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(B) Given: $E_0 = 150 	ext{ V/m}$,$f = 30 	ext{ MHz} = 30 	imes 10^6 	ext{ Hz}$.$1$. Calculate the magnetic field amplitude: $B_0 = \frac{E_0}{c}

Vedclass · Accepted Answer

$5 	imes 10^{-7} \sin \left[\pi \left(\frac{x}{5}-6 	imes 10^{+7} tight)ight] \hat{z} 	ext{ T}$

Vedclass · Answer

(B) Given: $E_0 = 150 	ext{ V/m}$,$f = 30 	ext{ MHz} = 30 	imes 10^6 	ext{ Hz}$.$1$. Calculate the magnetic field amplitude: $B_0 = \frac{E_0}{c} = \frac{150}{3 	imes 10^8} = 5 	imes 10^{-7} 	ext{ T}$.$2$. Calculate angular frequency: $\omega = 2\pi f = 2\pi 	imes 30 	imes 10^6 = 60\pi 	imes 10^6 	ext{ rad/s} = 6\pi 	imes 10^7 	ext{ rad/s}$.$3$. Calculate wave number: $k = \frac{\omega}{c} = \frac{6\pi 	imes 10^7}{3 	imes 10^8} = 0.2\pi = \frac{\pi}{5} 	ext{ rad/m}$.$4$. Direction: The wave propagates along $\hat{i}$ ($x$-axis) and $\vec{E}$ is along $\hat{j}$ ($y$-axis). Since $\vec{B}$ must be perpendicular to both $\vec{E}$ and the direction of propagation,$\vec{B}$ is along $\hat{j} 	imes \hat{i} = -\hat{k}$ or $\hat{k} 	imes \hat{i} = \hat{j}$. Specifically,$\vec{E} 	imes \vec{B}$ must be in the direction of propagation $(\hat{i})$. Thus,$\hat{j} 	imes \hat{k} = \hat{i}$,so $\vec{B}$ is along the $z$-axis.$5$. The wave equation is $\vec{B} = B_0 \sin(kx - \omega t) \hat{k} = 5 	imes 10^{-7} \sin \left[\pi \left(\frac{x}{5} - 6 	imes 10^7 tight)ight] \hat{z} 	ext{ T}$.

In a plane $EM$ wave,the electric field oscillates sinusoidally at a frequency of $30 \text{ MHz}$ and amplitude $150 \text{ V/m}$. Identify the correct expression for $\vec{B}$ assuming the wave is propagating along the $x$-axis and the electric field is oscillating along the $y$-axis.

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