A plane electromagnetic wave of frequency 25 | vedclass

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Question

(A) For an electromagnetic wave propagating in vacuum,the relationship between the magnitude of the electric field $(E)$ and the magnetic field $(B)$

Vedclass · Accepted Answer

$2.1 	imes 10^{-8} 	ext{ T}$

Vedclass · Answer

(A) For an electromagnetic wave propagating in vacuum,the relationship between the magnitude of the electric field $(E)$ and the magnetic field $(B)$ is given by $E = cB$,where $c$ is the speed of light in vacuum $(c \approx 3 	imes 10^8 	ext{ ms}^{-1})$.Given the electric field magnitude $E = 6.3 	ext{ Vm}^{-1}$.We need to find the magnitude of the magnetic field $B$.Using the formula $B = \frac{E}{c}$,we substitute the values:$B = \frac{6.3}{3 	imes 10^8} 	ext{ T}$.$B = 2.1 	imes 10^{-8} 	ext{ T}$.Thus,the magnitude of the magnetic field is $2.1 	imes 10^{-8} 	ext{ T}$.

$A$ plane electromagnetic wave of frequency $25 \text{ MHz}$ propagates in vacuum along the positive $x$-direction. At a particular point in space and time,if the electric field is $6.3 \hat{j} \text{ Vm}^{-1}$,then the magnitude of the magnetic field of the wave at this point at the same time is:

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