/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 43 A small helium-neon laser emits ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A small helium-neon laser emits red visible light with a power of \(4.60 \mathrm{~mW}\) in a beam that has a diameter of \(2.50 \mathrm{~mm}\). (a) What are the amplitudes of the electric and magnetic fields of the light? (b) What are the average energy densities associated with the electric field and with the magnetic field? (c) What is the total energy contained in a \(1.00-\mathrm{m}\) length of the beam?

Short Answer

Expert verified
(a) \(E_0 \approx 843 \; \text{V/m}, B_0 \approx 2.81 \times 10^{-6} \; \text{T}\). (b) \(U_e \approx 3.15 \times 10^{-3} \; \text{J/m}^3, U_m \approx 1.05 \times 10^{-9} \; \text{J/m}^3\). (c) \(U \approx 1.55 \times 10^{-8} \; \text{J}\).

Step by step solution

01

Calculate the intensity of the laser beam

The intensity (I) of the laser beam can be calculated using the formula: \( I = \frac{P}{A} \), where \( P \) is the power of the laser and \( A \) is the cross-sectional area of the beam. First, we find the area \( A \) using the diameter given:\[ A = \pi \left( \frac{d}{2} \right)^2 = \pi \left( \frac{2.50 \times 10^{-3}}{2} \right)^2 \]Now, insert the values into the formula:\[ A \approx 4.91 \times 10^{-6} \; \text{m}^2 \]Next, insert \( P = 4.60 \times 10^{-3} \; \text{W} \) into the intensity formula:\[ I = \frac{4.60 \times 10^{-3}}{4.91 \times 10^{-6}} \approx 937 \; \text{W/m}^2 \]
02

Find the electric field amplitude

The electric field amplitude (E_0) can be determined using the intensity calculated in Step 1:\[ I = \frac{1}{2} c \varepsilon_0 E_0^2 \]Rearrange the formula to solve for \( E_0 \):\[ E_0 = \sqrt{\frac{2I}{c \varepsilon_0}} \]Where \( c = 3.00 \times 10^8 \; \text{m/s} \) and \( \varepsilon_0 = 8.85 \times 10^{-12} \; \text{F/m} \).Substitute the values:\[ E_0 = \sqrt{\frac{2 \times 937}{3.00 \times 10^8 \times 8.85 \times 10^{-12}}} \approx 843 \; \text{V/m} \]
03

Find the magnetic field amplitude

Calculate the magnetic field amplitude (B_0) using the relation between the electric and magnetic fields in an electromagnetic wave:\[ B_0 = \frac{E_0}{c} \]Substitute the value of \( E_0 \) from Step 2:\[ B_0 = \frac{843}{3.00 \times 10^8} \approx 2.81 \times 10^{-6} \; \text{T} \]
04

Calculate average energy density of the electric field

The average energy density of the electric field (U_e) is given by:\[ U_e = \frac{1}{2} \varepsilon_0 E_0^2 \]Substitute \( E_0 = 843 \; \text{V/m} \) and \( \varepsilon_0 = 8.85 \times 10^{-12} \; \text{F/m} \):\[ U_e = \frac{1}{2} \times 8.85 \times 10^{-12} \times 843^2 \approx 3.15 \times 10^{-3} \; \text{J/m}^3 \]
05

Calculate average energy density of the magnetic field

The average energy density of the magnetic field (U_m) is given by:\[ U_m = \frac{1}{2} \frac{B_0^2}{\mu_0} \]Where \( \mu_0 = 4\pi \times 10^{-7} \; \text{T m/A} \).Substitute \( B_0 = 2.81 \times 10^{-6} \; \text{T} \):\[ U_m = \frac{1}{2} \times \frac{(2.81 \times 10^{-6})^2}{4\pi \times 10^{-7}} \approx 1.05 \times 10^{-9} \; \text{J/m}^3 \]
06

