JEE Mains · Physics · STD 12 - 8. Electromagnetic waves
For a plane electromagnetic wave, the magnetic field at a point \(x\) and time \(t\) is \(\overrightarrow{ B }( x , t )=\left[1.2 \times 10^{-7} \sin \left(0.5 \times 10^{3} x +1.5 \times 10^{11} t \right) \hat{ k }\right] T\) The instantaneous electric field \(\overrightarrow{ E }\) corresponding to \(\overrightarrow{ B }\) is : (speed of light \(\left.c=3 \times 10^{8} ms ^{-1}\right)\)
- A \(\overrightarrow{ E }( x , t )=\left[36 \sin \left(0.5 \times 10^{3} x +1.5 \times 10^{11} t \right) \hat{ k }\right] \frac{ v }{ m }\)
- B \(\overrightarrow{ E }( x , t )=\left[-36 \sin \left(0.5 \times 10^{3} x +1.5 \times 10^{11} t \right) \hat{ j }\right] \frac{ v }{ m }\)
- C \(\overrightarrow{ E }( x , t )=\left[-36 \sin \left(1 \times 10^{3} x +1.5 \times 10^{11} t \right) \hat{ j }\right] \frac{ v }{ m }\)
- D \(\overrightarrow{ E }( x , t )=\left[36 \sin \left(1 \times 10^{3} x +1.5 \times 10^{11} t \right) \hat{ j }\right] \frac{ v }{ m }\)
Answer & Solution
Correct Answer
(B) \(\overrightarrow{ E }( x , t )=\left[-36 \sin \left(0.5 \times 10^{3} x +1.5 \times 10^{11} t \right) \hat{ j }\right] \frac{ v }{ m }\)
Step-by-step Solution
Detailed explanation
\(\overrightarrow{ E }\) and \(\overrightarrow{ B }\) are perpendicular for \(EM\) wave \(E _{0}= CB _{0}\) \(=3 \times 10^{8} \times 1.2 \times 10^{-7}\) \(=36\) Having same phase Propagation is along \(-x-\)axis, \(\overrightarrow{ B }\) is along \(z-\)axis hence…
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