The electric field part of an electromagnetic wave in vacuum is
$E = 3.1 \, N C^{-1} \cos [ (1.8 \, rad \, m^{-1}) y + (5.4 \times 10^8 \, rad \, s^{-1}) t ] \hat{i}$
The wavelength of this part of the electromagnetic wave is ...... $m$.

About $20 \%$ of the power of a $100 \ W$ bulb is converted to visible radiation. Assuming that the radiation is emitted isotropically and neglecting reflection,the average intensity of visible radiation at a distance of $5 \ m$ is $\frac{\alpha}{25 \pi} \ W/m^2$. The value of $\alpha$ is

In an electromagnetic wave,at an instant and at a particular position,the electric field is along the negative $z$-axis and the magnetic field is along the positive $x$-axis. Then the direction of propagation of the electromagnetic wave is

$A$ plane electromagnetic wave of frequency $500\, MHz$ is travelling in vacuum along $y$-direction. At a particular point in space and time,$\overrightarrow{B} = 8.0 \times 10^{-8} \hat{z}\, T$. The value of electric field at this point is (speed of light $c = 3 \times 10^{8}\, m/s$). $\hat{x}, \hat{y}, \hat{z}$ are unit vectors along $x, y$ and $z$ directions.

$A$ microwave oven acts on the principle of:

In an electromagnetic wave,if $\overrightarrow{E}$ is the electric field and $\overrightarrow{B}$ is the magnetic field,then the direction of propagation of the wave is given by: