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A plane electromagnetic wave is incident on a material surface. If the wave delivers momentum \(p\) and energy \(E\), then (A) \(p=0, E=0\) (B) \(p \neq 0, E \neq 0\) (C) \(p \neq 0, E=0\) (D) \(p=0, E \neq 0\)

Short Answer

Expert verified
Based on our analysis of the different possibilities, we can conclude that the transfer of momentum implies the transfer of energy, and vice versa. Therefore, the correct option is (B) \(p \neq 0, E \neq 0\).

Step by step solution

01

Define a plane electromagnetic wave

A plane electromagnetic wave consists of oscillating electric and magnetic fields which are perpendicular to each other and to the direction of propagation of the wave. The energy carried by an electromagnetic wave is stored in these electric and magnetic fields.
02

Recognize the relationship between momentum and energy

When an electromagnetic wave strikes a material surface, it may be absorbed, reflected, or transmitted through the material. In any of these cases, there will be a transfer of energy and momentum from the wave to the material. Therefore, if the wave delivers momentum p and energy E, both cannot simultaneously be zero.
03

Consider the possibilities given in the options

We need to analyze the given options and find out which one aligns with our understanding of the transfer of energy and momentum in a plane electromagnetic wave: (A) This option states that both momentum and energy are equal to zero, which contradicts our understanding that a plane electromagnetic wave carries energy and momentum. (B) This option states that both momentum and energy are nonzero, which is in agreement with our understanding that an electromagnetic wave delivers both energy and momentum to a material surface. (C) This option states that momentum is nonzero, but energy is zero. However, we know that if the wave delivers momentum, it must also deliver energy to the material. (D) This option states that momentum is zero, but energy is nonzero. Again, we know that if the wave delivers energy, it must also deliver momentum to the material.
04

Choose the correct option

Based on our analysis of the different possibilities, we can conclude that the transfer of momentum implies the transfer of energy, and vice versa. Therefore, the correct option is (B) \(p \neq 0, E \neq 0\).

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Most popular questions from this chapter

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) \((1 / 4 \pi) \times 10^{-7}\)

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 \(\mathrm{B}^{-}\)

The wavelength of \(\mathrm{x}\) rays is of the order of (A) \(1 \mathrm{~cm}\) (B) \(1 \mathrm{~m}\) (C) Imicron (D) 1angstrom

When a plane electromagnetic wave travels in vacuum, the average electric energy density is given by \(\left(E_{0}\right.\) is the amplitude of the electric field) (A) \((1 / 4) \varepsilon_{0} E_{0}^{2}\) (B) \((1 / 2) \varepsilon_{0} E_{0}^{2}\) (C) \(2 e_{\mathrm{o}} \mathrm{E}_{\mathrm{O}}^{2}\) (D) \(4 \varepsilon_{0} E_{0}^{2}\)

If the wavelength of light is \(4000^{\circ} \mathrm{A}\) then the number of waves in \(1 \mathrm{~mm}\) length will be (A) \(2.5\) (B) 2500 (C) 250 (D) 25000

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