The magnetic field of a plane electromagnetic | vedclass

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Question

(B) Given the magnetic field: $\overrightarrow{B} = 3 	imes 10^{-8} \sin [200 \pi(y + ct)] \hat{i} \, T$.The amplitude of the electric field $E_0$ is

Vedclass · Accepted Answer

$\overrightarrow{E} = -9 \sin [200 \pi(y + ct)] \hat{k} \, V/m$

Vedclass · Answer

(B) Given the magnetic field: $\overrightarrow{B} = 3 	imes 10^{-8} \sin [200 \pi(y + ct)] \hat{i} \, T$.The amplitude of the electric field $E_0$ is related to the amplitude of the magnetic field $B_0$ by the relation $E_0 = c B_0$.Substituting the values: $E_0 = (3 	imes 10^{8} \, m/s) 	imes (3 	imes 10^{-8} \, T) = 9 \, V/m$.The direction of wave propagation is given by the vector $\overrightarrow{k}$. Since the argument is $(y + ct)$,the wave is propagating in the negative $y$-direction,so $\hat{k}_{prop} = -\hat{j}$.The direction of the electric field is given by the cross product relation $\hat{E} = \hat{k}_{prop} 	imes \hat{B}$.Here,$\hat{B} = \hat{i}$ and $\hat{k}_{prop} = -\hat{j}$.Therefore,$\hat{E} = (-\hat{j}) 	imes \hat{i} = -(-\hat{k}) = \hat{k}$.Wait,checking the cross product: $(-\hat{j}) 	imes \hat{i} = -(\hat{j} 	imes \hat{i}) = -(-\hat{k}) = \hat{k}$.However,standard convention for electromagnetic waves is $\vec{E} 	imes \vec{B} = \vec{S}$ (direction of propagation). Given $\vec{B} = B_0 \hat{i}$ and propagation is $-\hat{j}$,then $\vec{E} 	imes \hat{i} = -\hat{j}$. This implies $\vec{E} = -9 \sin [200 \pi(y + ct)] \hat{k} \, V/m$.

The magnetic field of a plane electromagnetic wave is $\overrightarrow{B} = 3 \times 10^{-8} \sin [200 \pi(y + ct)] \hat{i} \, T$,where $c = 3 \times 10^{8} \, m/s$ is the speed of light. The corresponding electric field is:

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