the magnetic field associated with this electromagnetic wave is . . . . . . $T$.

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