NEET · Physics · STD 12 - 8. Electromagnetic waves
The magnetic field of a plane electromagnetic wave is given by \(\overrightarrow{ B }=3 \times 10^{-8} \cos \left(1.6 \times 10^3 x +48 \times 10^{10} t \right) \hat{ j }\), then the associated electric field will be :
- A \(3 \times 10^{-8} \cos \left(1.6 \times 10^3 x +48 \times 10^{10} t \right) \hat{ i }\,V / m\)
- B \(3 \times 10^{-8} \sin \left(1.6 \times 10^3 x +48 \times 10^{10} t \right) \hat{ i }\,V / m\)
- C \(9 \sin \left(1.6 \times 10^3 x -48 \times 10^{10} t \right) \hat{ k}\,V / m\)
- D \(9 \cos \left(1.6 \times 10^3 x +48 \times 10^{10} t \right) \hat{ k }\, V / m\)
Answer & Solution
Correct Answer
(D) \(9 \cos \left(1.6 \times 10^3 x +48 \times 10^{10} t \right) \hat{ k }\, V / m\)
Step-by-step Solution
Detailed explanation
\(B=3 \times 10^{-8} \cos \left(1.6 \times 10^3 x +48 \times 10^{10} t \right)\) \(C=\frac{\omega}{ k }=\frac{48 \times 10^{10}}{1.6 \times 10^3}=3 \times 10^8\,m / s\) \(C=E_0 / B_0\) \(E=3 \times 10^{-8} \times 3 \times 10^8=9\,N / C\)
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