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(B) Given: Wavelength $\lambda = 8\,mm = 8 	imes 10^{-3}\,m$. Maximum electric field $E_{0} = 60\,Vm^{-1}$.Propagation direction is along $+x$-axis.

Vedclass · Accepted Answer

$E_{y}=60 \sin \left[\frac{\pi}{4} 	imes 10^{3}(x - 3 	imes 10^{8}t)ight] \hat{j}\,Vm^{-1}$,$B_{z}=2 	imes 10^{-7} \sin \left[\frac{\pi}{4} 	imes 10^{3}(x - 3 	imes 10^{8}t)ight] \hat{k}\,T$

Vedclass · Answer

(B) Given: Wavelength $\lambda = 8\,mm = 8 	imes 10^{-3}\,m$. Maximum electric field $E_{0} = 60\,Vm^{-1}$.Propagation direction is along $+x$-axis. Electric field is along $y$-axis.$1$. Calculate magnetic field amplitude: $B_{0} = \frac{E_{0}}{c} = \frac{60}{3 	imes 10^{8}} = 2 	imes 10^{-7}\,T$.$2$. Determine direction of magnetic field: Since the wave propagates in the direction of $\vec{E} 	imes \vec{B}$,and $\vec{E}$ is in $\hat{j}$ and propagation is in $\hat{i}$,$\vec{B}$ must be in $\hat{k}$ direction.$3$. Calculate wave number $k$: $k = \frac{2\pi}{\lambda} = \frac{2\pi}{8 	imes 10^{-3}} = \frac{\pi}{4} 	imes 10^{3}\,m^{-1}$.$4$. The wave equation is $E = E_{0} \sin(k(x - ct))\hat{j}$ and $B = B_{0} \sin(k(x - ct))\hat{k}$.Substituting the values,we get $E_{y} = 60 \sin \left[\frac{\pi}{4} 	imes 10^{3}(x - 3 	imes 10^{8}t)ight] \hat{j}\,Vm^{-1}$ and $B_{z} = 2 	imes 10^{-7} \sin \left[\frac{\pi}{4} 	imes 10^{3}(x - 3 	imes 10^{8}t)ight] \hat{k}\,T$.

An $EM$ wave propagating in $x$-direction has a wavelength of $8\,mm$. The electric field vibrating in $y$-direction has a maximum magnitude of $60\,Vm^{-1}$. Choose the correct equations for electric and magnetic fields if the $EM$ wave is propagating in vacuum.

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