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Electromagnetic Waves question

2023 · 11 Apr · Shift 2 · Q44
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  5. /2023 · 11 Apr · Shift 2 · Q44

Electromagnetic Waves question

2023 · 11 Apr · Shift 2 · Q44

JEE MainPhysicsElectromagnetic WavesMCQ+4 / −1
A plane electromagnetic wave of frequency 20 MHz20 ~\mathrm{MHz}20 MHz propagates in free space along x\mathrm{x}x-direction. At a particular space and time, E→=6.6j^ V/m\overrightarrow{\mathrm{E}}=6.6 \hat{j} \mathrm{~V} / \mathrm{m}E=6.6j^​ V/m. What is B→\overrightarrow{\mathrm{B}}B at this point?
  1. A
    −2.2×10−8i^T-2.2 \times 10^{-8} \hat{i} T−2.2×10−8i^T
  2. B
    2.2×10−8i^T2.2 \times 10^{-8} \hat{i} T2.2×10−8i^T
  3. C
    2.2×10−8k^T2.2 \times 10^{-8} \hat{k} T2.2×10−8k^T
  4. D
    −2.2×10−8k^T-2.2 \times 10^{-8} \hat{k} T−2.2×10−8k^T
View written solutionFree

Correct answer: C

  1. Use the relation between E⃗\vec EE, B⃗\vec BB and direction of propagation

For an electromagnetic wave in free space:

E=cBE = cBE=cB

So,

B=EcB = \frac{E}{c}B=cE​

Also, the propagation direction is along i^\hat ii^ (the positive xxx-axis), and for an EM wave:

E⃗×B⃗ gives the direction of propagation\vec E \times \vec B \text{ gives the direction of propagation}E×B gives the direction of propagation

Given:

E⃗=6.6 j^  V/m\vec E = 6.6\,\hat j\; \text{V/m}E=6.6j^​V/m

and propagation is along +i^+\hat i+i^.

We need B⃗\vec BB such that

j^×B⃗=i^\hat j \times \vec B = \hat ij^​×B=i^

Now,

j^×k^=i^\hat j \times \hat k = \hat ij^​×k^=i^

So B⃗\vec BB must be along +k^+\hat k+k^.

  1. Calculate magnitude of B⃗\vec BB
B=Ec=6.63×108B = \frac{E}{c} = \frac{6.6}{3\times 10^8}B=cE​=3×1086.6​ B=2.2×10−8  TB = 2.2 \times 10^{-8}\; \text{T}B=2.2×10−8T
  1. Write the vector form

Since direction is k^\hat kk^,

B⃗=2.2×10−8 k^  T\vec B = 2.2 \times 10^{-8}\, \hat k\; \text{T}B=2.2×10−8k^T
  1. Match with options

This corresponds to:

Option C:

2.2×10−8k^  T2.2 \times 10^{-8} \hat k\; \text{T}2.2×10−8k^T
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