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Chapter 21: Problem 84

Typical large values for electric and magnetic fields attained in laboratories are about \(1.0 \times 10^{4} \mathrm{V} / \mathrm{m}\) and \(2.0 \mathrm{T} .(a)\) Determine the energy density for each field and compare. \((b)\) What magnitude electric field would be needed to produce the same energy density as the \(2.0-\mathrm{T}\) magnetic field?

Short Answer

Expert verified
(a) Electric: \(4.425 \times 10^{-4} \mathrm{J/m^3}\), Magnetic: \(1.59 \times 10^{6} \mathrm{J/m^3}\); Magnetic is higher. (b) For equal energy density: \(5.33 \times 10^{6} \mathrm{V/m}\).

Step by step solution

01

Understand Energy Density Formula

The energy density for an electric field is given by the formula \( u_E = \frac{1}{2}\varepsilon_0 E^2 \), where \( E \) is the electric field magnitude and \( \varepsilon_0 \) is the vacuum permittivity, \( \varepsilon_0 = 8.85 \times 10^{-12} \ \mathrm{F/m} \). The energy density for a magnetic field is given by \( u_B = \frac{B^2}{2\mu_0} \), where \( B \) is the magnetic field magnitude and \( \mu_0 \) is the permeability of free space, \( \mu_0 = 4\pi \times 10^{-7} \ \mathrm{N/A^2} \).
02

Calculate Electric Field Energy Density

Substitute the given electric field value into the energy density equation for electric fields.\[ u_E = \frac{1}{2} \times 8.85 \times 10^{-12} \times (1.0 \times 10^{4})^2 \]Calculate the result:\[ u_E = \frac{1}{2} \times 8.85 \times 10^{-12} \times 10^{8} = 4.425 \times 10^{-4} \ \mathrm{J/m^3} \]
03

Calculate Magnetic Field Energy Density

Substitute the given magnetic field value into the energy density equation for magnetic fields.\[ u_B = \frac{(2.0)^2}{2 \times 4\pi \times 10^{-7}} \]Calculate the result:\[ u_B = \frac{4.0}{2\times 4\pi \times 10^{-7}} \approx 1.59 \times 10^{6} \ \mathrm{J/m^3} \]
04

Compare Energy Densities

Compare the calculated energy densities for electric and magnetic fields.\[ u_E = 4.425 \times 10^{-4} \ \mathrm{J/m^3} \]and\[ u_B = 1.59 \times 10^{6} \ \mathrm{J/m^3} \]The magnetic field has a significantly higher energy density than the electric field.
05

Find Electric Field for Equal Energy Density

To find the electric field magnitude \( E_{new} \) that makes the energy density equal to the magnetic field's energy density \( u_B \), set \( \frac{1}{2} \varepsilon_0 E_{new}^2 = u_B \). Solve for \( E_{new} \):\[ E_{new} = \sqrt{\frac{2u_B}{\varepsilon_0}} \]Substitute \( u_B = 1.59 \times 10^{6} \ \mathrm{J/m^3} \) and \( \varepsilon_0 = 8.85 \times 10^{-12} \):\[ E_{new} = \sqrt{\frac{2 \times 1.59 \times 10^{6}}{8.85 \times 10^{-12}}} \approx 5.33 \times 10^{6} \ \mathrm{V/m} \]

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Key Concepts

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

Electric Fields
Electric fields are a fascinating and vital concept in physics. They represent the force exerted by an electric charge and are crucial for understanding various electrical phenomena. An electric field is defined by the magnitude and direction it applies around a charged particle. It is measured in volts per meter (V/m).

Electric fields have several unique properties:
  • They originate from electric charges.
  • The field lines point away from positive charges and toward negative charges.
  • The strength of the field decreases with distance from the charge that generates it.
An understanding of electric fields provides insight into how electrical components like capacitors and circuits function. Studying these fields helps in grasping the essence of how electricity operates on both a macro and micro scale.
Magnetic Fields
Magnetic fields are another fundamental force in nature, complementing electric fields. They arise from moving electric charges and are typically denoted by the symbol \(B\). These fields are measured in teslas (T) and are essential in explaining magnetic forces.

Key aspects of magnetic fields include:
  • They form closed loops traveling from the north to the south pole.
  • Moving electric charges or currents generate them.
  • They can induce electric currents through a process called electromagnetic induction.
Understanding magnetic fields is crucial for various technologies, such as electric motors, generators, and transformers, which rely on magnetic principles to function efficiently. By studying magnetic fields, students can comprehend the interconnectedness of magnetic and electric phenomena.
Vacuum Permittivity
Vacuum permittivity, represented by \(\varepsilon_0\), is a measure of how much resistance an electric field encounters in a vacuum. It is a fundamental physical constant with a value of approximately \(8.85 \times 10^{-12} \ \mathrm{F/m}\) (farads per meter).

This concept is essential because it:
  • Influences how electric charges interact in a vacuum.
  • Appears in the formula for the energy density of an electric field: \( u_E = \frac{1}{2}\varepsilon_0 E^2 \).
  • Plays a critical role in defining the speed of light, as it relates to another constant, magnetic permeability.
Having a grasp of vacuum permittivity aids in understanding electrical interactions not just in a laboratory setting but also in various fields including telecommunications and space exploration.
Permeability of Free Space
Permeability of free space, denoted by \(\mu_0\), is a measure of how well a magnetic field can permeate a vacuum. Its standard value is \(4\pi \times 10^{-7} \ \mathrm{N/A^2}\) (newtons per ampere squared).

The role and importance of \(\mu_0\) include:
  • Determining the interaction between magnetic fields and materials.
  • Facilitating the calculation of energy density in magnetic fields: \( u_B = \frac{B^2}{2\mu_0} \).
  • Being a factor in the light speed expression in a vacuum, interlinked with vacuum permittivity.
Understanding permeability of free space is critical in various applications, such as the enhancement of magnetic devices and the development of magnetic resonance technologies. It allows scientists and engineers to predict and model how magnetic fields will behave in different contexts.

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