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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) Magnetic field energy density: 0.254 J/m鲁; (b) Electric field energy density: 2.00 脳 10鈦宦光伆 J/m鲁.

Step by step solution

01

Calculate the Magnetic Field at the Wire's Surface

The magnetic field at the surface of a long straight wire carrying a current \(I\) is given by the formula: \(B = \frac{\mu_0 I}{2 \pi r}\), where \(\mu_0 = 4\pi \times 10^{-7} \, \text{T}\cdot\text{m/A}\) is the permeability of free space, and \(r\) is the radius of the wire. First, we convert the wire diameter from millimeters to meters: \(d = 2.5 \, \text{mm} = 0.0025 \, \text{m}\), thus the radius \(r = \frac{d}{2} = 0.00125 \, \text{m}\). Substitute the values: \(B = \frac{4\pi \times 10^{-7} \times 10}{2 \pi \times 0.00125} \approx 8 \times 10^{-4} \, \text{T}\).
02

Calculate the Energy Density of the Magnetic Field

The energy density \(u_B\) of a magnetic field is given by the formula: \(u_B = \frac{B^2}{2\mu_0}\). Using the magnetic field found in Step 1, \(B = 8 \times 10^{-4} \, \text{T}\), substitute into the formula: \(u_B = \frac{(8 \times 10^{-4})^2}{2 \times 4\pi \times 10^{-7}} \approx 0.254 \, \text{J/m}^3\).
03

Calculate the Electric Field at the Surface of the Wire

The electric field \(E\) in a wire can be calculated using Ohm鈥檚 Law in terms of the electric field as \(E = \rho J\), where \(\rho\) is the resistivity and \(J\) is the current density. First, we find the resistivity \(\rho\) from \(R/L = 3.3 \, \Omega/\text{km} = 3.3 \times 10^{-3} \, \Omega/\text{m}\). \(A\) is the cross-sectional area of the wire: \(A = \pi r^2 = \pi (0.00125)^2 \, \text{m}^2 \approx 4.91 \times 10^{-6} \, \text{m}^2\). Current density \(J = \frac{I}{A} = \frac{10}{4.91 \times 10^{-6}} \, \text{A/m}^2\). Substituting values gives \(E = 3.3 \times 10^{-3} \cdot \frac{10}{4.91 \times 10^{-6}} \approx 6.72 \, \text{V/m}\).
04

Calculate the Energy Density of the Electric Field

The energy density \(u_E\) of an electric field is given by: \(u_E = \frac{1}{2} \varepsilon_0 E^2\), where \(\varepsilon_0 = 8.85 \times 10^{-12} \, \text{F/m}\) is the permittivity of free space. Substitute \(E = 6.72 \, \text{V/m}\) from Step 3: \(u_E = \frac{1}{2} \times 8.85 \times 10^{-12} \times (6.72)^2 \approx 2.00 \times 10^{-10} \, \text{J/m}^3\).

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Key Concepts

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

Magnetic Field Energy Density
Understanding the concept of magnetic field energy density is crucial for analyzing electromagnetic fields. In this specific exercise, you might initially wonder how a current carrying wire can possess energy. The answer lies in the magnetic field generated around it. When a current flows through a long straight wire, it produces a circular magnetic field around itself. The energy associated with this magnetic field, per unit volume, is known as the magnetic field energy density.

To calculate this, we refer to the formula:
  • \( u_B = \frac{B^2}{2\mu_0} \)
Here, \(B\) stands for the magnetic field, and \(\mu_0\) is the permeability of free space. The formula essentially tells us how much energy is stored in the magnetic field per volume unit. In this case, the magnetic field at the wire's surface was found to be approximately \(8 \times 10^{-4} \text{ T}\), and by using the formula, the magnetic field energy density calculated was \(0.254 \text{ J/m}^3\). This gives insight into how energy is distributed in the space around a conducting wire.
Electric Field Energy Density
Much like magnetic fields have energy associated with them, so do electric fields. This energy per unit volume in an electric field is termed as electric field energy density. In the exercise, we're interested in calculating this around the surface of a copper wire that carries a uniform current.

