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A length of copper wire carries a current of 10 A uniformly distributed through its cross section. Calculate the energy density of (a) the magnetic field and (b) the electric field at the surface of the wire. The wire diameter is \(2.5 \mathrm{~mm},\) and its resistance per unit length is \(3.3 \Omega / \mathrm{km}\).

Short Answer

Expert verified
(a) The magnetic energy density is approximately 0.0255 J/m鲁. (b) The electric energy density is approximately 4.8 x 10鈦宦光伒 J/m鲁.

Step by step solution

01

Calculate Magnetic Field Intensity

First, calculate the intensity of the magnetic field at the surface of the wire. Use the formula for the magnetic field around a long straight wire: \( B = \frac{\mu_0 I}{2 \pi r} \), where \( \mu_0 = 4\pi \times 10^{-7} \text{ Tm/A} \) is the permeability of free space, \( I = 10 \text{ A} \) is the current, and \( r = \frac{2.5 \text{ mm}}{2} = 1.25 \text{ mm} = 1.25 \times 10^{-3} \text{ m} \) is the radius of the wire. Substituting the values, \( B = \frac{(4\pi \times 10^{-7} \text{ Tm/A})(10 \text{ A})}{2\pi (1.25 \times 10^{-3} \text{ m})} = 8 \times 10^{-4} \text{ T} \).
02

Calculate Magnetic Energy Density

The magnetic energy density \( u_B \) can be found using the formula \( u_B = \frac{B^2}{2\mu_0} \). Substitute the value of \( B \) we found: \( u_B = \frac{(8 \times 10^{-4} \text{ T})^2}{2 \times 4\pi \times 10^{-7} \text{ Tm/A}} = \frac{6.4 \times 10^{-7}}{8\pi \times 10^{-7}} \approx 0.0255 \text{ J/m}^3 \).
03

Calculate Electric Field Intensity

The electric field intensity \( E \) at the surface can be calculated from the voltage per unit length using Ohm's law, \( V = IR \), where \( I = 10 \text{ A} \) and \( R = 3.3 \times 10^{-3} \text{ \(\Omega\)/m} \). So \( V = 10 \text{ A} \times 3.3 \times 10^{-3} \text{ \(\Omega\)/m} = 0.033 \text{ V/m} \). Thus, the electric field intensity \( E \) is \( 0.033 \text{ V/m} \).
04

Calculate Electric Energy Density

The electric energy density \( u_E \) can be calculated using the formula \( u_E = \frac{1}{2} \varepsilon_0 E^2 \), where \( \varepsilon_0 = 8.85 \times 10^{-12} \text{ F/m} \) is the permittivity of free space. Substitute the value of \( E \): \( u_E = \frac{1}{2} \times 8.85 \times 10^{-12} \text{ F/m} \times (0.033 \text{ V/m})^2 = 4.8 \times 10^{-15} \text{ J/m}^3 \).

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Key Concepts

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

Magnetic Field
When a current flows through a wire, it generates a magnetic field around the wire. This magnetic field, represented by the symbol \( B \), is a vector field that surrounds the current-carrying conductor. To calculate the magnetic field at the surface of a wire, we use the formula:
  • \( B = \frac{\mu_0 I}{2 \pi r} \)
In this formula:
  • \( \mu_0 \) is the permeability of free space, approximately \( 4\pi \times 10^{-7} \text{ Tm/A} \).
  • \( I \) is the current through the wire, measured in amperes (A).
  • \( r \) is the radius of the wire, which is half of the diameter.
By substituting the known values, you can calculate the magnetic field intensity at the surface of the wire. This magnetic field influences charged particles moving within its vicinity and plays a crucial role in electromagnetic interactions.
Electric Field
The electric field, often denoted as \( E \), represents the force per unit charge exerted on electric charges within a field. It is directly related to the voltage and resistance in a conductive material using Ohm's Law.
  • Ohm's Law states: \( V = IR \)
  • \( V \) is the voltage, \( I \) is the current, and \( R \) is the resistance per unit length of the conductor.
  • The electric field can be found as \( E = \frac{V}{ ext{length}} \).
For a wire carrying a steady current, the electric field intensity helps us understand how charges move across the wire. In our calculation, we used known current and resistance values to compute the electric field at the surface. This electric field creates potential differences across the materials it influences and plays a role in everything from electrical circuits to electromagnetic waves.
Energy Density
Energy density refers to the amount of energy stored per unit volume within a field, whether magnetic or electric.