Calculate total energy in a 1 m length of the beam

The total energy \( U \) contained in a 1.00 m length of the beam can be found by multiplying the average energy density by the volume of the cylinder formed by the beam:\[ U = (U_e + U_m) \times A \times L \]Where \( L = 1.00 \; \text{m} \), \( A \approx 4.91 \times 10^{-6} \; \text{m}^2 \), and add \( U_e + U_m \) from Steps 4 and 5:\[ U = (3.15 \times 10^{-3} + 1.05 \times 10^{-9}) \times 4.91 \times 10^{-6} \times 1.00 \approx 1.55 \times 10^{-8} \; \text{J} \]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Electric Field Amplitude
The electric field amplitude is a crucial measure of the strength of the electric component in an electromagnetic wave, like a laser beam. It's denoted as \(E_0\), and it's directly related to the intensity of the beam. The intensity \(I\) of the laser is given by the power per unit area. To find \(E_0\), we employ the formula: \[ I = \frac{1}{2} c \varepsilon_0 E_0^2 \] Here, \(c\) is the speed of light and \(\varepsilon_0\) is the permittivity of free space. Rearranging the formula gives us the amplitude: \[ E_0 = \sqrt{\frac{2I}{c \varepsilon_0}} \] Upon substitution of the known values, the calculated electric field amplitude for our helium-neon laser is \(843\; \text{V/m}\). This defines how strongly charged particles will be influenced by the laser's electric field.
Magnetic Field Amplitude
An electromagnetic wave includes both electric and magnetic fields. The amplitude of the magnetic field, represented by \(B_0\), can be derived from the electric field amplitude. The relationship between the electric field \(E_0\) and the magnetic field \(B_0\) in an electromagnetic wave is given by: \[ B_0 = \frac{E_0}{c} \] Where \(c\) is the velocity of light in vacuum. Using our previously calculated \(E_0\) for the laser, we find: \[ B_0 = \frac{843}{3.00 \times 10^8} \approx 2.81 \times 10^{-6}\; \text{T} \] This tiny value reflects how the magnetic field in an electromagnetic wave is much weaker compared to its electric counterpart in terms of amplitude.
Energy Density
Energy density in an electromagnetic field indicates the energy stored in the field per unit volume. We look at both the electric and magnetic contributions. For the electric field, the energy density \(U_e\) is computed as: \[ U_e = \frac{1}{2} \varepsilon_0 E_0^2 \] Plugging the electric field values gives: \[ U_e \approx 3.15 \times 10^{-3}\; \text{J/m}^3 \] For the magnetic field, the energy density \(U_m\) is: \[ U_m = \frac{1}{2} \frac{B_0^2}{\mu_0} \] Where \(\mu_0\) is the permeability of free space. By substituting \(B_0\), we find: \[ U_m \approx 1.05 \times 10^{-9}\; \text{J/m}^3 \] These values tell us how much energy is stored by each field component in the laser beam, reflecting the balance between electric and magnetic energies.
Laser Beam Power
Laser beam power is a fundamental property that tells us the energy output per second by the laser, measured here in milliwatts (mW). In our example, the beam power \(P\) is \(4.60\, \text{mW}\), a relatively low power typical for helium-neon lasers used in laboratories.
The power determines the beam intensity, critical for understanding energy transfer in applications. Calculating power helps us express intensity \(I\) using:
\[ I = \frac{P}{A} \]
where \(A\) is the cross-sectional area of the beam. Solving for intensity is the first step in assessing how a laser beam interacts with materials or how effective it might be in various applications.
Cross-sectional Area
The cross-sectional area of a laser beam is vital because it determines how the beam's power is distributed over a surface. Calculating this area involves using the beam diameter, as the beam is typically circular. The formula is:
\[ A = \pi \left( \frac{d}{2} \right)^2 \]
For our laser, with a beam diameter of \(2.50 \, \text{mm}\), converting to meters and calculating gives:
\[ A \approx 4.91 \times 10^{-6} \, \text{m}^2 \]
This small area suggests a concentrated beam, typical in laser setups, ensuring high precision and allowing control over the beam's impact in applications such as reading barcodes or precision cutting. The efficiency and potential damage of the beam are partly determined by this area calculation.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Consider electromagnetic waves propagating in air. (a) Determine the frequency of a wave with a wavelength of (i) \(5.0 \mathrm{~km}\), (ii) \(5.0 \mu \mathrm{m}\), (iii) \(5.0 \mathrm{~nm}\). (b) What is the wavelength (in meters and nanometers) of (i) gamma rays of frequency \(6.50 \times 10^{21} \mathrm{~Hz}\) and (ii) an AM station radio wave of frequency \(590 \mathrm{kHz}\) ?

A cylindrical conductor with a circular cross section has a radius \(a\) and a resistivity \(\rho\) and carries a constant current \(I\). (a) What are the magnitude and direction of the electric-field vector \(\overrightarrow{\boldsymbol{E}}\) at a point just inside the wire at a distance \(a\) from the axis? (b) What are the magnitude and direction of the magneticfield vector \(\overrightarrow{\boldsymbol{B}}\) at the same point? (c) What are the magnitude and direction of the Poynting vector \(\overrightarrow{\boldsymbol{S}}\) at the same point? (The direction of \(\vec{S}\) is the direction in which electromagnetic energy flows into or out of the conductor.) (d) Use the result in part (c) to find the rate of flow of energy into the volume occupied by a length \(l\) of the conductor. (Hint: Integrate \(\vec{S}\) over the surface of this volume.) Compare your result to the rate of generation of thermal energy in the same volume. Discuss why the energy dissipated in a currentcarrying conductor, due to its resistance, can be thought of as entering through the cylindrical sides of the conductor.

Fields from a Light Bulb. We can reasonably model a \(75-\mathrm{W}\) incandescent light bulb as a sphere \(6.0 \mathrm{~cm}\) in diameter. Typically, only about \(5 \%\) of the energy goes to visible light; the rest goes largely to nonvisible infrared radiation. (a) What is the visible-light intensity (in \(\mathrm{W} / \mathrm{m}^{2}\) ) at the surface of the bulb? (b) What are the amplitudes of the electric and magnetic fields at this surface, for a sinusoidal wave with this intensity?

A satellite \(575 \mathrm{~km}\) above the earth's surface transmits sinusoidal electromagnetic waves of frequency \(92.4 \mathrm{MHz}\) uniformly in all directions, with a power of \(25.0 \mathrm{~kW}\). (a) What is the intensity of these waves as they reach a receiver at the surface of the earth directly below the satellite? (b) What are the amplitudes of the electric and magnetic fields at the receiver? (c) If the receiver has a totally absorbing panel measuring \(15.0 \mathrm{~cm}\) by \(40.0 \mathrm{~cm}\) oriented with its plane perpendicular to the direction the waves travel, what average force do these waves exert on the panel? Is this force large enough to cause significant effects?

Laboratory Lasers. He-Ne lasers are often used in physics demonstrations. They produce light of wavelength \(633 \mathrm{~nm}\) and a power of \(0.500 \mathrm{~mW}\) spread over a cylindrical beam \(1.00 \mathrm{~mm}\) in diameter (although these quantities can vary). (a) What is the intensity of this laser beam? (b) What are the maximum values of the electric and magnetic fields? (c) What is the average energy density in the laser beam?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.