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Chapter 24: Problem 22

An industrial laser is used to burn a hole through a piece of metal. The average intensity of the light is \(\bar{S}=1.23 \times 10^{9} \mathrm{~W} / \mathrm{m}^{2}\). What is the rms value of (a) the electric field and (b) the magnetic field in the electromagnetic wave emitted by the laser?

Short Answer

Expert verified
(a) The rms electric field is approximately \(9.62 \times 10^5 \mathrm{~V/m}\). (b) The rms magnetic field is approximately \(3.21 \times 10^{-3} \mathrm{~T}\).

Step by step solution

01

Understanding the relationship between intensity and the electric field

The average intensity \( \bar{S} \) of an electromagnetic wave is related to the root mean square (rms) value of the electric field \( E_{rms} \) by the formula: \[ \bar{S} = \frac{1}{2} \epsilon_0 c E_{rms}^2 \] where \( \epsilon_0 = 8.85 \times 10^{-12} \mathrm{~C}^2/\mathrm{N}\cdot\mathrm{m}^2 \) is the permittivity of free space, and \( c = 3 \times 10^8 \mathrm{~m/s} \) is the speed of light in vacuum.
02

Solving for the rms value of the electric field

Rearrange the formula to solve for \( E_{rms} \): \[ E_{rms} = \sqrt{\frac{2 \bar{S}}{\epsilon_0 c}} \] Substituting the given values: \[ E_{rms} = \sqrt{\frac{2 \times 1.23 \times 10^9}{8.85 \times 10^{-12} \times 3 \times 10^8}} \] Calculate \( E_{rms} \).
03

Calculating the rms value for the electric field

Carry out the computation: \[ E_{rms} = \sqrt{\frac{2.46 \times 10^9}{2.655 \times 10^{-3}}} = \sqrt{9.26 \times 10^{11}} \] \[ E_{rms} \approx 9.62 \times 10^5 \mathrm{~V/m} \]
04

Understanding the relationship between electric and magnetic fields

The rms value of the magnetic field \( B_{rms} \) is related to the rms value of the electric field \( E_{rms} \) by the equation: \[ B_{rms} = \frac{E_{rms}}{c} \] Use \( c = 3 \times 10^8 \mathrm{~m/s} \) to solve for \( B_{rms} \).
05

Solving for the rms value of the magnetic field

Substitute the value we found for \( E_{rms} \) into the equation: \[ B_{rms} = \frac{9.62 \times 10^5}{3 \times 10^8} \] Calculate \( B_{rms} \).
06

Calculating the rms value for the magnetic field

Perform the calculation: \[ B_{rms} = 3.21 \times 10^{-3} \mathrm{~T} \]

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Key Concepts

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

Electric Field
Whenever you encounter an electromagnetic wave, like the laser used to burn through metal, it comes with an electric field component. The electric field in electromagnetic waves is an essential factor since it carries energy.

To find the root mean square (rms) value of the electric field, we use the formula:
  • Average intensity, \( \bar{S} \), is related to \( E_{rms} \) by: \[ \bar{S} = \frac{1}{2} \epsilon_0 c E_{rms}^2 \]
  • \( \epsilon_0 \) is the permittivity of free space, and \( c \) is the speed of light.
By rearranging this formula, we can solve for \( E_{rms} \):
  • \( E_{rms} = \sqrt{\frac{2 \bar{S}}{\epsilon_0 c}} \)
After plugging in the known values, we find that the rms value of the electric field is approximately \( 9.62 \times 10^5 \mathrm{~V/m} \).

This means that on average, the electric field strength oscillates with this magnitude in the space the wave occupies.
Magnetic Field
In an electromagnetic wave, wherever there's an electric field, there鈥檚 also a magnetic field. These fields support and sustain each other as they travel through space.

The relationship between the electric and magnetic fields is given by the formula:
  • \( B_{rms} = \frac{E_{rms}}{c} \)
  • Where \( B_{rms} \) is the root mean square value of the magnetic field, and \( c \) is the speed of light.
By substituting the found value of \( E_{rms} \) into this equation, we calculate \( B_{rms} \), and find it to be approximately \( 3.21 \times 10^{-3} \mathrm{~T} \).

This magnetic field is perpendicular to the electric field in the wave, working in harmony to propagate electromagnetic energy.
Intensity of Electromagnetic Waves
The intensity of an electromagnetic wave, such as the laser in this scenario, represents the power per unit area it carries.

In mathematical terms, the intensity \( \bar{S} \) connects to the electric and magnetic fields as a measure of how much energy the wave delivers.
  • High intensity means the wave can carry more energy to do work, like burning through metal.
  • It plays a key role in determining the electric field鈥檚 strength using the formula: \[ \bar{S} = \frac{1}{2} \epsilon_0 c E_{rms}^2 \]
Understanding intensity helps in grasping how energetic a wave is. High intensity influences the effectiveness of actions such as cutting or burning in industrial applications.

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Most popular questions from this chapter

Interactive Solution \(\underline{24.53}\) at provides one model for problems like this one. The drawing shows an edge-on view of the solar panels on a communications satellite. The dashed line specifies the normal to the panels. Sunlight strikes the panels at an angle \(\theta\) with respect to the normal. If the solar power impinging on the panels is \(2600 \mathrm{~W}\) when \(\theta=65^{\circ}\), what is it when \(\theta=25^{\circ} ?\)

The drawing shows light incident on a polarizer whose transmission axis is parallel to the \(z\) axis. The polarizer is rotated clockwise through an angle \(\alpha\) between 0 and \(90^{\circ}\). While the polarizer is being rotated, does the intensity of the transmitted light increase, decrease, or remain the same if the incident light is (a) unpolarized, (b) polarized parallel to the \(z\) axis, and (c) polarized parallel to the \(y\) axis? Provide a reason for each of your answers.

An argon-ion laser produces a cylindrical beam of light whose average power is \(0.750 \mathrm{~W}\). How much energy is contained in a 2.50 -m length of the beam?

At one time television sets used "rabbit-ears" antennas. Such an antenna consists of a pair of metal rods. The length of each rod can be adjusted to be one-quarter of a wavelength of an electromagnetic wave whose frequency is \(60.0 \mathrm{MHz}\). How long is each rod?

The mean distance between earth and the sun is \(1.50 \times 10^{11} \mathrm{~m} .\) The average intensity of solar radiation incident on the upper atmosphere of the earth is \(1390 \mathrm{~W} / \mathrm{m}^{2}\). Assuming the sun emits radiation uniformly in all directions, determine the total power radiated by the sun.
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