$A$ radio receiver antenna that is $4 \, m$ long is oriented along the direction of the electromagnetic wave and it receives a signal of intensity $8 \times 10^{-16} \, W/m^2$. The maximum instantaneous potential difference across the two ends of the antenna is (in $ \, \mu V$)

If the electric field intensity of a uniform plane electromagnetic wave is given as $E = -301.6 \sin (kz - \omega t) \hat{a}_{x} + 452.4 \sin (kz - \omega t) \hat{a}_{y} \text{ V/m}$. Then,the magnetic intensity $H$ of this wave in $\text{A/m}$ will be (Given: Speed of light in vacuum $c = 3 \times 10^{8} \text{ m/s}$,permeability of vacuum $\mu_{0} = 4\pi \times 10^{-7} \text{ N/A}^{2}$)

An electromagnetic wave has its electric and magnetic fields given by $\vec{E}(t) = \vec{E}_m \sin(kx - \omega t)$ and $\vec{B}(t) = \vec{B}_m \sin(kx - \omega t)$. If the directions of $\vec{E}_m$ and $\vec{B}_m$ are in the direction of $\hat{i} + \hat{j}$ and $\hat{i} - \hat{j}$ respectively,the unit vector that gives the direction of propagation of the wave is:

If the peak value of the magnetic field of an electromagnetic wave is $30 \times 10^{-9} \ T$,then the peak value of the electric field is (in $Vm^{-1}$)

If the average power per unit area delivered by an electromagnetic wave is $9240 \ W \ m^{-2}$,then the amplitude of the oscillating magnetic field in the $EM$ wave is: (in $\mu T$)

What is the displacement variable in a simple pendulum and the propagation of light?