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Question

(A) The direction of propagation of the wave is given by the wave vector $\vec{k}$, which is along the $z$-axis ($+\hat{k}$ direction). The relationsh

Vedclass · Accepted Answer

$-1.8 	imes 10^{-7} \sin(kz - \omega t) \hat{i} \ T$

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(A) The direction of propagation of the wave is given by the wave vector $\vec{k}$, which is along the $z$-axis ($+\hat{k}$ direction). The relationship between the electric field $\vec{E}$, magnetic field $\vec{B}$, and the direction of propagation $\hat{c}$ is given by $\vec{B} = \frac{1}{c} (\hat{c} 	imes \vec{E})$. Here, $\hat{c} = \hat{k}$ and $\vec{E} = E_0 \sin(kz - \omega t) \hat{j}$. The direction of $\vec{B}$ is $\hat{k} 	imes \hat{j} = -\hat{i}$. The magnitude of the magnetic field is $B_0 = \frac{E_0}{c} = \frac{54}{3 	imes 10^8} = 18 	imes 10^{-8} = 1.8 	imes 10^{-7} \ T$. Therefore, $\vec{B} = -1.8 	imes 10^{-7} \sin(kz - \omega t) \hat{i} \ T$.

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

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