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(B) The relationship between the maximum electric field $(E_0)$ and maximum magnetic field $(B_0)$ is given by $E_0 = B_0 c$.Given $E_0 = 60 	ext{ Vm

Vedclass · Accepted Answer

$B_z = 2 	imes 10^{-7} \sin \left[ \frac{\pi}{2} 	imes 10^3 (x - 3 	imes 10^8 t) ight] \hat{k} 	ext{ T}$

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(B) The relationship between the maximum electric field $(E_0)$ and maximum magnetic field $(B_0)$ is given by $E_0 = B_0 c$.Given $E_0 = 60 	ext{ Vm}^{-1}$ and $c = 3 	imes 10^8 	ext{ ms}^{-1}$,we have $B_0 = \frac{E_0}{c} = \frac{60}{3 	imes 10^8} = 2 	imes 10^{-7} 	ext{ T}$.The wave number $k$ is given by $k = \frac{2\pi}{\lambda}$. Given $\lambda = 4 	ext{ mm} = 4 	imes 10^{-3} 	ext{ m}$,we have $k = \frac{2\pi}{4 	imes 10^{-3}} = \frac{\pi}{2} 	imes 10^3 	ext{ rad/m}$.The angular frequency $\omega$ is given by $\omega = ck = (3 	imes 10^8) 	imes (\frac{\pi}{2} 	imes 10^3) = \frac{3\pi}{2} 	imes 10^{11} 	ext{ rad/s}$.The wave propagates in the $+x$ direction and the electric field is in the $+y$ direction. Since the direction of propagation is $\vec{E} 	imes \vec{B}$,the magnetic field must be in the $+z$ direction.Thus,the equation for the magnetic field is $B_z = B_0 \sin(kx - \omega t) = 2 	imes 10^{-7} \sin \left[ \frac{\pi}{2} 	imes 10^3 (x - 3 	imes 10^8 t) ight] \hat{k} 	ext{ T}$.

$A$ plane electromagnetic $(EM)$ wave is propagating along the $x$-direction. It has a wavelength of $4 \text{ mm}$. If the electric field is in the $y$-direction with a maximum magnitude of $60 \text{ Vm}^{-1}$,the equation for the magnetic field is:

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