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Electromagnetic Waves question

2020 · 7 Jan · Shift 1 · Q48
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Electromagnetic Waves question

2020 · 7 Jan · Shift 1 · Q48

JEE MainPhysicsElectromagnetic WavesMCQ+4 / −1
If the magnetic field in a plane electromagnetic wave is given by B→\overrightarrow BB= 3 ×\times× 10-8 sin(1.6 ×\times× 103x + 48 ×\times× 1010t) j^\widehat jj​ T, then what will be expression for electric field ?
  1. A
    E→\overrightarrow EE= (9sin(1.6 ×\times× 103x + 48 ×\times× 1010t) k^\widehat kk V/m)
  2. B
    E→\overrightarrow EE= (60sin(1.6 ×\times× 103x + 48 ×\times× 1010t) k^\widehat kk V/m)
  3. C
    E→\overrightarrow EE= (3 ×\times× 10-8 sin(1.6 ×\times× 103x + 48 ×\times× 1010t) i^\widehat ii V/m)
  4. D
    E→\overrightarrow EE= (3 ×\times× 10-8 sin(1.6 ×\times× 103x + 48 ×\times× 1010t) j^\widehat jj​ V/m)
View written solutionFree

Correct answer: A

  1. Given magnetic field
B⃗=3×10−8sin⁡(1.6×103x+48×1010t) j^  T\vec B = 3\times 10^{-8}\sin(1.6\times 10^3 x + 48\times 10^{10} t)\,\hat j\;\text{T}B=3×10−8sin(1.6×103x+48×1010t)j^​T

So the magnetic field amplitude is

B0=3×10−8  TB_0 = 3\times 10^{-8}\;\text{T}B0​=3×10−8T

and it is along j^\hat jj^​.

  1. Relation between electric and magnetic fields in an electromagnetic wave

For a plane electromagnetic wave in free space,

E0=cB0E_0 = cB_0E0​=cB0​

where

c=3×108  m/sc = 3\times 10^8\;\text{m/s}c=3×108m/s

Thus,

E0=(3×108)(3×10−8)=9  V/mE_0 = (3\times 10^8)(3\times 10^{-8}) = 9\;\text{V/m}E0​=(3×108)(3×10−8)=9V/m
  1. Direction of electric field

In an electromagnetic wave, E⃗\vec EE, B⃗\vec BB, and direction of propagation are mutually perpendicular, and

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

The phase is

sin⁡(kx+ωt)\sin(kx + \omega t)sin(kx+ωt)

which corresponds to propagation along the negative xxx-direction.

So propagation is along −i^-\hat i−i^.

Given

B⃗∥j^\vec B \parallel \hat jB∥j^​

we need E⃗\vec EE such that

E⃗×j^=−i^\vec E \times \hat j = -\hat iE×j^​=−i^

Now,

k^×j^=−i^\hat k \times \hat j = -\hat ik^×j^​=−i^

Hence,

E⃗∥k^\vec E \parallel \hat kE∥k^
  1. Write the electric field expression

Since E⃗\vec EE and B⃗\vec BB are in phase,

E⃗=9sin⁡(1.6×103x+48×1010t) k^  V/m\vec E = 9\sin(1.6\times 10^3 x + 48\times 10^{10} t)\,\hat k\;\text{V/m}E=9sin(1.6×103x+48×1010t)k^V/m
  1. Check options
  • A: E⃗=9sin⁡(1.6×103x+48×1010t) k^  V/m\vec E= 9\sin(1.6 \times 10^3x + 48 \times 10^{10}t)\,\hat k\;\text{V/m}E=9sin(1.6×103x+48×1010t)k^V/m ✅
  • B: amplitude is wrong ❌
  • C: magnitude and direction both wrong ❌
  • D: direction and magnitude wrong ❌

Therefore, the correct option is A.

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