JEE Mains · Physics · STD 12 - 8. Electromagnetic waves
A plane electromagnetic wave is moving in free space with velocity \(c =3 \times 10^8 m / s\) and its electric field is given as \(\overrightarrow{ E }=54 \sin ( kz -\omega t ) \hat{ j } V / m\), where \(\hat{j}\) is the unit vector along \(y\)-axis. The magnetic field vector \(\overrightarrow{ B }\) of the wave is :
- A \(-1.8\times10^{-7}sin(kz-\omega t)\hat{i}T\)
- B \(1.4\times10^{-7}sin(kz-ot)\hat{k}T\)
- C \(1.4\times10^{-7}sin(kz-\omega t)\hat{i}T\)
- D \(+1.8\times10^{-7}sin(kz-\omega t)\hat{i}T\)
Answer & Solution
Correct Answer
(A) \(-1.8\times10^{-7}sin(kz-\omega t)\hat{i}T\)
Step-by-step Solution
Detailed explanation
\(\hat{B}=\hat{C}\times\hat{E}=\hat{k}\times\hat{j}=-\hat{i}\) .. \(\vec{B}=\frac{54}{3\times10^{8}}sin(kz-cot)(-\hat{i})\) \(=-1.8\times10^{-7}sin(kz-\omega t)\hat{i}\)
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