JEE Mains · Physics · STD 12 - Electromagnetic waves
- JEE Mains 2025
- Medium
- MCQ
The electric field of an electromagnetic wave in free space is \(\overrightarrow{\mathrm{E}}=57 \cos \left[7.5 \times 10^6 \mathrm{t}-5 \times 10^{-3}(3 x+4 y)\right](4 \hat{i}-3 \hat{j}) N / C\). The associated magnetic field in Tesla is _______.
Pick an option to check your answer.
Answer & Solution
Correct Answer
(C) \(\overrightarrow{\mathrm{B}}=-\frac{57}{3 \times 10^8} \cos \left[7.5 \times 10^6 \mathrm{t}-5 \times 10^{-3}(3 x+4 y)\right](5 \hat{k})\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \overrightarrow{\mathrm{K}}=3 \hat{\mathrm{i}}+4 \hat{\mathrm{j}} \\ & \hat{\mathrm{K}}=\frac{3 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}}{5} \\ & \hat{\mathrm{E}}=\frac{4 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}}{5} \\ & \hat{\mathrm{~B}}=\hat{\mathrm{K}} \times \hat{\mathrm{E}} \\ & \hat{\mathrm{B}}=-\hat{\mathrm{Z}} \\ & \mathrm{B}_0=\frac{\mathrm{E}_0}{\mathrm{C}}=\frac{57}{3 \times 10^8}\end{aligned}\)
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