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 answerC$\vec{E}=E_0 \cos \left(\omega t-\frac{2 \pi}{\lambda} y\right) \hat{z}$
- 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'.