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According to electromagnetic wave theory, the electric and magnetic field vector associated with the wave is given by \[\] $|E| = E_{max} \cos(kx – \omega t)$ \[\] $|B| = B_{max} \cos(kx – \omega t)$ \[\] The energy of electromagnetic wave is confined in the form of electric and magnetic fields. The energy per unit volume in an electric field is given by $U_E = \large\frac{1}{2} \in_oE^2$ and the energy per unit volume in a magnetic field is given by $ U_M =\large\frac{1}{2} B^2/ \mu_0$ . The energy density associated with the magnetic field is equal to the energy density associated with the electric field. The total energy is equally shared by the electromagnetic as well as magnetic field. The rate of flow of energy in an electromagnetic wave is defined as pointing vector $S = 1/\mu_o (E \times B)$. A plane electromagnetic wave varies sinusoidally at $90\: MHz $ as it travels along the $+ve\: x$ direction. The peak value of the electric field is $2\: millivolt/m$ and it is directed along $ \pm y$ direction \[\] The average energy density in the radiation is

$\begin {array} {1 1} (a)\;1.5 \times 10^{-7} J/m^3 & \quad (b)\;1.75 \times 10^{-14} J/m^3 \\ (c)\;3 \times 10^{-12}J/m^3 & \quad (d)\;3 \times 10^{-14} J/m^3 \end {array}$

1 Answer

$u_E = \large\frac{1}{2}$$ \in_oE^2 = \large\frac{1}{2}$$ \times 8.85 × 10^{-12} \times (2 \times 10^{-3} )^2 = 1.75 \times 10^{-14} J/m^3$
Ans : (b)
answered Mar 5, 2014 by thanvigandhi_1

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