A plane electromagnetic wave of wavelength $\lambda $ has an intensity $I.$ It is propagating along the positive $Y-$ direction. The allowed expressions for the electric and magnetic fields are given by
$\vec E\, = \,\sqrt {\frac{I}{{{\varepsilon _0}C}}} \cos \left[ {\frac{{2\pi }}{\lambda }(y - ct)} \right]\,\hat i\,;\,\vec B\, = \,\frac{1}{c}E\hat k$
$\vec E\, = \,\sqrt {\frac{I}{{{\varepsilon _0}C}}} \cos \left[ {\frac{{2\pi }}{\lambda }(y - ct)} \right]\,\hat k\,;\,\vec B\, = - \,\frac{1}{c}E\hat i$
$\vec E\, = \,\sqrt {\frac{{2I}}{{{\varepsilon _0}C}}} \cos \left[ {\frac{{2\pi }}{\lambda }(y - ct)} \right]\,\hat k\,;\,\vec B\, = + \frac{1}{c}E\hat i$
$\vec E\, = \,\sqrt {\frac{{2I}}{{{\varepsilon _0}C}}} \cos \left[ {\frac{{2\pi }}{\lambda }(y + ct)} \right]\,\hat k\,;\,\vec B\, = \frac{1}{c}E\hat i$
The amplitude of the magnetic field part of a harmonic electromagnetic wave in vacuum is $B_0 = 510 \;nT$.What is the amplitude of the electric field (in $N/C$) part of the wave?
For plan electromagnetic waves propagating in the $z-$ direction, which one of the following combination gives the correct possible direction for $\vec E$ and $\vec B$ field respectively?
Even though an electric field $E$ exerts a force $qE$ on a charged particle yet the electric field of an $EM$ wave does not contribute to the radiation pressure (but transfers energy). Explain.
The kinetic energy possessed by a body of mass $m$ moving with a velocity $ v$ is equal to $\frac{1}{2}m{v^2}$, provided
For a plane electromagnetic wave, the magnetic field at a point $x$ and time $t$ is
$\overrightarrow{ B }( x , t )=\left[1.2 \times 10^{-7} \sin \left(0.5 \times 10^{3} x +1.5 \times 10^{11} t \right) \hat{ k }\right] T$
The instantaneous electric field $\overrightarrow{ E }$ corresponding to $\overrightarrow{ B }$ is : (speed of light $\left.c=3 \times 10^{8} ms ^{-1}\right)$