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In a plane EM wave the electric field oscillates sinusoidally at a frequency of $30 \mathrm{MHz}$ and amplitude $150 \mathrm{~V} / \mathrm{m}$. Identify the correct expression of $\overline{\mathrm{B}}$ assuming the wave is propagating along $x$-axis and is oscillating along $y$-axis.
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Verified Answer
The correct answer is:
$5 \times 10^{-7} \sin \left[\pi\left(\frac{x}{5}-6 \times 10^{+7} t\right)\right] \hat{z} \mathrm{~T}$
$\begin{aligned}
& B_0=\frac{E_0}{C}=\frac{150}{3 \times 10^8}=5 \times 10^{-7} \\
& f=30 \mathrm{MHz}=30 \times 10^6 \mathrm{~Hz} \\
& \omega=60 \pi \times 10^6 \mathrm{~Hz}, \mathrm{~K}=\frac{\omega}{\mathrm{C}}=\frac{\mathrm{x}}{5}
\end{aligned}$
as $\hat{\mathrm{C}}=\overrightarrow{\mathrm{E}} \times \overline{\mathrm{B}}$
So, direction of $\overrightarrow{\mathrm{B}}$ is along z-axis.
$\text {Thus, } \overrightarrow{\mathrm{B}}=\left(5 \times 10^{-7}\right) \sin \left[\frac{\pi}{5} \mathrm{x}-6 \times 10^7 \pi \mathrm{t}\right] \hat{\mathrm{z}}$
& B_0=\frac{E_0}{C}=\frac{150}{3 \times 10^8}=5 \times 10^{-7} \\
& f=30 \mathrm{MHz}=30 \times 10^6 \mathrm{~Hz} \\
& \omega=60 \pi \times 10^6 \mathrm{~Hz}, \mathrm{~K}=\frac{\omega}{\mathrm{C}}=\frac{\mathrm{x}}{5}
\end{aligned}$
as $\hat{\mathrm{C}}=\overrightarrow{\mathrm{E}} \times \overline{\mathrm{B}}$
So, direction of $\overrightarrow{\mathrm{B}}$ is along z-axis.
$\text {Thus, } \overrightarrow{\mathrm{B}}=\left(5 \times 10^{-7}\right) \sin \left[\frac{\pi}{5} \mathrm{x}-6 \times 10^7 \pi \mathrm{t}\right] \hat{\mathrm{z}}$
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