Intensity of sunlight is observed as $0.092\, {Wm}^{-2}$ at a point in free space. What will be the peak value of magnetic field at that point? $\left(\sigma_{0}=8.85 \times 10^{-12}\, {C}^{2} \,{N}^{-1} \,{m}^{-2}\right.$ )

A particle of mass $\mathrm{m}$ and charge $\mathrm{q}$ has an initial velocity $\overline{\mathrm{v}}=\mathrm{v}_{0} \hat{\mathrm{j}} .$ If an electric field $\overrightarrow{\mathrm{E}}=\mathrm{E}_{0} \hat{\mathrm{i}}$ and magnetic field $\overrightarrow{\mathrm{B}}=\mathrm{B}_{0} \hat{\mathrm{i}}$ act on the particle, its speed will double after a time:

A plane electromagnetic wave with frequency of $30 {MHz}$ travels in free space. At particular point in space and time, electric field is $6 {V} / {m}$. The magnetic field at this point will be ${x} \times 10^{-8} {T}$. The value of ${x}$ is ….. .

In a plane electromagnetic wave, the electric field oscillates sinusoidally at a frequency of $2.0 \times 10^{10}\; Hz$ and amplitude $48\; Vm ^{-1}$

$(a)$ What is the wavelength of the wave?

$(b)$ What is the amplitude of the oscillating magnetic field?

$(c)$ Show that the average energy density of the $E$ field equals the average energy density of the $B$ field. $\left[c=3 \times 10^{8} \;m s ^{-1} .\right]$

The $rms$ value of the electric field of the light coming from the Sun is $720\;N/C$. The average total energy density of the electromagnetic wave is