A plane electromagnetic wave of frequency 20 | vedclass

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(D) For an electromagnetic wave in free space, the relationship between the magnitude of the electric field $(E)$ and the magnetic field $(B)$ is give

Vedclass · Accepted Answer

$B_z = 3.1 	imes 10^{-8} \ T$

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(D) For an electromagnetic wave in free space, the relationship between the magnitude of the electric field $(E)$ and the magnetic field $(B)$ is given by $E = Bc$, where $c$ is the speed of light in vacuum $(c = 3 	imes 10^8 \ ms^{-1})$.Given $E_y = 9.3 \ Vm^{-1}$.Using the relation $B_z = \frac{E_y}{c}$, we get:$B_z = \frac{9.3}{3 	imes 10^8} \ T$.$B_z = 3.1 	imes 10^{-8} \ T$.Since the wave travels in the $+x$ direction and the electric field is in the $+y$ direction, the magnetic field must be in the $+z$ direction to satisfy the direction of propagation ($\vec{E} 	imes \vec{B}$ direction).

$A$ plane electromagnetic wave of frequency $20 \ MHz$ travels in free space along the $+x$ direction. At a particular point in space and time, the electric field vector of the wave is $E_y = 9.3 \ Vm^{-1}$. Then, the magnetic field vector of the wave at that point is:

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