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Electromagnetic Waves question

2023 · 13 Apr · Shift 2 · Q51
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  5. /2023 · 13 Apr · Shift 2 · Q51

Electromagnetic Waves question

2023 · 13 Apr · Shift 2 · Q51

JEE MainPhysicsElectromagnetic WavesMCQ+4 / −1
In an electromagnetic wave, at an instant and at particular position, the electric field is along the negative zzz-axis and magnetic field is along the positive xxx-axis. Then the direction of propagation of electromagnetic wave is:
  1. A
    at 45∘45^{\circ}45∘ angle from positive y-axis
  2. B
    positive yyy-axis
  3. C
    negative y\mathrm{y}y-axis
  4. D
    positive z-axis
View written solutionFree

Correct answer: C

  1. For an electromagnetic wave, the direction of propagation is given by the Poynting vector:
S⃗=1μ0(E⃗×B⃗)\vec{S} = \frac{1}{\mu_0}(\vec{E} \times \vec{B})S=μ0​1​(E×B)

So we need the direction of:

E⃗×B⃗\vec{E} \times \vec{B}E×B
  1. Given directions:
  • Electric field is along negative zzz-axis: E⃗=−k^\vec{E} = -\hat{k}E=−k^
  • Magnetic field is along positive xxx-axis: B⃗=i^\vec{B} = \hat{i}B=i^
  1. Compute the cross product:
E⃗×B⃗=(−k^)×i^\vec{E} \times \vec{B} = (-\hat{k}) \times \hat{i}E×B=(−k^)×i^

Using cyclic relations:

i^×j^=k^,j^×k^=i^,k^×i^=j^\hat{i} \times \hat{j} = \hat{k}, \quad \hat{j} \times \hat{k} = \hat{i}, \quad \hat{k} \times \hat{i} = \hat{j}i^×j^​=k^,j^​×k^=i^,k^×i^=j^​

Therefore,

(−k^)×i^=−(k^×i^)=−j^(-\hat{k}) \times \hat{i} = -(\hat{k} \times \hat{i}) = -\hat{j}(−k^)×i^=−(k^×i^)=−j^​
  1. Thus the propagation direction is along:
−j^-\hat{j}−j^​

which is the negative yyy-axis.

  1. Checking options:
  • A: at 45∘45^\circ45∘ from positive yyy-axis ❌
  • B: positive yyy-axis ❌
  • C: negative yyy-axis ✅
  • D: positive zzz-axis ❌

Hence, the correct answer is Option C.

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