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Electromagnetic Waves question

2019 · 12 Apr · Shift 1 · Q49
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Electromagnetic Waves question

2019 · 12 Apr · Shift 1 · Q49

JEE MainPhysicsElectromagnetic WavesMCQ+4 / −1
An electromagnetic wave is represented by the electric field E→=E0n^sin⁡[ωt+(6y−8z)]\overrightarrow E = {E_0}\widehat n\sin \left[ {\omega t + \left( {6y - 8z} \right)} \right]E=E0​nsin[ωt+(6y−8z)]. Taking unit vectors in x, y and z directions to be i^,j^,k^\widehat i,\widehat j,\widehat ki,j​,k, the direction of propagation s^\widehat ss, is :
  1. A
    s^=3i^−4j^5\widehat s = {{3\widehat i - 4\widehat j} \over 5}s=53i−4j​​
  2. B
    s^=−4k^+3j^5\widehat s = {{ - 4\widehat k + 3\widehat j} \over 5}s=5−4k+3j​​
  3. C
    s^=(−3j^+4k^5)\widehat s = \left( {{{ - 3\widehat j + 4\widehat k} \over 5}} \right)s=(5−3j​+4k​)
  4. D
    s^=4j^−3k^5\widehat s = {{4\widehat j - 3\widehat k} \over 5}s=54j​−3k​
View written solutionFree

Correct answer: C

  1. Identify the phase of the wave

The electric field is given as

E⃗=E0n^ sin⁡[ωt+(6y−8z)].\vec E = E_0\hat n\,\sin\left[\omega t + (6y-8z)\right].E=E0​n^sin[ωt+(6y−8z)].

For a plane electromagnetic wave, the standard phase form is

sin⁡(ωt−k⃗⋅r⃗)\sin(\omega t - \vec k\cdot \vec r)sin(ωt−k⋅r)

or equivalently

sin⁡(ωt+k⃗′⋅r⃗),\sin(\omega t + \vec k'\cdot \vec r),sin(ωt+k′⋅r),

where the direction of propagation is determined from the spatial part.

  1. Compare with the standard propagation form

Given phase:

ωt+6y−8z.\omega t + 6y - 8z.ωt+6y−8z.

Write the spatial part as

6y−8z=−(−6y+8z).6y-8z = -( -6y+8z).6y−8z=−(−6y+8z).

So,

ωt+6y−8z=ωt−(−6y+8z).\omega t + 6y - 8z = \omega t -(-6y+8z).ωt+6y−8z=ωt−(−6y+8z).

Hence,

k⃗⋅r⃗=−6y+8z.\vec k\cdot \vec r = -6y+8z.k⋅r=−6y+8z.

Therefore the wave vector is

k⃗=−6j^+8k^.\vec k = -6\hat j + 8\hat k.k=−6j^​+8k^.

The wave propagates in the direction of k⃗\vec kk.

  1. Find the unit vector along propagation

Magnitude of k⃗\vec kk:

∣k⃗∣=(−6)2+82=36+64=100=10.|\vec k| = \sqrt{(-6)^2 + 8^2} = \sqrt{36+64} = \sqrt{100}=10.∣k∣=(−6)2+82​=36+64​=100​=10.

Thus the unit vector is

s^=k⃗∣k⃗∣=−6j^+8k^10=−3j^+4k^5.\hat s = \frac{\vec k}{|\vec k|} = \frac{-6\hat j+8\hat k}{10} = \frac{-3\hat j+4\hat k}{5}.s^=∣k∣k​=10−6j^​+8k^​=5−3j^​+4k^​.
  1. Match with the options
s^=−3j^+4k^5\hat s = \frac{-3\hat j+4\hat k}{5}s^=5−3j^​+4k^​

which is exactly Option C.

  1. Conclusion

The direction of propagation is

s^=−3j^+4k^5\boxed{\hat s = \frac{-3\hat j+4\hat k}{5}}s^=5−3j^​+4k^​​

So the correct option is C.

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