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Chapter 15: Problem 2146

Energy density of an electromagnetic wave of intensity \(0.02 \mathrm{Wm}^{-2}\) is (A) \(6.67 \times 10^{-11} \mathrm{Jm}^{-3}\) (B) \(6 \times 10^{6} \mathrm{Jm}^{-3}\) (C) \(1.5 \times 10^{10} \mathrm{Jm}^{-3}\) (D) none of the above

Short Answer

Expert verified
The energy density (u) of an electromagnetic wave with an intensity of \(0.02 \,\mathrm{Wm}^{-2}\) can be found using the formula \(I = u \times c\), where I is the intensity and c is the speed of light. Plugging in the given values, we get \(u = \frac{0.02}{3 \times 10^8}\), resulting in an energy density of \(u = 6.67 \times 10^{-11} \, \mathrm{Jm}^{-3}\). Therefore, the correct answer is (A) \(6.67 \times 10^{-11} \, \mathrm{Jm}^{-3}\).

Step by step solution

01

Write the formula relating intensity, energy density, and the speed of light

The formula relating intensity (I), energy density (u), and the speed of light (c) is: \[ I = u \times c \]
02

Plug in given values and solve for energy density

We are given the intensity \(I = 0.02 \,\mathrm{Wm}^{-2}\) and the speed of light \(c = 3 \times 10^8 \, \mathrm{m/s}\). Plugging in the values into the formula, we have: \[ 0.02 = u \times (3 \times 10^8) \] To solve for energy density (u), we divide both sides by \(3 \times 10^8\): \[ u = \frac{0.02}{3 \times 10^8} \]
03

Calculate the value of energy density

Performing the division, we get the value of energy density: \[ u = 6.67 \times 10^{-11} \, \mathrm{Jm}^{-3} \]
04

Choose the correct option

Comparing the calculated value of energy density with the provided options, we find that the correct answer is: (A) \(6.67 \times 10^{-11} \, \mathrm{Jm}^{-3}\)

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Most popular questions from this chapter

Electromagnetic wave is produced by oscillating electric and magnetic fields \(E^{-}\) and \(B^{-}\). Choose only the incorrect statement from the following (A) \(\mathrm{E}^{-}\) is perpendicular to \(\mathrm{B}^{-}\). (B) \(E^{-}\) is perpendicular to the direction of propagation of the wave (C) \(\mathrm{B}^{-}\) is perpendicular to the direction of propagation of the wave (D) \(E^{-}\) is parallel to \(\mathrm{B}^{-}\)

Infrared radiations are detected by (A) spectrometer (B) bolometer (C) photocells (D) geiger tubes

Maxwell's modified form of Ampere's circuital law is (A) \(\oint \mathrm{B}^{-} \cdot \mathrm{dS}^{-}\) (B) \(\phi \mathrm{B}^{-} \cdot \mathrm{dS}^{-}=\mu_{\mathrm{o}} \mathrm{i}\) (C) $\oint \mathrm{B}^{-} \cdot \mathrm{d} \ell^{-}=\mu_{\mathrm{o}} \mathrm{i}+\mu_{0} \in_{0}\left(\mathrm{~d} \Phi_{\mathrm{E}} / \mathrm{dt}\right)$ (D) $\oint \mathrm{B}^{-} \cdot \mathrm{d} \mathcal{\ell}^{-}=\mu_{0} \mathrm{i}+\left(1 / \in_{0}\right)\left(\mathrm{d}_{\mathrm{q}} / \mathrm{dt}\right)$

If the electric field associated with a radiation of frequency $10 \mathrm{MH} z\( is \)\mathrm{E}=10 \sin (\mathrm{kx}-\omega \mathrm{t}) \mathrm{mV} / \mathrm{m}$ then its energy density is $\mathrm{Jm}^{-3}\left(\varepsilon_{0}=8.85 \times 10^{-12} \mathrm{C}^{2} \mathrm{~N}^{-1} \mathrm{~m}^{-2}\right)$ (A) \(4.425 \times 10^{-10}\) (B) \(6.26 \times 10^{-14}\) (C) \(8.85 \times 10^{-16}\) (D) \(8.85 \times 10^{-14}\)

In a plane electromagnetic wave, the electric field oscillates sinusoidaly at a frequency of \(2.0 \times 10^{10} \mathrm{~Hz}\). if the peak value of electric field is \(60 \mathrm{Vm}^{-1}\) the average energy density (in \(\mathrm{Jm}^{-3}\) ) of the magnetic field of the wave will be (given \(\left.\mu_{0}=4 \pi \times 10^{-7} \mathrm{Tm} / \mathrm{A}\right)\) (A) \(2 \pi \times 10^{-7}\) (B) \((1 / 2 \pi) \times 10^{-7}\) (C) \(4 \pi \times 10^{-7}\) (D) \((1 / 4 \pi) \times 10^{-7}\)
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