KCET · Physics · Electromagnetic Waves
An electromagnetic wave is travelling in \( x \)-direction with electric field vector given by, \( \vec{E}_{y}=E_{0} \sin (k x-\omega t) \hat{j} \). The correct expression for magnetic field vector is
- A \( \vec{B}_{y}=\frac{E_{0}}{C} \sin (k x-\omega t) \hat{j} \)
- B \(\vec{B}_{y}=E_{0} C \sin (k x-\omega t) \hat{j} \)
- C \( \vec{B}_{z}=\frac{E_{0}}{C} \sin (k x-\omega t) \hat{k} \)
- D \({\vec{B}_{z}}{=E_{0} C \sin (k x-\omega t) \hat{k}} \)
Answer & Solution
Correct Answer
(C) \( \vec{B}_{z}=\frac{E_{0}}{C} \sin (k x-\omega t) \hat{k} \)
Step-by-step Solution
Detailed explanation
\(\vec{E}_{y}=E_{0} \sin (k x-\omega t) \hat{j} \)
\( C=\frac{E_{0}}{B_{0}} \Rightarrow B_{0}=\frac{E_{0}}{C} \)
\( B_{z}=\frac{E_{0}}{C} \sin (k x-\omega t) \hat{k}\)
\( C=\frac{E_{0}}{B_{0}} \Rightarrow B_{0}=\frac{E_{0}}{C} \)
\( B_{z}=\frac{E_{0}}{C} \sin (k x-\omega t) \hat{k}\)
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