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Electromagnetic Waves question

2021 · 20 Jul · Shift 2 · Q51
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  5. /2021 · 20 Jul · Shift 2 · Q51

Electromagnetic Waves question

2021 · 20 Jul · Shift 2 · Q51

JEE MainPhysicsElectromagnetic WavesMCQ+4 / −1
In an electromagnetic wave the electric field vector and magnetic field vector are given as E→=E0i^\overrightarrow E = {E_0}\widehat iE=E0​i and B→=B0k^\overrightarrow B = {B_0}\widehat kB=B0​k respectively. The direction of propagation of electromagnetic wave is along :
  1. A
    (k^)\left( {\widehat k} \right)(k)
  2. B
    j^\widehat jj​
  3. C
    (−k^)\left( { - \widehat k} \right)(−k)
  4. D
    (−j^)\left( { - \widehat j} \right)(−j​)
View written solutionFree

Correct answer: D

  1. For an electromagnetic wave, the direction of propagation is given by the Poynting vector:
S⃗∝E⃗×B⃗\vec S \propto \vec E \times \vec BS∝E×B

So we must compute E⃗×B⃗\vec E \times \vec BE×B.

  1. Given:
E⃗=E0i^,B⃗=B0k^\vec E = E_0\hat i, \qquad \vec B = B_0\hat kE=E0​i^,B=B0​k^

Therefore,

E⃗×B⃗=(E0i^)×(B0k^)\vec E \times \vec B = (E_0\hat i) \times (B_0\hat k)E×B=(E0​i^)×(B0​k^) =E0B0(i^×k^)= E_0B_0 (\hat i \times \hat k)=E0​B0​(i^×k^)
  1. Using unit vector cross-product relations:
i^×j^=k^,j^×k^=i^,k^×i^=j^\hat i \times \hat j = \hat k, \qquad \hat j \times \hat k = \hat i, \qquad \hat k \times \hat i = \hat ji^×j^​=k^,j^​×k^=i^,k^×i^=j^​

Hence,

i^×k^=−j^\hat i \times \hat k = -\hat ji^×k^=−j^​

So,

E⃗×B⃗=−E0B0j^\vec E \times \vec B = -E_0B_0\hat jE×B=−E0​B0​j^​
  1. Therefore, the electromagnetic wave propagates along:
−j^-\hat j−j^​
  1. Checking options:
  • A: k^\hat kk^ ❌
  • B: j^\hat jj^​ ❌
  • C: −k^-\hat k−k^ ❌
  • D: −j^-\hat j−j^​ ✅

Hence the correct option is D.

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