JEE Mains · Physics · STD 12 - 8. Electromagnetic waves
The electric field of a plane electromagnetic wave is given by \(\overrightarrow{ E }= E _{0}(\hat{ x }+\hat{ y }) \sin ( kz -\omega t )\) Its magnetic field will be given by
- A \(\frac{ E _{0}}{ c }(\hat{ x }-\hat{ y }) \cos ( kz -\omega t )\)
- B \(\frac{ E _{0}}{ c }(-\hat{ x }+\hat{ y }) \sin ( kz -\omega t )\)
- C \(\frac{ E _{0}}{ c }(\hat{ x }-\hat{ y }) \sin ( kz -\omega t )\)
- D \(\frac{ E _{0}}{ c }(\hat{ x }+\hat{ y }) \sin ( kz -\omega t )\)
Answer & Solution
Correct Answer
(B) \(\frac{ E _{0}}{ c }(-\hat{ x }+\hat{ y }) \sin ( kz -\omega t )\)
Step-by-step Solution
Detailed explanation
\(\overrightarrow{ E }= E _{0}(\hat{ x }+\hat{ y }) \sin ( kz -\omega t )\) direction of propagation \(=+\hat{ k }\) \(\hat{ E }=\frac{\hat{ i }+\hat{ j }}{\sqrt{2}}\) \(\hat{ k }=\hat{ E } \times \hat{ B }\)…
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