Question 8

MCQMEDIUM

The electric field associated with an electromagnetic wave travelling in vacuum is given by E0sin⁑(3y+4z+Ο‰t)i^E_0 \sin(3y + 4z + \omega t)\hat{i}, where Ο‰\omega is the angular frequency. All quantities are in SI units. The correct statement(s) about this wave is/are: [Given: speed of light in vacuum c=3Γ—108Β msβˆ’1c = 3 \times 10^8 \text{ ms}^{-1}.]

(A)

The wave is travelling in βˆ’15(3j^+4k^)-\frac{1}{5}(3\hat{j} + 4\hat{k}) direction.

(B)

The magnitude of the wave vector is 0.5Β mβˆ’10.5 \text{ m}^{-1}.

(C)

The value of Ο‰\omega is 1.5Γ—109Β radΒ sβˆ’11.5 \times 10^9 \text{ rad s}^{-1}.

(D)

The magnetic field associated with this wave is given by E0csin⁑(3y+4z+Ο‰t)(4j^βˆ’3k^)\frac{E_0}{c} \sin(3y + 4z + \omega t)(4\hat{j} - 3\hat{k}).

Detailed Solution

  1. The phase of the wave is given by Ο•=3y+4z+Ο‰t\phi = 3y + 4z + \omega t.

Comparing this with the general form (k⃗⋅r⃗+ωt)(\vec{k} \cdot \vec{r} + \omega t), we identify the wave vector k⃗=3j^+4k^\vec{k} = 3\hat{j} + 4\hat{k}.

Since the signs of the spatial term and the temporal term are the same, the wave propagates in the βˆ’kβƒ—-\vec{k} direction. The direction of propagation is n^=βˆ’kβƒ—βˆ£kβƒ—βˆ£=βˆ’3j^+4k^32+42=βˆ’15(3j^+4k^)\hat{n} = -\frac{\vec{k}}{|\vec{k}|} = -\frac{3\hat{j} + 4\hat{k}}{\sqrt{3^2 + 4^2}} = -\frac{1}{5}(3\hat{j} + 4\hat{k}). Thus, option (A) is correct.

  1. The magnitude of the wave vector is k=∣kβƒ—βˆ£=32+42=5Β mβˆ’1k = |\vec{k}| = \sqrt{3^2 + 4^2} = 5 \text{ m}^{-1}. Thus, option (B) is incorrect.

  2. The angular frequency Ο‰\omega is related to kk and cc by Ο‰=ck=(3Γ—108Β msβˆ’1)(5Β mβˆ’1)=1.5Γ—109Β radΒ sβˆ’1\omega = ck = (3 \times 10^8 \text{ ms}^{-1})(5 \text{ m}^{-1}) = 1.5 \times 10^9 \text{ rad s}^{-1}. Thus, option (C) is correct.

  3. The magnetic field B⃗\vec{B} is given by B⃗=k⃗×E⃗ω\vec{B} = \frac{\vec{k} \times \vec{E}}{\omega}. Substituting the values:

Bβƒ—=(3j^+4k^)Γ—(E0sin⁑ϕi^)Ο‰=E0sin⁑ϕω[3(j^Γ—i^)+4(k^Γ—i^)]=E0sin⁑ϕω[4j^βˆ’3k^]\vec{B} = \frac{(3\hat{j} + 4\hat{k}) \times (E_0 \sin \phi \hat{i})}{\omega} = \frac{E_0 \sin \phi}{\omega} [3(\hat{j} \times \hat{i}) + 4(\hat{k} \times \hat{i})] = \frac{E_0 \sin \phi}{\omega} [4\hat{j} - 3\hat{k}]

Since Ο‰=5c\omega = 5c, Bβƒ—=E05csin⁑(3y+4z+Ο‰t)(4j^βˆ’3k^)\vec{B} = \frac{E_0}{5c} \sin(3y + 4z + \omega t)(4\hat{j} - 3\hat{k}). Option (D) is incorrect because it lacks the factor of 1/51/5.

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