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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).

For a sinusoidal electromagnetic wave in vacuum, such as that described by Eq. (32.16), show that the \(average\) energy density in the electric field is the same as that in the magnetic field.

In the 25-ft Space Simulator facility at NASA's Jet Propulsion Laboratory, a bank of overhead arc lamps can produce light of intensity 2500 W/m\(^2\) at the floor of the facility. (This simulates the intensity of sunlight near the planet Venus.) Find the average radiation pressure (in pascals and in atmospheres) on (a) a totally absorbing section of the floor and (b) a totally reflecting section of the floor. (c) Find the average momentum density (momentum per unit volume) in the light at the floor.

Electromagnetic waves propagate much differently in conductors than they do in dielectrics or in vacuum. If the resistivity of the conductor is sufficiently low (that is, if it is a sufficiently good conductor), the oscillating electric field of the wave gives rise to an oscillating conduction current that is much larger than the displacement current. In this case, the wave equation for an electric field \(\vec{E} (x, t) = E_y(x, t)\hat{\jmath}\) en propagating in the +\(x\)-direction within a conductor is $${\partial^2E_y(x, t)\over \partial x^2} = {\mu \over \rho} {\partial Ey(x, t)\over \partial t}$$ where \(\mu\) is the permeability of the conductor and \(\rho\) is its resistivity. (a) A solution to this wave equation is \(E_y(x, t) = E_{max} e^{-k_C x} cos(k_Cx - \omega t)\), where \(k_C = \sqrt{(\omega \mu/2\rho}\). Verify this by substituting E_y(x, t) into the above wave equation. (b) The exponential term shows that the electric field decreases in amplitude as it propagates. Explain why this happens. (\(Hint\): The field does work to move charges within the conductor. The current of these moving charges causes \(i^2R\) heating within the conductor, raising its temperature. Where does the energy to do this come from?) (c) Show that the electric-field amplitude decreases by a factor of 1/\(e\) in a distance \(1/k_C = \sqrt{2\rho/\omega\mu}\), and calculate this distance for a radio wave with frequency \(f\) = 1.0 MHz in copper (resistivity 1.72 \(\times\) 10\(^{-8 } \Omega \bullet m\); permeability \(\mu = \mu_0\)). Since this distance is so short, electromagnetic waves of this frequency can hardly propagate at all into copper. Instead, they are reflected at the surface of the metal. This is why radio waves cannot penetrate through copper or other metals, and why radio reception is poor inside a metal structure.

A source of sinusoidal electromagnetic waves radiates uniformly in all directions. At a distance of 10.0 m from this source, the amplitude of the electric field is measured to be 3.50 N/C. What is the electric-field amplitude 20.0 cm from the source?

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