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Chapter 32: Problem 21

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

Short Answer

Expert verified
The maximum electric field strength \(E_{\max}\) is 24.1 N/C, and the maximum magnetic field strength \(B_{\max}\) is \(8.03 脳 10^{-8} T\).

Step by step solution

01

Calculate the intensity of light

The intensity \(I\) of light at a distance \(r\) from a point source radiating uniformly in all directions is given by the equation \(I = \frac{P}{4蟺r^2}\) where \(P\) is the power of the source. We are given that \(P = 60.0 W\) and \(r = 5.00 m\). Substituting these values in, we get:\(I = \frac{60.0 W}{4蟺(5.00 m)^2} = 0.955 W/m^2.\)
02

Find the magnetic field strength

The amplitude of the magnetic field \(B_{\max}\) can be found using the equation \(B_{\max} = \sqrt{\frac{2I}{\mu_0 c}}\) where \(\mu_0\) is the permeability of free space and \(c\) is the speed of light. Substituting \(I = 0.955 W/m^2\), \(\mu_0 = 4蟺 脳 10^{-7} Tm/A\), and \(c = 3 脳 10^8 m/s\), we get:\(B_{\max} = \sqrt{\frac{2(0.955 W/m^2)}{4蟺 脳 10^{-7} Tm/A 脳 3 脳 10^8 m/s}} = 8.03 脳 10^{-8} T\)
03

Find the electric field strength

The amplitude of the electric field \(E_{\max}\) can be found using the equation \(E_{\max} = cB_{\max}\) where \(c\) is the speed of light. Substituting \(c = 3 脳 10^8 m/s\) and \(B_{\max} = 8.03 脳 10^{-8} T\), we get:\(E_{\max} = 3 脳 10^8 m/s 脳 8.03 脳 10^{-8} T = 24.1 N/C\).

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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 is a fundamental concept in optics and refers to the amount of energy that a light source emits per unit area. When we say a light source is intense, we mean it emits a lot of energy across a surface area in a short period of time.
For point sources, like light bulbs or stars, the intensity decreases as you move further away from the source. This decrease is because the light spreads out over a larger area. A good way to remember this is to imagine throwing paint evenly on a canvas; the further you go from the source, the thinner the paint layer becomes.To calculate intensity, you can use the formula:
  • \( I = \frac{P}{4\pi r^2} \)
Where
  • \( I \) is the intensity
  • \( P \) is the power or the total amount of energy per second that the source emits in watts (W)
  • \( r \) is the distance from the source in meters (m)
  • \( 4\pi r^2 \) represents the surface area of a sphere, as the light spreads out in all directions
By understanding and using this formula, you can determine how intense the light is at any given distance from a source.
Magnetic Field Strength
Magnetic field strength in light is linked to how light waves interact. Every light wave has two components, electric and magnetic, which oscillate perpendicular to each other. The strength of the magnetic field component determines how strongly the magnetic aspects of the wave interact with its surroundings. To find the magnetic field strength, we can use the equation:
  • \( B_{\max} = \sqrt{\frac{2I}{\mu_0 c}} \)
Where:
  • \( B_{\max} \) is the maximum magnetic field strength
  • \( I \) is the light's intensity
  • \( \mu_0 \) is the permeability of free space (approximately \(4\pi \times 10^{-7} \) Tm/A)
  • \( c \) is the speed of light in a vacuum (approximately \(3 \times 10^8 \) m/s)
The permeability of free space is a measure of how much resistance the vacuum has to the formation of a magnetic field. So, when calculating magnetic field strength, this value is crucial. Light's magnetic component may be weaker compared to its electric counterpart (since values of \( B_{\max} \) are often small), but it's pivotal in electromagnetic wave dynamics.
Electric Field Strength
The electric field strength of monochromatic light is another key aspect of light waves. Just like magnetic fields, all light waves have an electric component. The electric field strength refers to how strongly this component can affect other charges in its path. The formula to find the electric field strength when you know the magnetic field strength is:
  • \( E_{\max} = cB_{\max} \)
Where:
  • \( E_{\max} \) is the maximum electric field strength
  • \( c \) is the speed of light \(3 \times 10^8\) m/s
  • \( B_{\max} \) is the magnetic field strength
Electric fields are vectors, meaning they have both magnitude and direction. They "push" or "pull" charges within the field. The equation above shows that electric and magnetic fields are intertwined; they're just two sides of the same coin in an electromagnetic wave.By understanding the electric field, we gain insight into how light interacts with other charges, influencing things like voltage in a circuit or the polarization of light.

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Most popular questions from this chapter

A sinusoidal electromagnetic wave is propagating in vacuum in the \(+z\) -direction. If at a particular instant and at a certain point in space the electric field is in the \(+x\) -direction and has magnitude \(4.00 \mathrm{~V} / \mathrm{m}\) what are the magnitude and direction of the magnetic field of the wave at this same point in space and instant in time?

An electromagnetic wave of wavelength \(435 \mathrm{nm}\) is traveling in vacuum in the \(-z\) -direction. The electric field has amplitude \(2.70 \times 10^{-3} \mathrm{~V} / \mathrm{m}\) and is parallel to the \(x\) -axis. What are (a) the frequency and (b) the magnetic-field amplitude? (c) Write the vector equations for \(\overrightarrow{\boldsymbol{E}}(z, t)\) and \(\overrightarrow{\boldsymbol{B}}(z, t)\)

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 \(590 \mathrm{kHz} ?\)

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 \(\vec{E}(x, t)=E_{y}(x, t) \hat{\jmath}\) propagating in the \(+x\) -direction within a conductor is $$ \frac{\partial^{2} E_{y}(x, t)}{\partial x^{2}}=\frac{\mu}{\rho} \frac{\partial E_{y}(x, t)}{\partial t} $$ where \(\mu\) is the permeability of the conductor and \(\rho\) is its resistivity. (a) \(\mathrm{A}\) solution to this wave equation is \(E_{y}(x, t)=E_{\max } e^{-k_{\mathrm{C}} x} \cos \left(k_{\mathrm{C}} x-\omega t\right)\) where \(k_{\mathrm{C}}=\sqrt{\omega \mu / 2 \rho}\). 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 \(i^{2} R\) 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 \(1 / k_{\mathrm{C}}=\sqrt{2 \rho / \omega \mu},\) and calculate this distance for a radio wave with frequency \(f=1.0 \mathrm{MHz}\) in copper (resistivity \(1.72 \times 10^{-8} \Omega \cdot \mathrm{m} ;\) 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 sinusoidal electromagnetic wave from a radio station passes perpendicularly through an open window that has area \(0.500 \mathrm{~m}^{2}\). At the window, the electric field of the wave has rms value \(0.0400 \mathrm{~V} / \mathrm{m}\). How much energy does this wave carry through the window during a 30.0 s commercial?
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