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JEE Main 201219 MayOnline
Exam
JEE Main
Year
2012
Date
19 May
Mode
Online
Subject
Physics
Chapter
Electromagnetic Waves

JEE Main 2012 · 19 May · Online · Physics · Electromagnetic Waves

An electromagnetic wave with frequency ω and wavelength λ travels...

An electromagnetic wave with frequency ω\omega and wavelength λ\lambda travels in the +y+y direction. Its magnetic field is along +x+x-axis. The vector equation for the associated electric field (of amplitude E0E_0 ) is

Options

  1. AE⃗=−E0cos⁡(ωt+2πλy)x^\vec{E}=-E_0 \cos \left(\omega t+\frac{2 \pi}{\lambda} y\right) \hat{x}
  2. BE⃗=E0cos⁡(ωt−2πλy)x^\vec{E}=E_0 \cos \left(\omega t-\frac{2 \pi}{\lambda} y\right) \hat{x}
  3. CE⃗=E0cos⁡(ωt−2πλy)z^\vec{E}=E_0 \cos \left(\omega t-\frac{2 \pi}{\lambda} y\right) \hat{z}
  4. DE⃗=−E0cos⁡(ωt+2πλy)z^\vec{E}=-E_0 \cos \left(\omega t+\frac{2 \pi}{\lambda} y\right) \hat{z}

Correct answer

C. E⃗=E0cos⁡(ωt−2πλy)z^\vec{E}=E_0 \cos \left(\omega t-\frac{2 \pi}{\lambda} y\right) \hat{z}

Step-by-step solution

In an electromagnetic wave electric field and magnetic field are perpendicular to the direction of propagation of wave. The vector equation for the electric field is E⃗=E0cos⁡(ωt−2πλy)z^\vec{E}=E_0 \cos \left(\omega t-\frac{2 \pi}{\lambda} y\right) \hat{z}
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