A plane electromagnetic wave of frequency 50, | vedclass

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(B) For an electromagnetic wave, the relationship between the electric field magnitude $E$ and the magnetic field magnitude $B$ is given by $E = cB$,

Vedclass · Accepted Answer

$2.1 	imes 10^{-8}\,\hat k\,T$

Vedclass · Answer

(B) For an electromagnetic wave, the relationship between the electric field magnitude $E$ and the magnetic field magnitude $B$ is given by $E = cB$, where $c$ is the speed of light in free space $(c = 3 	imes 10^8\, m/s)$.Therefore, $B = \frac{E}{c} = \frac{6.3}{3 	imes 10^8} = 2.1 	imes 10^{-8}\, T$.The direction of the wave propagation is given by the vector $\vec{E} 	imes \vec{B}$. Since the wave travels in the $+x$ direction $(\hat i)$ and the electric field is in the $+y$ direction $(\hat j)$, we have $\hat i = \hat j 	imes \hat B$.This implies that the direction of the magnetic field $\vec{B}$ must be along the $+z$ direction $(\hat k)$, because $\hat j 	imes \hat k = \hat i$.Thus, $\vec{B} = 2.1 	imes 10^{-8}\,\hat k\,T$.

$A$ plane electromagnetic wave of frequency $50\, MHz$ travels in free space along the positive $x-$ direction. At a particular point in space and time, $\vec E = 6.3\,\hat j\,V/m$. The corresponding magnetic field $\vec B$ at that point will be:

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