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A monochromatic light source with power output 60.0 W radiates light of wavelength 700 nm uniformly in all directions. Calculate \(E_{max}\) and \(B_{max}\) for the 700-nm light at a distance of 5.00 m from the source.

Short Answer

Expert verified
E_{max} \approx 120.54 \text{ V/m}, B_{max} \approx 4.02 \times 10^{-7} \text{ T}.

Step by step solution

01

Understand the Problem

We need to find the maximum values of electric field ( E_{max} ) and magnetic field ( B_{max} ) at a specific distance from a light source emitting light at a certain wavelength. Given are the power output (60.0 W), wavelength (700 nm), and distance (5.00 m).
02

Calculate Intensity

First, determine the intensity of light at a distance of 5.00 m from the source. Intensity (I) is defined as the power per unit area, and for a spherical surface, it is given by:\[ I = \frac{P}{4\pi r^2} \]where P = 60.0 W and r = 5.00 m.\[ I = \frac{60.0}{4\pi (5.00)^2} = \frac{60.0}{100\pi} \approx 0.191 \text{ W/m}^2 \]
03

Relation Between Intensity and Electric Field

Intensity is related to the electric field by the equation:\[ I = \frac{1}{2} \epsilon_0 c E_{max}^2 \]Solving for E_{max} gives:\[ E_{max} = \sqrt{\frac{2I}{\epsilon_0 c}} \]Here, \epsilon_0 is the permittivity of free space (8.85 \times 10^{-12} \text{ C}^2/\text{N} \cdot m}^2) and c is the speed of light (3 \times 10^8\ \text{m/s}).
04

Calculate E_{max}

Substitute the values into the formula for E_{max}:\[ E_{max} = \sqrt{\frac{2 \times 0.191}{8.85 \times 10^{-12} \times 3 \times 10^8}} \]Calculate the value:\[ E_{max} \approx 120.54 \text{ V/m} \]
05

Calculate B_{max} Using Electric Field

The maximum magnetic field (B_{max}) is related to the maximum electric field by the speed of light:\[ E_{max} = c B_{max} \]Thus,\[ B_{max} = \frac{E_{max}}{c} \]\[ B_{max} = \frac{120.54}{3 \times 10^8} \approx 4.02 \times 10^{-7} \, \text{T} \]

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Key Concepts

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

Intensity of Light
Intensity of light describes how much power is delivered by light per unit area. It measures the concentration of light energy falling on a surface. The formula for intensity when light spreads uniformly in all directions from a point source is: \[ I = \frac{P}{4\pi r^2} \]where:
  • \( P \) is the total power emitted by the source (in Watts, W).
  • \( r \) is the distance from the light source (in meters, m).
The intensity decreases as you move away from the source, since the energy gets distributed over a larger area. This concept is crucial in understanding how much energy a surface will receive at a certain distance from the light source.
Electric Field
The electric field, in the context of electromagnetic waves, signifies the force field created by charged particles. It is a vector field, meaning it has both magnitude and direction.In relation to light, the electric field contributes to the wave's energy. The maximum electric field, \( E_{max} \), can be found using the relationship:\[ I = \frac{1}{2} \epsilon_0 c E_{max}^2 \] Solving for \( E_{max} \), we have:\[ E_{max} = \sqrt{\frac{2I}{\epsilon_0 c}} \] where:
  • \( \epsilon_0 \) is the permittivity of free space (\( 8.85 \times 10^{-12} \text{ C}^2/\text{N} \cdot \text{m}^2\)).
  • \( c \) is the speed of light (\( 3 \times 10^8\ \text{m/s} \)).
This field plays a central role in how light interacts with other charges and materials.
Magnetic Field
The magnetic field is another component of an electromagnetic wave, complementing the electric field. It also has both direction and magnitude. The relationship between the electric field and the magnetic field in a light wave is defined by the speed of light, \( c \):\[ E_{max} = c B_{max} \]Rearranging to solve for \( B_{max} \), the maximum magnetic field, gives:\[ B_{max} = \frac{E_{max}}{c} \] This shows that the magnetic field is directly proportional to the electric field and inversely proportional to the speed of light. In practice, the magnetic field component of light is many times weaker than the electric field component, but it's crucial for the propagation of electromagnetic waves.
Monochromatic Light
Monochromatic light refers to light of a single wavelength or color. It is pure and consists of only one frequency. An example of monochromatic light is that emitted by a laser. Monochromatic light is significant in experiments and technologies that require precise control of light's wavelength, such as spectroscopy. Characteristics of monochromatic light include:
  • Uniform frequency throughout.
  • Consistent wavelength.
These properties make monochromatic light predictable and highly useful in scientific research.
Wavelength
The wavelength is a fundamental property of waves defining the distance between successive peaks of a wave, often measured in meters or nanometers. It dictates many of the light's properties such as color.For visible light, different wavelengths correspond to different colors. In the context of the exercise, a wavelength of \( 700 \ \text{nm} \) corresponds to red light. Light's wavelength also determines how it interacts with various substances. For instance:
  • Shorter wavelengths (like blue light) scatter more than longer wavelengths (like red light).
  • Different wavelengths can have different energy levels.
This concept is vital in understanding phenomena like diffraction, interference, and the Doppler effect in waves.

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Most popular questions from this chapter

A sinusoidal electromagnetic wave having a magnetic field of amplitude 1.25 \(\mu\)T and a wavelength of 432 nm is traveling in the +\(x\)-direction through empty space. (a) What is the frequency of this wave? (b) What is the amplitude of the associated electric field? (c) Write the equations for the electric and magnetic fields as functions of \(x\) and t in the form of Eqs. (32.17).

A sinusoidal electromagnetic wave emitted by a cellular phone has a wavelength of 35.4 cm and an electric-field amplitude of 5.40 \(\times\) 10\(^{-2}\) V/m at a distance of 250 m from the phone. Calculate (a) the frequency of the wave; (b) the magnetic-field amplitude; (c) the intensity of the wave.

A satellite 575 km above the earth's surface transmits sinusoidal electromagnetic waves of frequency 92.4 MHz uniformly in all directions, with a power of 25.0 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 cm by 40.0 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?

Consider each of the electric- and magnetic-field orientations given next. In each case, what is the direction of propagation of the wave? (a) \(\vec{E}\) in the +\(x\)-direction, \(\vec{B}\) in the +\(y\)-direction; (b) \(\vec{E}\) in the -\(y\)-direction, \(\vec{B}\) in the +\(x\)-direction; (c) \(\vec{E}\) in the +\(z\)-direction, \(\vec{B}\) in the -\(x\)-direction; (d) \(\vec{E}\) in the +\(y\)-direction, \(\vec{B}\) in the -\(z\)-direction.

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

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