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In an $EM$ wave propagating along $X-$ direction magnetic field oscillates at a frequency of $3 \times 10^{10}\, Hz$ along $Y-$ direction and has an amplitude of $10^{-7}\, T$. The expression for electric field will be
$E_z = 30\, sin 2\pi\,(100x -3 \times 10^{10}\, t)\, V/m$
$E_z = 300\, sin 2\pi\,(100x -3 \times 10^{10}\, t)\, V/m$
$E_y = 30\, sin 2\pi\,(100x -3 \times 10^{10}\, t)\, V/m$
$E_z = 300\, sin 2\pi\,(100x - 3 \times 10^{10}\, t)\, V/m$
Solution
$\mathrm{f}=3 \times 10^{10} \mathrm{H}_{2} \quad \mathrm{B}_{0}=10^{-7} \mathrm{T}$
$\mathrm{E}_{0}=\mathrm{CB}_{0}=3 \times 10^{8} \times 10^{-7}=30 \mathrm{\,V} / \mathrm{m}$
$\lambda=\frac{\mathrm{C}}{\mathrm{f}}=10^{-2} \mathrm{\,m}$
$ \mathrm{E}_{2} =\mathrm{E}_{0} \sin (\mathrm{Kx}-\mathrm{wt}) $
$=30 \sin \left(\frac{2 \pi}{\lambda} \mathrm{x}-\frac{2 \pi}{\mathrm{T}} \mathrm{t}\right) $
$\boxed{{{\text{E}}_2} = 30\sin 2\pi \left( {100{\text{x}} – 3 \times {{10}^{10}}{\text{t}}} \right)}$
