The energies required to set up in a cube of side $10 \ cm$,$(a)$ a uniform electric field of $10^7 \ V \ m^{-1}$ and $(b)$ a uniform magnetic field of $0.25 \ Wb \ m^{-2}$,are respectively about $(\mu_0 = 4\pi \times 10^{-7} \ H \ m^{-1}, \epsilon_0 = 8.9 \times 10^{-12} \ F \ m^{-1})$.

The graph between radiation power $P$ from a linear antenna and the square of the length of the antenna $l^2$ is shown in the figure for two antennas $A$ and $B$. If all other parameters are the same,find the ratio of the wavelengths of $A$ and $B$,i.e.,$\frac{\lambda_A}{\lambda_B}$.

$A$ parallel plate capacitor has a capacitance $C = 200 \ pF$. It is connected to a $230 \ V$ $AC$ supply with an angular frequency $\omega = 300 \ rad/s$. The $rms$ value of the conduction current in the circuit and the displacement current in the capacitor,respectively,are:

The equation of the electric field of an electromagnetic wave propagating through free space is given by:
$E = \sqrt{377} \sin(6.27 \times 10^3 t - 2.09 \times 10^{-5} x) \text{ N/C}$.
The average power (intensity) of the electromagnetic wave is $\left(\frac{1}{\alpha}\right) \text{ W/m}^2$. The value of $\alpha$ is . . . . . . .
(Take $\sqrt{\frac{\mu_0}{\varepsilon_0}} = 377$ in $SI$ units)

In an electromagnetic wave,the electric field and magnetising field are $100\,V\,m^{-1}$ and $0.265\,A\,m^{-1}$ respectively. The maximum energy flow is.......$W\,m^{-2}$.

The energy that will be ideally radiated by a $100\,kW$ transmitter in $1$ hour is: