JEE Mains · Physics · STD 12 - 8. Electromagnetic waves
An \(EM\) wave propagating in \(x\)-direction has a wavelength of \(8\,mm\). The electric field vibrating \(y\) direction has maximum magnitude of \(60\,Vm ^{-1}\). Choose the correct equations for electric and magnetic fields if the \(EM\) wave is propagating in vacuum
- A \(E_{y}=60 \sin \left[\frac{\pi}{4} \times 10^{3}\left( x -3 \times 10^{8} t \right)\right] \hat{ j }\,Vm ^{-1}\) \(B _{z}=2 \sin \left[\frac{\pi}{4} \times 10^{3}\left( x -3 \times 10^{8} t \right)\right] \hat{ k }\,T\)
- B \(E_{y}=60 \sin \left[\frac{\pi}{4} \times 10^{3}\left( x -3 \times 10^{8} t \right)\right] \hat{ j }\,Vm ^{-1}\) \(B _{z}=2 \times 10^{-7} \sin \left[\frac{\pi}{4} \times 10^{3}\left( x -3 \times 10^{8} t \right)\right] \hat{ k }\,T\)
- C \(E _{y}=2 \times 10^{-7} \sin \left[\frac{\pi}{4} \times 10^{3}\left( x -3 \times 10^{8} t \right)\right] \hat{ j }\,Vm ^{-1}\) \(B _{z}=60 \sin \left[\frac{\pi}{4} \times 10^{3}\left( x -3 \times 10^{8} t \right)\right] \hat{ k }\, T\)
- D \(E _{ y }=2 \times 10^{-7} \sin \left[\frac{\pi}{4} \times 10^{4}\left( x -4 \times 10^{8} t \right)\right] \hat{ j }\,Vm ^{-1}\) \(B _{z}=60 \sin \left[\frac{\pi}{4} \times 10^{4}\left( x -4 \times 10^{8} t \right)\right] \hat{ k } \,T\)
Answer & Solution
Correct Answer
(B) \(E_{y}=60 \sin \left[\frac{\pi}{4} \times 10^{3}\left( x -3 \times 10^{8} t \right)\right] \hat{ j }\,Vm ^{-1}\) \(B _{z}=2 \times 10^{-7} \sin \left[\frac{\pi}{4} \times 10^{3}\left( x -3 \times 10^{8} t \right)\right] \hat{ k }\,T\)
Step-by-step Solution
Detailed explanation
\(B _{0}=\frac{ E _{0}}{ c }=\frac{60}{3 \times 10^{8}}=2 \times 10^{-7}\,T\) \(\widehat{ E } \times \widehat{ B }\) must be direction of propagation So, \(\widehat{B} \rightarrow z\)-axis \(k =\frac{2 \pi}{\lambda}=\frac{\pi}{4} \times 10^{3}\,m ^{-1}\)…
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