A plane electromagnetic wave of frequency 25 | vedclass

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Question

(B) The relationship between the magnitudes of the electric field $(E)$ and the magnetic field $(B)$ in an electromagnetic wave is given by $E = cB$,w

Vedclass · Accepted Answer

$6.3 \hat{j} 	ext{ Vm}^{-1}$

Vedclass · Answer

(B) The relationship between the magnitudes of the electric field $(E)$ and the magnetic field $(B)$ in an electromagnetic wave is given by $E = cB$,where $c$ is the speed of light in free space $(c = 3 	imes 10^8 	ext{ ms}^{-1})$.Substituting the given values: $E = (3 	imes 10^8 	ext{ ms}^{-1}) 	imes (2.1 	imes 10^{-8} 	ext{ T}) = 6.3 	ext{ Vm}^{-1}$.The direction of propagation of the electromagnetic wave is given by the direction of the vector $\overrightarrow{E} 	imes \overrightarrow{B}$.Given that the wave travels along the $X$-direction $(\hat{i})$ and $\overrightarrow{B}$ is along the $Z$-direction $(\hat{k})$,we have $\overrightarrow{E} 	imes \hat{k} = \hat{i}$.Since $\hat{j} 	imes \hat{k} = \hat{i}$,the electric field $\overrightarrow{E}$ must be in the $Y$-direction $(\hat{j})$.Therefore,$\overrightarrow{E} = 6.3 \hat{j} 	ext{ Vm}^{-1}$.

$A$ plane electromagnetic wave of frequency $25 \text{ MHz}$ travels in free space along the $X$-direction. At a particular point in space and time,the magnetic field is $\overrightarrow{B} = 2.1 \times 10^{-8} \hat{k} \text{ T}$. Find the electric field $\overrightarrow{E}$ at this point.

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