The electric field for an electromagnetic wave in free space is $E = i 30 \cos (k z - 5 \times 10^8 t)$,where the magnitude of $E$ is in $V/m$. The magnitude of the wave vector $k$ is (velocity of electromagnetic wave in free space $= 3 \times 10^8 \ m/s$).
- A$0.46 \ rad \ m^{-1}$
- B$3 \ rad \ m^{-1}$
- C$1.66 \ rad \ m^{-1}$
- D$0.83 \ rad \ m^{-1}$
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Similar Questions
$A$ plane electromagnetic wave has a frequency of $2.0 \times 10^{10} \ Hz$ and its energy density is $1.02 \times 10^{-8} \ J/m^3$ in vacuum. The amplitude of the magnetic field of the wave is close to $....nT$. (Given: $\frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \ Nm^2/C^2$ and speed of light $c = 3 \times 10^8 \ m/s$)
DifficultJEE MAIN 2020
View SolutionThe intensity of the light from a bulb incident on a surface is $0.22 \, W/m^2$. The amplitude of the magnetic field in this light wave is . . . . . . $\times 10^{-9} \, T$. (Given: Permittivity of vacuum $\epsilon_0 = 8.85 \times 10^{-12} \, C^2 N^{-1} m^{-2}$,speed of light in vacuum $c = 3 \times 10^8 \, m/s$)
The electric field strength in an $EM$ wave is $10^4 \, V/m$. The magnitude of the magnetic field strength (in tesla) will be:
Medium
View SolutionThe electric field of a plane electromagnetic wave in a medium is given by $\vec{E}(x, y, z, t) = E_0 \hat{n} e^{i k_0[(x+y+z)-ct]}$, where $c$ is the speed of light in free space. The $\vec{E}$ field is polarized in the $x-z$ plane. If the speed of the wave in the medium is $v$, then:
An electromagnetic wave is travelling in $x$-direction with electric field vector given by $\vec{E}_{y} = E_{0} \sin(kx - \omega t) \hat{j}$. The correct expression for the magnetic field vector is:
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