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Electromagnetic Waves question

2024 · 9 Apr · Shift 1 · Q61
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  5. /2024 · 9 Apr · Shift 1 · Q61

Electromagnetic Waves question

2024 · 9 Apr · Shift 1 · Q61

JEE MainPhysicsElectromagnetic WavesMCQ+4 / −1
A plane EM wave is propagating along xxx direction. It has a wavelength of 4 mm4 \mathrm{~mm}4 mm. If electric field is in yyy direction with the maximum magnitude of 60 Vm−160 \mathrm{~Vm}^{-1}60 Vm−1, the equation for magnetic field is :
  1. A
    Bz=2×10−7sin⁡[π2(x−3×108t)]k^T\mathrm{B}_z=2 \times 10^{-7} \sin \left[\frac{\pi}{2}\left(x-3 \times 10^8 \mathrm{t}\right)\right] \hat{\mathrm{k}} \mathrm{T}Bz​=2×10−7sin[2π​(x−3×108t)]k^T
  2. B
    Bz=2×10−7sin⁡[π2×103(x−3×108t)]k^T\mathrm{B}_z=2 \times 10^{-7} \sin \left[\frac{\pi}{2} \times 10^3\left(x-3 \times 10^8 \mathrm{t}\right)\right] \hat{\mathrm{k}} \mathrm{T}Bz​=2×10−7sin[2π​×103(x−3×108t)]k^T
  3. C
    Bz=60sin⁡[π2(x−3×108t)]k^T\mathrm{B}_z=60 \sin \left[\frac{\pi}{2}\left(x-3 \times 10^8 \mathrm{t}\right)\right] \hat{\mathrm{k}} \mathrm{T}Bz​=60sin[2π​(x−3×108t)]k^T
  4. D
    Bx=60sin⁡[π2(x−3×108t)]i^T\mathrm{B}_x=60 \sin \left[\frac{\pi}{2}\left(x-3 \times 10^8 \mathrm{t}\right)\right] \hat{\mathrm{i}} \mathrm{T}Bx​=60sin[2π​(x−3×108t)]i^T
View written solutionFree

Correct answer: B

  1. Write the general form of a plane EM wave

For an EM wave propagating in the +x+x+x direction,

E⃗=E0sin⁡(kx−ωt) j^,\vec E = E_0 \sin(kx-\omega t)\,\hat j,E=E0​sin(kx−ωt)j^​,

then the magnetic field must be perpendicular to both i^\hat ii^ and j^\hat jj^​, so it is along k^\hat kk^:

B⃗=B0sin⁡(kx−ωt) k^.\vec B = B_0 \sin(kx-\omega t)\,\hat k.B=B0​sin(kx−ωt)k^.

So immediately:

  • magnetic field cannot be along xxx direction,
  • hence option D is wrong.
  1. Find the magnitude of magnetic field

For an electromagnetic wave in vacuum,

E0=cB0E_0 = cB_0E0​=cB0​

so

B0=E0c=603×108=2×10−7 T.B_0 = \frac{E_0}{c} = \frac{60}{3\times 10^8} = 2\times 10^{-7}\,\text{T}.B0​=cE0​​=3×10860​=2×10−7T.

Thus options C and D are wrong because their amplitude is incorrect.

  1. Find the wave number kkk from wavelength

Given

λ=4 mm=4×10−3 m.\lambda = 4\,\text{mm} = 4\times 10^{-3}\,\text{m}.λ=4mm=4×10−3m.

Now,

k=2πλ=2π4×10−3=π2×103 m−1.k = \frac{2\pi}{\lambda} = \frac{2\pi}{4\times 10^{-3}} = \frac{\pi}{2}\times 10^3\,\text{m}^{-1}.k=λ2π​=4×10−32π​=2π​×103m−1.

So the phase should be

k(x−ct)=π2×103 (x−3×108t).k(x-ct)=\frac{\pi}{2}\times 10^3\,(x-3\times 10^8 t).k(x−ct)=2π​×103(x−3×108t).
  1. Write the complete magnetic field equation

Therefore,

B⃗=2×10−7sin⁡[π2×103(x−3×108t)]k^ T.\vec B = 2\times 10^{-7}\sin\left[\frac{\pi}{2}\times 10^3\left(x-3\times 10^8 t\right)\right]\hat k\,\text{T}.B=2×10−7sin[2π​×103(x−3×108t)]k^T.

This matches Option B.

  1. Check all options systematically
  • A: amplitude correct, direction correct, but wave number missing factor 10310^3103 ⇒\Rightarrow⇒ wrong.
  • B: amplitude correct, direction correct, phase correct ⇒\Rightarrow⇒ correct.
  • C: wrong amplitude ⇒\Rightarrow⇒ wrong.
  • D: wrong direction and wrong amplitude ⇒\Rightarrow⇒ wrong.

Final Answer: B\boxed{\text{B}}B​

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