An em wave is propagating in a medium with a velocity $\vec v =v\hat i.$ The instantaneous oscillating electric field of this em wave is along $+y$ axis. Then the direction of oscillating magnetic field of the em wave will be along

A mathematical representation of electromagnetic wave is given by the two equations $E = E_{max}\,\, cos (kx -\omega\,t)$ and $B = B_{max} cos\, (kx -\omega\,t),$ where $E_{max}$ is the amplitude of the electric field and $B_{max}$ is the amplitude of the magnetic field. What is the intensity in terms of $E_{max}$ and universal constants $μ_0, \in_0.$

The electric field of a plane electromagnetic wave is given by $\overrightarrow{ E }= E _{0}(\hat{ x }+\hat{ y }) \sin ( kz -\omega t )$ Its magnetic field will be given by

The electric field of a plane electromagnetic wave is given by

$\overrightarrow{\mathrm{E}}=\mathrm{E}_{0} \frac{\hat{\mathrm{i}}+\hat{\mathrm{j}}}{\sqrt{2}} \cos (\mathrm{kz}+\omega \mathrm{t})$ At $\mathrm{t}=0,$ a positively charged particle is at the point $(\mathrm{x}, \mathrm{y}, \mathrm{z})=\left(0,0, \frac{\pi}{\mathrm{k}}\right) .$ If its instantaneous velocity at $(t=0)$ is $v_{0} \hat{\mathrm{k}},$ the force acting on it due to the wave is

The magnetic field in a plane electromagnetic wave is $\mathrm{B}_y=\left(3.5 \times 10^{-7}\right) \sin \left(1.5 \times 10^3 \mathrm{x}+0.5\right.$ $\left.\times 10^{11} \mathrm{t}\right) \mathrm{T}$. The corresponding electric field will be

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? $\left(\sigma_{0}=8.85 \times 10^{-12}\, {C}^{2} \,{N}^{-1} \,{m}^{-2}\right.$ )