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
$\vec E = {E_{0\,}}\,\cos \left( {\omega t - \frac{{2\pi }}{\lambda }y} \right)\hat x$
$\vec E = -{E_{0\,}}\,\cos \left( {\omega t + \frac{{2\pi }}{\lambda }y} \right)\hat x$
$\vec E = -{E_{0\,}}\,\cos \left( {\omega t + \frac{{2\pi }}{\lambda }y} \right)\hat z$
$\vec E = {E_{0\,}}\,\cos \left( {\omega t - \frac{{2\pi }}{\lambda }y} \right)\hat z$
In an electromagnetic wave, the electric and magnetising fields are $100\,V\,{m^{ - 1}}$ and $0.265\,A\,{m^{ - 1}}$. The maximum energy flow is.......$W/{m^2}$
What physical quantity is the same for $X-$rays of wavelength $10^{-10} \;m ,$ $red$ light of wavelength $6800\; \mathring A$ and radiowaves of wavelength $500 \;m ?$
If the magnetic field of a plane electromagnetic wave is given by (The speed of light $ = 3 \times {10^8}\,m/s$ )
$B = 100 \times {10^{ - 6}}\,\sin \,\left[ {2\pi \times 2 \times {{10}^{15}}\,\left( {t - \frac{x}{c}} \right)} \right]$
then the maximum electric field associated with it is
A radio receiver antenna that is $2 \,m$ long is oriented along the direction of the electromagnetic wave and receives a signal of intensity $5 \times {10^{ - 16}}W/{m^2}$. The maximum instantaneous potential difference across the two ends of the antenna is
The magnetic field of a plane electromagnetic wave is given by $\overrightarrow{ B }=3 \times 10^{-8} \cos \left(1.6 \times 10^3 x +48 \times 10^{10} t \right) \hat{ j }$, then the associated electric field will be :