The electric field energy density can be calculated using:
  • \( u_E = \frac{1}{2} \varepsilon_0 E^2 \)
Where \(E\) is the electric field strength, and \(\varepsilon_0\) represents the permittivity of free space, a constant that characterizes the ability of a field to permeate through vacuum. For the given problem, the electric field was evaluated to be \(6.72 \text{ V/m}\), leading to an energy density of \(2.00 \times 10^{-10} \text{ J/m}^3\). This small value suggests that electric fields, while crucial, store less energy compared to magnetic fields per volume in our setup.
Current Density
Current density is a measure of how an electric current is distributed across a cross-sectional area. It's a foundational concept that helps us understand how current flows through materials, and it's vital for calculating properties like electric fields.

The current density \(J\) is given by the formula:
  • \( J = \frac{I}{A} \)
Where \(I\) is the current, and \(A\) is the cross-sectional area. In the copper wire scenario, the current magnitude was \(10\text{ A}\), and with a cross-sectional area computed to be about \(4.91 \times 10^{-6} \text{ m}^2\), the derived current density was found to be quite high. Such calculations are pivotal for designing wiring materials capable of carrying high currents without overheating.
Ohm鈥檚 Law
Ohm鈥檚 Law is a fundamental principle in the study of electrical circuits and electromagnetic fields. It relates the voltage across a conductor to the current through it, using the relation:
  • \( V = IR \)
But it can also be expressed in terms of electric field \(E\) and current density \(J\), particularly in resistive materials, as \(E = \rho J\). Here, \(\rho\) is the resistivity of the material, a measure of how much it resists the flow of current.

In this exercise, understanding Ohm鈥檚 Law in terms of \(E\) and \(J\) allowed for the calculation of the electric field using the resistivity and the current density in the wire. Such relationships simplify the analysis of circuits massively, showing their indispensability in electrical engineering and physics.
Resistivity Calculation
Resistivity is a key concept when dealing with electrical properties of materials. It's basically an intrinsic property that quantifies how strongly a given material opposes the flow of electric current. Its role is crucial when computing electric field and related characteristics.

The exercise gives the resistivity per unit length of the copper wire as \(3.3 \Omega/\text{km}\), which can be converted to the SI unit \(\Omega/\text{m}\) for easier computation. Given the wire's dimensions, resistivity plays a determining role in identifying how thick or long the wire needs to be to support desired electric currents effectively. Through resistivity, engineers decide on material suitability for various electrical applications. This characteristic fundamentally determines how conductive or resistive a material will be, significantly influencing practical applications in circuit design and hardware engineering.

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Most popular questions from this chapter

At \(t=0\), a battery is connected to a series arrangement of a resistor and an inductor. If the inductive time constant is \(37.0\) \(\mathrm{ms}\), at what time is the rate at which energy is dissipated in the resistor equal to the rate at which energy is stored in the inductor's magnetic field?

A metal rod is forced to move with constant velocity \(\vec{v}\) along two parallel metal rails, connected with a strip of metal at one end. A magnetic field of magnitude \(B=0.350 \mathrm{~T}\) points out of the page. (a) If the rails are separated by \(L=25.0 \mathrm{~cm}\) and the speed of the rod is \(55.0 \mathrm{~cm} / \mathrm{s}\), what emf is generated? (b) If the rod has a resistance of \(18.0 \Omega\) and the rails and connector have negligible resistance, what is the current in the rod? (c) At what rate is energy being transferred to thermal energy?

The flux linkage through a certain coil of \(0.75 \Omega\) resistance would be \(26 \mathrm{mWb}\) if there were a current of \(5.5 \mathrm{~A}\) in it. (a) Calculate the inductance of the coil. (b) If a \(6.0 \mathrm{~V}\) ideal battery were suddenly connected across the coil, how long would it take for the current to rise from 0 to \(2.5 \mathrm{~A}\) ?

A square loop of wire is held in a uniform \(0.24 \mathrm{~T}\) magnetic field directed perpendicular to the plane of the loop. The length of each side of the square is decreasing at a constant rate of \(5.0 \mathrm{~cm} / \mathrm{s}\). What emf is induced in the loop when the length is \(12 \mathrm{~cm}\) ?

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