A plane electromagnetic wave travels in free | vedclass

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Question

(A) We know that the ratio of the magnitudes of the electric field $(E)$ and the magnetic field $(B)$ in an electromagnetic wave is equal to the speed

Vedclass · Accepted Answer

$2.9 	imes 10^{-8} \ T$

Vedclass · Answer

(A) We know that the ratio of the magnitudes of the electric field $(E)$ and the magnetic field $(B)$ in an electromagnetic wave is equal to the speed of light $(c)$ in free space, given by the relation: $c = \frac{E}{B}$.Given: $E = 8.7 \ V \ m^{-1}$ and $c = 3 	imes 10^8 \ m \ s^{-1}$.Rearranging the formula to solve for the magnetic field $B$: $B = \frac{E}{c}$.Substituting the values: $B = \frac{8.7}{3 	imes 10^8} = 2.9 	imes 10^{-8} \ T$.Since the wave travels along the $z$-axis and the electric field is along the $x$-axis, the magnetic field must be along the $y$-axis to satisfy the direction of propagation ($\vec{E} 	imes \vec{B}$ direction).

$A$ plane electromagnetic wave travels in free space along the $z$-axis. At a particular point in space, the electric field along the $x$-axis is $8.7 \ V \ m^{-1}$. The magnetic field along the $y$-axis is:

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The correct statement among the following is:

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