The peak value of the magnetic field of an electromagnetic wave moving in a magnetic field is $20 \, nT$. The peak value of the electric field intensity is ..... $V/m$.

$A$ plane polarized monochromatic $EM$ wave is travelling in a vacuum along the $z$-direction such that at $t = t_1$,it is found that the electric field is zero at a spatial point $z_1$. The next zero that occurs in its neighborhood is at $z_2$. The frequency of the electromagnetic wave is:

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

$A$ plane electromagnetic wave of frequency $100 \, MHz$ is travelling in vacuum along the $x$-direction. At a particular point in space and time,$\overrightarrow{B} = 2.0 \times 10^{-8} \hat{k} \, T$ (where $\hat{k}$ is the unit vector along the $z$-direction). What is $\overrightarrow{E}$ at this point?

For a plane electromagnetic wave,the magnetic field at a point $x$ and time $t$ is $\overrightarrow{ B }( x , t ) = [1.2 \times 10^{-7} \sin (0.5 \times 10^{3} x + 1.5 \times 10^{11} t) \hat{ k }] \text{ T}$. The instantaneous electric field $\overrightarrow{ E }$ corresponding to $\overrightarrow{ B }$ is: (speed of light $c = 3 \times 10^{8} \text{ m/s}$)

For an $EM$ wave,the electric field intensity due to a bulb of $100\ W$ at a distance of $3\ m$ is $E$. What is the electric field intensity produced by radiations coming from a $50\ W$ bulb at the same distance?