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Electromagnetic Waves question

2021 · 16 Mar · Shift 1 · Q53
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  5. /2021 · 16 Mar · Shift 1 · Q53

Electromagnetic Waves question

2021 · 16 Mar · Shift 1 · Q53

JEE MainPhysicsElectromagnetic WavesMCQ+4 / −1
A plane electromagnetic wave of frequency 500 MHz is travelling in vacuum along y-direction. At a particular point in space and time, B→\overrightarrow BB= 8.0 ×\times× 10 −-− 8 z^\widehat zz T. The value of electric field at this point is : (speed of light = 3 ×\times× 108 ms −-− 1) x^\widehat xx, y^\widehat yy​, z^\widehat zz are unit vectors along x, y and z directions.
  1. A
    2.6 x^\widehat xx V/m
  2. B
    −-− 24 x^\widehat xx V/m
  3. C
    24 x^\widehat xx V/m
  4. D
    −-− 2.6 y^\widehat yy​ V/m
View written solutionFree

Correct answer: C: $24\HAT X\,\TEXT{V/M}$

  1. Use the direction relation for an electromagnetic wave

For a plane electromagnetic wave in vacuum,

E⃗⊥B⃗,E⃗×B⃗ gives direction of propagation.\vec E \perp \vec B, \qquad \vec E \times \vec B \text{ gives direction of propagation.}E⊥B,E×B gives direction of propagation.

The wave is travelling along the positive yyy-direction, so

E⃗×B⃗=+y^.\vec E \times \vec B = +\hat y.E×B=+y^​.

Given magnetic field at the point:

B⃗=8.0×10−8(−z^) T=−8.0×10−8z^ T.\vec B = 8.0 \times 10^{-8}(-\hat z)\,\text{T} = -8.0\times 10^{-8}\hat z\,\text{T}.B=8.0×10−8(−z^)T=−8.0×10−8z^T.

We need a vector E⃗\vec EE such that

E⃗×(−z^)=+y^.\vec E \times (-\hat z) = +\hat y.E×(−z^)=+y^​.

Now,

x^×z^=−y^\hat x \times \hat z = -\hat yx^×z^=−y^​

so

x^×(−z^)=+y^.\hat x \times (-\hat z) = +\hat y.x^×(−z^)=+y^​.

Hence, E⃗\vec EE must be along +x^+\hat x+x^.

  1. Use the magnitude relation in vacuum

For electromagnetic waves in vacuum,

E=cB.E = cB.E=cB.

Given:

c=3×108 m/s,B=8.0×10−8 Tc = 3\times 10^8\,\text{m/s}, \qquad B = 8.0\times 10^{-8}\,\text{T}c=3×108m/s,B=8.0×10−8T

Therefore,

E=(3×108)(8.0×10−8)=24 V/m.E = (3\times 10^8)(8.0\times 10^{-8}) = 24\,\text{V/m}.E=(3×108)(8.0×10−8)=24V/m.
  1. Write the electric field vector

Since the direction is +x^+\hat x+x^ and magnitude is 24 V/m24\,\text{V/m}24V/m,

E⃗=24 x^ V/m\boxed{\vec E = 24\,\hat x\,\text{V/m}}E=24x^V/m​
  1. Match with options

This corresponds to:

  • Option C: 24x^ V/m24\hat x\,\text{V/m}24x^V/m
  1. Compare with stored correct answer

Stored correct answer is B: −24x^ V/m-24\hat x\,\text{V/m}−24x^V/m.

But if E⃗=−24x^\vec E = -24\hat xE=−24x^, then

(−x^)×(−z^)=x^×z^=−y^,(-\hat x) \times (-\hat z) = \hat x \times \hat z = -\hat y,(−x^)×(−z^)=x^×z^=−y^​,

which would mean the wave travels along −y^-\hat y−y^​, not along the given +y^+\hat y+y^​.

So the stored answer is inconsistent with the propagation direction.

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