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JEE Mains · Physics · STD 12 - 8. Electromagnetic waves
The electric field of a plane electromagnetic wave propagating along the \(x\) direction in vacuum is \(\overrightarrow{ E }= E _{0} \hat{ j } \cos (\omega t - kx )\). The magnetic field \(\overrightarrow{ B },\) at the moment \(t =0\) is :
- A \(\overrightarrow{ B }= E _{0} \sqrt{\mu_{0} \epsilon_{0}} \cos ( kx ) \hat{ j }\)
- B \(\overrightarrow{ B }=\frac{ E _{0}}{\sqrt{\mu_{0} \epsilon_{0}}} \cos ( kx ) \hat{ k }\)
- C \(\overrightarrow{ B }= E _{0} \sqrt{\mu_{0} \epsilon_{0}} \cos ( kx ) \hat{ k }\)
- D \(\overrightarrow{ B }=\frac{ E _{0}}{\sqrt{\mu_{0} \in_{0}}} \cos ( kx ){\hat{j}}\)
Answer & Solution
Correct Answer
(C) \(\overrightarrow{ B }= E _{0} \sqrt{\mu_{0} \epsilon_{0}} \cos ( kx ) \hat{ k }\)
Step-by-step Solution
Detailed explanation
\(\therefore \overrightarrow{ B }(\hat{ k })\) \(\Rightarrow \overrightarrow{ B }= B _{0} \cos ( wt - kx ) \hat{ k }\) Now put \(t=0\)
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