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 $\left.c=3 \times 10^{8} ms ^{-1}\right)$
$\overrightarrow{ E }( x , t )=\left[36 \sin \left(0.5 \times 10^{3} x +1.5 \times 10^{11} t \right) \hat{ k }\right] \frac{ v }{ m }$
$\overrightarrow{ E }( x , t )=\left[-36 \sin \left(0.5 \times 10^{3} x +1.5 \times 10^{11} t \right) \hat{ j }\right] \frac{ v }{ m }$
$\overrightarrow{ E }( x , t )=\left[-36 \sin \left(1 \times 10^{3} x +1.5 \times 10^{11} t \right) \hat{ j }\right] \frac{ v }{ m }$
$\overrightarrow{ E }( x , t )=\left[36 \sin \left(1 \times 10^{3} x +1.5 \times 10^{11} t \right) \hat{ j }\right] \frac{ v }{ m }$
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
If electromagnetic wave is propagating in $x-$ direction and electric and magnetic field are in $y$ and $z-$ direction respectively then write equation of $Ey$ and $Bz$.
An electromagnetic wave is represented by the electric field $\vec E = {E_0}\hat n\,\sin \,\left[ {\omega t + \left( {6y - 8z} \right)} \right]$. Taking unit vectors in $x, y$ and $z$ directions to be $\hat i,\hat j,\hat k$ ,the direction of propogation $\hat s$, is
An electromagnetic wave of frequency $\nu = 3.0\,MHz$ passes from vacuum into a dielectric medium with permitivity $\varepsilon = 4.0$. Then
The monoenergetic beam of electrons moving along $+ y$ direction enters a region of uniform electric and magnetic fields. If the beam goes straight undeflected, then fields $B$ and $E$ are directed respectively along