The ratio of amplitude of magnetic field to the amplitude of electric field for an electromagnetic wave propagating in vacuum is equal to 

Plane microwaves from a transmitter are directed normally towards a plane reflector. $A$ detector moves along the normal to the reflector. Between positions of $14$ successive maxima, the detector travels a distance $0.13\, m$. If the velocity of light is $3 \times 10^8 m/s$, find the frequency of the transmitter.

A plane electromagnetic wave of frequency $28 \,MHz$ travels in free space along the positive $x$-direction. At a particular point in space and time, electric field is $9.3 \,V / m$ along positive $y$-direction. The magnetic field (in $T$ ) at that point is

For an electromagnetic wave travelling in free space, the relation between average energy densities due to electric $\left( U _{ e }\right)$ and magnetic $\left( U _{ m }\right)$ fields is

The magnetic field of a plane electromagnetic wave is given by $\overrightarrow{ B }=3 \times 10^{-8} \cos \left(1.6 \times 10^3 x +48 \times 10^{10} t \right) \hat{ j }$, then the associated electric field will be :

The electric field of a plane polarized electromagnetic wave in free space at time $t = 0$ is given by an expression

$\vec E(x,y) = 10\hat j\, cos[(6x + 8z)]$

The magnetic field $\vec B (x,z, t)$ is given by : ($c$ is the velocity of light)