Magnetic Energy Density

The magnetic energy density, denoted \( u_B \), can be calculated using the formula:
  • \( u_B = \frac{B^2}{2\mu_0} \)
This formula shows how the square of the magnetic field intensity \( B \) contributes to the stored magnetic energy.

Electric Energy Density

Correspondingly, the electric energy density, denoted \( u_E \), is given by:
  • \( u_E = \frac{1}{2} \varepsilon_0 E^2 \)
Here, \( \varepsilon_0 \) is the permittivity of free space, which quantifies the ability to store electric field energy. Both energy densities highlight how fields store energy within a given volume, influencing how fields interact with materials and contribute to electromagnetic phenomena.
Copper Wire
Copper wire is a commonly used electrical conductor due to its excellent conductivity, flexibility, and durability. In the context of electromagnetic fields, copper wires are essential because:
  • They efficiently conduct electric current with minimal resistance.
  • The resistance per unit length is typically low, making it suitable for long-distance transmission.
  • The uniform distribution of current across the wire simplifies calculations for fields and forces.
In our exercises, we consider a copper wire with specific dimensions and resistance, which directly influences the calculated electric and magnetic fields. These properties of copper wire are vital in electrical engineering, informing designs for power lines, circuits, and electronic devices.

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Most popular questions from this chapter

Two long, parallel copper wires of diameter \(2.5 \mathrm{~mm}\) carry currents of 10 A in opposite directions. (a) Assuming that their central axes are \(20 \mathrm{~mm}\) apart, calculate the magnetic flux per meter of wire that exists in the space between those axes. (b) What percentage of this flux lies inside the wires? (c) Repeat part (a) for parallel currents.

A long cylindrical solenoid with 100 turns/cm has a radius of \(1.6 \mathrm{~cm} .\) Assume that the magnetic field it produces is parallel to its axis and is uniform in its interior. (a) What is its inductance per meter of length? (b) If the current changes at the rate of \(13 \mathrm{~A} / \mathrm{s}\) what emf is induced per meter?

A rectangular coil of \(N\) turns and of length \(a\) and width \(b\) is rotated at frequency \(f\) in a uniform magnetic field \(\vec{B}\), as indicated in Fig. \(30-40 .\) The coil is connected to co-rotating cylinders, against which metal brushes slide to make contact. (a) Show that the emf induced in the coil is given (as a function of time \(t\) ) by \(\mathscr{C}=2 \pi f N a b B \sin (2 \pi f t)=\mathscr{E}_{0} \sin (2 \pi f t)\) This is the principle of the commercial alternating-current generator. (b) What value of \(\mathrm{Nab}\) gives an emf with \(\mathscr{E}_{0}=150 \mathrm{~V}\) when the loop is rotated at \(60.0 \mathrm{rev} / \mathrm{s}\) in a uniform magnetic field of \(0.500 \mathrm{~T} ?\)

A solenoid that is \(85.0 \mathrm{~cm}\) long has a cross-sectional area of \(17.0 \mathrm{~cm}^{2}\). There are 950 turns of wire carrying a current of 6.60 A. (a) Calculate the energy density of the magnetic field inside the solenoid. (b) Find the total energy stored in the magnetic field there (neglect end effects).

The inductance of a closely wound coil is such that an emf of \(3.00 \mathrm{mV}\) is induced when the current changes at the rate of \(5.00 \mathrm{~A} / \mathrm{s}\). A steady current of 8.00 A produces a magnetic flux of \(40.0 \mu \mathrm{Wb}\) through each turn. (a) Calculate the inductance of the coil. (b) How many turns does the coil have?

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