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Home  >>  JEEMAIN and NEET  >>  Physics  >>  Class12  >>  Electromagnetic Waves

A plane electromagnetic wave in the visible region is moving along the $z$ direction. The frequency of the wave is $0.5 \times 10^{15}Hz$ and the electric field at any point is varying sinusoidally with time with amplitude of $1\: V/m$. Calculate the average values of densities of the magnetic field.

$\begin {array} {1 1} (a)\;2.21 \times 10^{-12}J/m^2 & \quad (b)\;2.21 \times 10^{-2}J/m^2 \\ (c)\;22.1 \times 10^{-12}J/m^2 & \quad (d)\;22.1 \times 10^{-2}J/m^2 \end {array}$

1 Answer

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The average value of magnetic field density is given by
$ = \large\frac{B_o^2}{ 4 \mu_o}$$ = \large\frac{1}{4}$$ \large\frac{E_o^2}{ \mu_oc^2}$$ = \large\frac{0.25 \times1}{ 4 \pi \times 10^{-7} \times (3 \times 10^8)^2} $$= 2.21 \times 10^{-12} J/m^2$
Ans : (a)


answered Mar 1, 2014 by thanvigandhi_1
edited Oct 8, 2014 by thagee.vedartham

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