JEE Mains · Physics · STD 12 - Electromagnetic waves
- JEE Mains 2019
- Medium
- MCQ
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
Pick an option to check your answer.
Answer & Solution
Correct Answer
(A) \(\vec E = 4.8 \times {10^{ 2}}\,\cos \,\left( {2 \times {{10}^7}z + 6 \times {{10}^{15}}t} \right)\left( { - \hat i + 2\hat j} \right)\frac{V}{m}\)
Step-by-step Solution
Detailed explanation
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 \((-\hat{\mathrm{i}}+2 \hat{\mathrm{j}})\)
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