Q82 / 90 · JEE Main 2012 (19 May Online) · Physics

An electromagnetic wave with frequency $\omega$ and wavelength $\lambda$ travels in the $+y$ direction. Its magnetic field is along $+x$-axis. The vector equation for the associated electric field (of amplitude $E_0$ ) is

  • A
    $\vec{E}=-E_0 \cos \left(\omega t+\frac{2 \pi}{\lambda} y\right) \hat{x}$
  • B
    $\vec{E}=E_0 \cos \left(\omega t-\frac{2 \pi}{\lambda} y\right) \hat{x}$
  • Correct answer
    C
    $\vec{E}=E_0 \cos \left(\omega t-\frac{2 \pi}{\lambda} y\right) \hat{z}$
  • D
    $\vec{E}=-E_0 \cos \left(\omega t+\frac{2 \pi}{\lambda} y\right) \hat{z}$

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 $$ \vec{E}=E_0 \cos \left(\omega t-\frac{2 \pi}{\lambda} y\right) \hat{z} $$

Hence, the correct answer is option: 'C'.

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