- Home
- Standard 12
- Physics
The magnetic field of an electromagnetic wave is given by
$\vec B = 1.6 \times {10^{ - 6}}\,\cos \,\left( {2 \times {{10}^7}z + 6 \times {{10}^{15}}t} \right)\left( {2\hat i + \hat j} \right)\frac{{Wb}}{{{m^2}}}$ The associated electric field will be
$\vec E = 4.8 \times {10^{ 2}}\,\cos \,\left( {2 \times {{10}^7}z + 6 \times {{10}^{15}}t} \right)\left( { - \hat i + 2\hat j} \right)\frac{V}{m}$
$\vec E = 4.8 \times {10^{ 2}}\,\cos \,\left( {2 \times {{10}^7}z + 6 \times {{10}^{15}}t} \right)\left( { - 2\hat j + 2\hat i} \right)\frac{V}{m}$
$\vec E = 4.8 \times {10^{ 2}}\,\cos \,\left( {2 \times {{10}^7}z + 6 \times {{10}^{15}}t} \right)\left( {\hat i + 2\hat j} \right)\frac{V}{m}$
$\vec E = 4.8 \times {10^{ 2}}\,\cos \,\left( {2 \times {{10}^7}z + 6 \times {{10}^{15}}t} \right)\left( {2\hat i + \hat j} \right)\frac{V}{m}$
Solution
If we use that direction of light propagation will be along $\overrightarrow{\mathrm{E}} \times \overrightarrow{\mathrm{B}}$. Then $(\mathrm{A})$ option is correct.
Magnitude of $E=C B$
$E=3 \times 10^{8} \times 1.6 \times 10^{-6} \times \sqrt{5}$
$E=4.8 \times 10^{2} \times \sqrt{5}$
$\overrightarrow{\mathrm{E}}$ and $\overrightarrow{\mathrm{B}}$ are perpendicular to each other $\Rightarrow \overrightarrow{\mathrm{E}} \cdot \overrightarrow{\mathrm{B}}=0$
$\Rightarrow$ Either direction of $\bar{E}$ is $\hat{i}-2 \hat{j}$ or $-\hat{i}+2 \hat{j}$ from given option
Also wave propagation direction is parallel to $\overrightarrow{\mathrm{E}} \times \overrightarrow{\mathrm{B}}$ which is $-\overrightarrow{\mathrm{k}}$
$\Rightarrow \overrightarrow{\mathrm{E}}$ is along $(-\hat{\mathrm{i}}+2 \hat{\mathrm{j}})$
