The electric field of a plane electromagnetic | vedclass

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Question

(A) The electric field is $\vec E = E_0 \hat i \cos(kz) \cos(\omega t)$.Since the wave propagates in the $+z$ direction and $\vec E$ is along the $x$-

Vedclass · Accepted Answer

$\vec B = \frac{E_0}{c} \hat j \sin(kz) \sin(\omega t)$

Vedclass · Answer

(A) The electric field is $\vec E = E_0 \hat i \cos(kz) \cos(\omega t)$.Since the wave propagates in the $+z$ direction and $\vec E$ is along the $x$-axis,the magnetic field $\vec B$ must be along the $y$-axis.Using Faraday's Law in differential form: $
abla 	imes \vec E = -\frac{\partial \vec B}{\partial t}$.For a wave propagating in the $z$-direction,$\frac{\partial E_x}{\partial z} = -\frac{\partial B_y}{\partial t}$.Calculating the derivative of $E_x$ with respect to $z$: $\frac{\partial}{\partial z} [E_0 \cos(kz) \cos(\omega t)] = -k E_0 \sin(kz) \cos(\omega t)$.Thus,$\frac{\partial B_y}{\partial t} = k E_0 \sin(kz) \cos(\omega t)$.Integrating with respect to $t$: $B_y = \int k E_0 \sin(kz) \cos(\omega t) dt = \frac{k E_0}{\omega} \sin(kz) \sin(\omega t)$.Since $c = \frac{\omega}{k}$,we have $\frac{k}{\omega} = \frac{1}{c}$.Therefore,$\vec B = \frac{E_0}{c} \hat j \sin(kz) \sin(\omega t)$.

The electric field of a plane electromagnetic wave is given by $\vec E = E_0 \hat i \cos(kz) \cos(\omega t)$. The corresponding magnetic field $\vec B$ is then given by:

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