JEE Mains · Physics · STD 12 - Electromagnetic waves
- JEE Mains 2022
- Medium
- MCQ
The magnetic field of a plane electromagnetic wave is given by \(\overrightarrow{ B }=2 \times 10^{-8} \sin \left(0.5 \times 10^{3} x +1.5 \times 10^{11} t \right) \hat{ j } T\) The amplitude of the electric field would be.
Pick an option to check your answer.
Answer & Solution
Correct Answer
(C) \(6 Vm ^{-1}\) along \(z\)-axis
Step-by-step Solution
Detailed explanation
\(c =\frac{ E _{0}}{ B _{0}} \Rightarrow E _{0}= cB _{0}\)
\(E _{0}=\left(3 \times 10^{8}\right)\left(2 \times 10^{-8}\right)\)
\(E _{0}=6 Vm ^{-1}\)
As, \(\overrightarrow{ B }=\) along \(y\)-axis
\(\overrightarrow{ v }\) =along negative \(x\)-axis
hence \(\quad \overrightarrow{ E }_{0}=\) along \(z\)-axis
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