JEE Mains · Physics · STD 12 - Electromagnetic waves
- JEE Mains 2024
- Hard
- MCQ
The electric field in an electromagnetic wave is given by \(\overrightarrow{\mathrm{E}}=\hat{\mathrm{i}} 40 \cos \omega\left(\mathrm{t}-\frac{\mathrm{z}}{\mathrm{c}}\right) N \mathrm{NC}^{-1}\). The magnetic field induction of this wave is (in SI unit) _______.
Pick an option to check your answer.
Answer & Solution
Correct Answer
(D) \(\vec{B}=\hat{j} \frac{40}{c} \cos \omega\left(t-\frac{z}{c}\right)\)
Step-by-step Solution
Detailed explanation
\(\overrightarrow{\mathrm{E}}=\hat{\mathrm{i}} 40 \cos \omega\left(\mathrm{t}-\frac{\mathrm{z}}{\mathrm{c}}\right)\)
\(\overrightarrow{\mathrm{E}}\) is along \(+\mathrm{x}\) direction
\(\overrightarrow{\mathrm{v}}\) is along \(+\mathrm{z}\) direction
So direction of \(\vec{B}\) will be along +y and magnitude of \(B\) will be \(\frac{E}{C}\)
So answer is \(\frac{40}{c} \cos \omega\left(t-\frac{z}{c}\right) \hat{j}\)
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