The magnetic field in a plane electromagnetic | vedclass

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Question

(A) In an electromagnetic wave,the relationship between the amplitudes of the electric field $(E_0)$ and the magnetic field $(B_0)$ is given by $E_0 =

Vedclass · Accepted Answer

$\vec E = B_0 c \sin(kx + \omega t) \hat k \ V/m$

Vedclass · Answer

(A) In an electromagnetic wave,the relationship between the amplitudes of the electric field $(E_0)$ and the magnetic field $(B_0)$ is given by $E_0 = c B_0$. Since the wave propagates in the negative $x$-direction (indicated by $kx + \omega t$),the direction of propagation is given by the cross product $\vec E 	imes \vec B$. The direction of propagation is $-\hat i$. Given $\vec B$ is in the $\hat j$ direction,we have $\vec E 	imes (B_0 \hat j) \propto -\hat i$. Since $\hat k 	imes \hat j = -\hat i$,the electric field must be in the $\hat k$ direction. Thus,$\vec E = E_0 \sin(kx + \omega t) \hat k = B_0 c \sin(kx + \omega t) \hat k \ V/m$.

The magnetic field in a plane electromagnetic wave is given by $\vec B = B_0 \sin(kx + \omega t) \hat j \ T$. The expression for the corresponding electric field will be (where $c$ is the speed of light).

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