For a plane electromagnetic wave, the magnetic field at a point

Question:

For a plane electromagnetic wave, the magnetic field at a point $x$ and time $t$ is

$\overrightarrow{\mathrm{B}}(\mathrm{x}, \mathrm{t})=\left[1.2 \times 10^{-7} \sin \left(0.5 \times 10^{3} \mathrm{x}+1.5 \times 10^{11} \mathrm{t}\right) \hat{\mathrm{k}}\right] \mathrm{T}$

The instantaneous electric field $\overrightarrow{\mathrm{E}}$ corresponding to $\overrightarrow{\mathrm{B}}$ is: (speed of light $\mathrm{c}=3 \times 10^{8} \mathrm{~ms}^{-1}$ )

  1. $\overrightarrow{\mathrm{E}}(\mathrm{x}, \mathrm{t})=\left[36 \sin \left(0.5 \times 10^{3} \mathrm{x}+1.5 \times 10^{11} \mathrm{t}\right) \hat{\mathrm{k}}\right] \frac{\mathrm{v}}{\mathrm{m}}$

  2. $\overrightarrow{\mathrm{E}}(\mathrm{x}, \mathrm{t})=\left[-36 \sin \left(0.5 \times 10^{3} \mathrm{x}+1.5 \times 10^{11} \mathrm{t}\right) \hat{\mathrm{j}}\right] \frac{\mathrm{v}}{\mathrm{m}}$

  3. $\overrightarrow{\mathrm{E}}(\mathrm{x}, \mathrm{t})=\left[36 \sin \left(1 \times 10^{3} \mathrm{x}+0.5 \times 10^{11} \mathrm{t}\right) \hat{\mathrm{j}}\right] \frac{\mathrm{v}}{\mathrm{m}}$

  4. $\overrightarrow{\mathrm{E}}(\mathrm{x}, \mathrm{t})=\left[36 \sin \left(1 \times 10^{3} \mathrm{x}+1.5 \times 10^{11} \mathrm{t}\right) \hat{\mathrm{j}}\right] \frac{\mathrm{v}}{\mathrm{m}}$

Correct Option: , 2

Solution:

$\overrightarrow{\mathrm{E}}$ and $\overrightarrow{\mathrm{B}}$ are perpendicular for $\mathrm{EM}$ wave

$\mathrm{E}_{0}=\mathrm{CB}_{0}$

$=3 \times 10^{8} \times 1.2 \times 10^{-7}$

$=36$

Having same phase

Propagation is along - $\mathrm{x}$-axis, $\overrightarrow{\mathrm{B}}$ is along z-axis hence $\overrightarrow{\mathrm{E}}$ must be along $\mathrm{y}$-axis.

So, option (2) is correct

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