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A plane EM wave is propagating along $\mathrm{x}$ direction. It has a wavelength of $4 \mathrm{~mm}$. If electric field is in y direction with the maximum magnitude of $60 \mathrm{Vm}^{-1}$, the equation for magnetic field is:$7$
$\mathrm{B}_z=60 \sin \left[\frac{\pi}{2}\left(\mathrm{x}-3 \times 10^8 \mathrm{t}\right)\right] \hat{\mathrm{kT}}$
$\mathrm{B}_z=2 \times 10^{-7} \sin \left[\frac{\pi}{2} \times 10^3\left(\mathrm{x}-3 \times 10^8 \mathrm{t}\right)\right] \hat{\mathrm{kT}}$
$\mathrm{B}_{\mathrm{x}}=60 \sin \left[\frac{\pi}{2}\left(\mathrm{x}-3 \times 10^8 \mathrm{t}\right)\right]$ i $\mathrm{i} \mathrm{T}$
$\mathrm{B}_z=2 \times 10^{-7} \sin \left[\frac{\pi}{2}\left(\mathrm{x}-3 \times 10^8 \mathrm{t}\right)\right] \hat{\mathrm{k} T}$
Solution
$\mathrm{E}=\mathrm{BC} \Rightarrow 60=\mathrm{B} \times 3 \times 10^8$
$\Rightarrow \mathrm{B}=2 \times 10^{-7}$
$\text { Also } \mathrm{C}=\mathrm{f} \lambda$
$\Rightarrow 3 \times 10^8=\mathrm{f} \times 4 \times 10^{-3}$
$\Rightarrow \mathrm{f}=\frac{3}{4} \times 10^{11}$
$\Rightarrow \omega=2 \pi \mathrm{f}=\frac{3}{4} \times 2 \pi \times 10^{11}$
$\Rightarrow \omega=\frac{\pi}{2} \times 10^3 \mathrm{C}$
$\Rightarrow \quad \text { Electric field } \Rightarrow \mathrm{y} \text { direction }$
$\text { Propagation } \Rightarrow \mathrm{x} \text { direction }$
$\text { Magnetic field } \Rightarrow z \text {-direction }$
