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If the magnetic field of a plane electromagnetic wave is given by
$5 \times 10^{-6} \sin \left(0.6 \times 10^2 x+0.5 \times 10^{10} t\right)$, then the speed of the wave is
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$5 \times 10^{-6} \sin \left(0.6 \times 10^2 x+0.5 \times 10^{10} t\right)$, then the speed of the wave is
Solution:
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Verified Answer
The correct answer is:
$0.83 \times 10^8 \mathrm{~m} / \mathrm{s}$
$\begin{aligned}
\text { Given, } k & =\frac{2 \pi}{\lambda}=0.6 \times 10^2 \mathrm{~m}^{-1} \\
\omega & =\frac{2 \pi}{T}=0.5 \times 10^{10} \mathrm{~s}^{-1} \\
\Rightarrow \text { Velocity, } \quad v & =\omega / k=\frac{2 \pi}{T} \times \frac{\lambda}{2 \pi}=\frac{0.5 \times 10^{10}}{0.6 \times 10^2} \\
v & =0.83 \times 10^8 \mathrm{~m} / \mathrm{s}
\end{aligned}$
\text { Given, } k & =\frac{2 \pi}{\lambda}=0.6 \times 10^2 \mathrm{~m}^{-1} \\
\omega & =\frac{2 \pi}{T}=0.5 \times 10^{10} \mathrm{~s}^{-1} \\
\Rightarrow \text { Velocity, } \quad v & =\omega / k=\frac{2 \pi}{T} \times \frac{\lambda}{2 \pi}=\frac{0.5 \times 10^{10}}{0.6 \times 10^2} \\
v & =0.83 \times 10^8 \mathrm{~m} / \mathrm{s}
\end{aligned}$
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