The magnetic field of a plane electromagnetic wave is \overrightarrow{{B}}=3 \\times10^{-8} \sin [200 \\pi( {y}+ {ct})] \stackrel{\^}{{i}} {T} Where c=3 \\times10^{8} {ms}^{-1} is the speed of light. The corresponding electric field is: [3 Sep 2020 Shift 1]

Correct Answer is : →E=−9sin[200π(y+ct)]^k V∕M

\overrightarrow{{E}}=-10^{-6} \sin [200 \\pi( {y}+ {ct})]\stackrel{\^}{{k}} V\/M

\overrightarrow{{E}}=-9\sin [200 \\pi( {y}+ {ct})]{{k}}^{\^} V\/M

\overrightarrow{{E}}=9\sin [200 \\pi( {y}+ {ct})]{{k}}^{\^} V\/M

\overrightarrow{{E}}=3\times10^{-8} \sin [200 \\pi( {y}+ {ct})]\stackrel{\^}{{k}} V\/M

JEE Mains 03-Sep-2020 Shift 1 Solved Paper

Section: Physics

Question : 5 of 75

Marks: +1, -0

The magnetic field of a plane electromagnetic wave is

→

=3×10−8‌sin[200π(y+ct)]‌

‌T
Where c=3×108 ms−1 is the speed of light.
The corresponding electric field is:

[3 Sep 2020 Shift 1]

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