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(D) Given: Amplitude of electric field $E_0 = 66\, Vm^{-1}$,Wavelength $\lambda = 3\, mm = 3 	imes 10^{-3}\, m$,Wave speed $c = 3 	imes 10^8\, ms^{-

Vedclass · Accepted Answer

$E_y = 66\cos(2\pi 	imes 10^{11}(t - x/c))$,$B_z = 2.2 	imes 10^{-7}\cos(2\pi 	imes 10^{11}(t - x/c))$

Vedclass · Answer

(D) Given: Amplitude of electric field $E_0 = 66\, Vm^{-1}$,Wavelength $\lambda = 3\, mm = 3 	imes 10^{-3}\, m$,Wave speed $c = 3 	imes 10^8\, ms^{-1}$.$1$. Calculate the amplitude of the magnetic field $B_0$:$B_0 = E_0 / c = 66 / (3 	imes 10^8) = 22 	imes 10^{-8} = 2.2 	imes 10^{-7}\, T$.$2$. Calculate the angular frequency $\omega$:$\omega = 2\pi f = 2\pi (c / \lambda) = 2\pi (3 	imes 10^8 / 3 	imes 10^{-3}) = 2\pi 	imes 10^{11}\, rad/s$.$3$. Determine the directions:The wave travels along $X$,and $E$ is along $Y$. Since the direction of propagation is given by $\vec{E} 	imes \vec{B}$,the magnetic field $B$ must be along the $Z-$ direction.$4$. Write the equations:$E_y = E_0 \cos(\omega(t - x/c)) = 66 \cos(2\pi 	imes 10^{11}(t - x/c))$.$B_z = B_0 \cos(\omega(t - x/c)) = 2.2 	imes 10^{-7} \cos(2\pi 	imes 10^{11}(t - x/c))$.Comparing with the options,option $D$ matches these values.

$A$ plane electromagnetic wave travelling along the $X-$ direction has a wavelength of $3\, mm$. The variation in the electric field occurs in the $Y-$ direction with an amplitude of $66\, Vm^{-1}$. The equations for the electric and magnetic fields as a function of $x$ and $t$ are respectively:

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