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JEE Mains · Physics · STD 12 - Electromagnetic waves
- JEE Mains 2020
- Hard
- MCQ
The magnetic field of a plane electromagnetic Wave is \(\overrightarrow{ B }=3 \times 10^{-8} \sin [200 \pi( y + ct )] \hat{ i }\, T\) Where \(c=3 \times 10^{8} \,ms ^{-1}\) is the speed of light. The corresponding electric field is
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) \(\overrightarrow{ E }=-9 \sin [200 \pi( y + ct )] \hat{ k }\, \;V / m\)
Step-by-step Solution
Detailed explanation
\(\overrightarrow{ B }=3 \times 10^{-8} \sin [200 \pi( y + ct )] \hat{ i } T\)
\(E _{0}= CB _{0} \Rightarrow E _{0}=3 \times 10^{8} \times 3 \times 10^{-8}\)
\(=9 V / m\)
and direction of wave propagation is given as
\((\overrightarrow{ E } \times \overrightarrow{ B }) \| \overrightarrow{ C }\)
\(\hat{ B }=\hat{ i } \quad \& \quad \hat{ C }=-\hat{ j }\)
so \(\hat{ E }=-\hat{ k }\)
\(\therefore \overrightarrow{ E }= E _{0} \sin [200 \pi( y + ct )](-\hat{ k }) V / m\)
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