(B) The average intensity $I_{av}$ of an electromagnetic wave is given by the formula:$I_{av} = \frac{1}{2} c \epsilon_0 E_0^2$Given:$E_0 = 36 \, V/m$

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(B) The average intensity $I_{av}$ of an electromagnetic wave is given by the formula:$I_{av} = \frac{1}{2} c \epsilon_0 E_0^2$Given:$E_0 = 36 \, V/m$$c = 3 \times 10^8 \, m/s$$\epsilon_0 = 8.85 \times 10^{-12} \, C^2/(N \cdot m^2)$Substituting the values:$I_{av} = \frac{1}{2} \times (3 \times 10^8) \times (8.85 \times 10^{-12}) \times (36)^2$$I_{av} = 0.5 \times 3 \times 10^8 \times 8.85 \times 10^{-12} \times 1296$$I_{av} = 1.5 \times 8.85 \times 1296 \times 10^{-4}$$I_{av} = 17208.6 \times 10^{-4} \approx 1.72 \, W/m^2$

Class 12 Physics Electromagnetic waves Properties of Electromagnetic Waves

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$A$ flood light is covered with a filter that transmits red light. The electric field of the emerging beam is represented by a sinusoidal plane wave
$E_x=36\,\sin\,(1.20 \times 10^7z -3.6 \times 10^{15}\,t)\,V/m$
The average intensity of the beam will be.....$W/m^2$

A
$0.86$
B
$1.72$
C
$3.44$
D
$6.88$

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Electromagnetic waves

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Properties of Electromagnetic Waves

If the direction of propagation of an electromagnetic wave is along the $z$-axis,then the directions of the electric field $(E)$ and magnetic field $(B)$ are:

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$A$ point source of electromagnetic radiation has an average power output of $960 \,W$. The peak value of the electric field at a distance $400 \,cm$ from the source is (in $\,Vm^{-1}$)

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$A$ plane polarized monochromatic $EM$ wave is travelling in a vacuum along the $z$-direction such that at $t = t_1$,it is found that the electric field is zero at a spatial point $z_1$. The next zero that occurs in its neighborhood is at $z_2$. The frequency of the electromagnetic wave is:

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If ${\varepsilon _0}$ and ${\mu _0}$ are respectively the electric permittivity and the magnetic permeability of free space,and ${\varepsilon}$ and ${\mu}$ are the corresponding quantities in a medium,the refractive index of the medium is:

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$A$ plane electromagnetic wave of frequency $25 \text{ MHz}$ travels in free space along the $X$-direction. At a particular point in space and time,the magnetic field is $\overrightarrow{B} = 2.1 \times 10^{-8} \hat{k} \text{ T}$. Find the electric field $\overrightarrow{E}$ at this point.

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$A$ flood light is covered with a filter that transmits red light. The electric field of the emerging beam is represented by a sinusoidal plane wave$E_x=36\,\sin\,(1.20 \times 10^7z -3.6 \times 10^{15}\,t)\,V/m$The average intensity of the beam will be.....$W/m^2$

Similar Questions

If the direction of propagation of an electromagnetic wave is along the $z$-axis,then the directions of the electric field $(E)$ and magnetic field $(B)$ are:

$A$ point source of electromagnetic radiation has an average power output of $960 \,W$. The peak value of the electric field at a distance $400 \,cm$ from the source is (in $\,Vm^{-1}$)

$A$ plane polarized monochromatic $EM$ wave is travelling in a vacuum along the $z$-direction such that at $t = t_1$,it is found that the electric field is zero at a spatial point $z_1$. The next zero that occurs in its neighborhood is at $z_2$. The frequency of the electromagnetic wave is:

If ${\varepsilon _0}$ and ${\mu _0}$ are respectively the electric permittivity and the magnetic permeability of free space,and ${\varepsilon}$ and ${\mu}$ are the corresponding quantities in a medium,the refractive index of the medium is:

$A$ plane electromagnetic wave of frequency $25 \text{ MHz}$ travels in free space along the $X$-direction. At a particular point in space and time,the magnetic field is $\overrightarrow{B} = 2.1 \times 10^{-8} \hat{k} \text{ T}$. Find the electric field $\overrightarrow{E}$ at this point.

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$A$ flood light is covered with a filter that transmits red light. The electric field of the emerging beam is represented by a sinusoidal plane wave
$E_x=36\,\sin\,(1.20 \times 10^7z -3.6 \times 10^{15}\,t)\,V/m$
The average intensity of the beam will be.....$W/m^2$