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JEE Mains · Physics · STD 12 - 8. Electromagnetic waves
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
- A \(\overrightarrow{ E }=-10^{-6} \sin [200 \pi( y + ct )] \hat{ k }\, \;V / m\)
- B \(\overrightarrow{ E }=-9 \sin [200 \pi( y + ct )] \hat{ k }\, \;V / m\)
- C \(\overrightarrow{ E }=9 \sin [200 \pi( y + ct )] \hat{ k }\, \;V / m\)
- D \(\overrightarrow{ E }=3 \times 10^{-8} \sin [200 \pi( y + ct )] \hat{ k }\, \;V / m\)
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…
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