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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).

Short Answer

Expert verified
(a) The energy density is approximately 34.5 J/m³. (b) The total energy stored is approximately 0.0498 J.

Step by step solution

01

Convert Unit Measurements

First, convert the given measurements into standard SI units. The length of the solenoid is initially given in centimeters, which we convert to meters: \(85.0 \text{ cm} = 0.85 \text{ m}\). Similarly, convert the cross-sectional area from \(17.0 \text{ cm}^2 \) to square meters: \(17.0 \text{ cm}^2 = 1.7 \times 10^{-3} \text{ m}^2\).
02

Calculate the Magnetic Field in the Solenoid

The magnetic field \( B \) inside a solenoid can be calculated using the formula \( B = \mu_0 \cdot n \cdot I \), where \( \mu_0 \) (the permeability of free space) equals \(4\pi \times 10^{-7} \text{ T} \cdot \text{m/A} \), \( n \) is the number of turns per unit length, and \( I \) is the current. Calculate \( n = \frac{950}{0.85} \approx 1117.65 \text{ turns/m}\), then \( B = 4\pi \times 10^{-7} \cdot 1117.65 \cdot 6.60 \approx 0.0093 \text{ T}\).
03

Calculate the Energy Density of the Magnetic Field

The energy density \( u \) of the magnetic field is given by the formula \( u = \frac{1}{2} \cdot \frac{B^2}{\mu_0} \). Substituting the values, \( u = \frac{0.0093^2}{2 \times 4\pi \times 10^{-7}} \approx 34.5 \text{ J/m}^3\).
04

Calculate the Volume of the Solenoid

The volume \( V \) of the solenoid is calculated as \( V = \text{cross-sectional area} \times \text{length} = 1.7 \times 10^{-3} \text{ m}^2 \times 0.85 \text{ m} = 1.445 \times 10^{-3} \text{ m}^3\).
05

Calculate the Total Energy Stored

The total energy \( U \) stored in the magnetic field is \( U = \text{energy density} \times \text{volume} = 34.5 \text{ J/m}^3 \times 1.445 \times 10^{-3} \text{ m}^3 \approx 0.0498 \text{ J}\).

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Key Concepts

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

solenoid
A solenoid is a type of electromagnetic device that generates a magnetic field when electric current passes through it. It is essentially a coil of wire, often wrapped tightly around a cylindrical core. Typically, this configuration results in a uniform magnetic field inside the coil, especially when the length of the solenoid is much larger than its diameter. This uniformity of the magnetic field is crucial in many practical applications, such as electromagnets, inductors, and solenoids used in car engines and electrical circuits.

Understanding the structure of a solenoid helps us see why they are efficient at converting electrical energy into magnetic fields. Their design, featuring multiple loops of wire, ensures that the magnetic lines of force are concentrated in the interior of the coil. This concentration is pivotal for effectiveness in controlling magnetic fields.
energy density
Energy density in the context of a magnetic field refers to the amount of energy stored in a given volume of space due to the magnetic field. To compute the energy density, we use the formula: \[ u = \frac{1}{2} \cdot \frac{B^2}{\mu_0} \] where \( B \) is the magnetic field strength, and \( \mu_0 \) is the permeability of free space. This formula shows us how stronger magnetic fields and lower permeability settings increase the energy density.

Energy density is an important concept because it tells us how much magnetic energy is contained in a space, and it relates to how much potential work the magnetic field can perform. It's like a measure of the magnetic field's potential to exert forces or induce currents. This concept is crucial in designing devices such as transformers and inductors, where energy efficiency and storage are key requirements.
permeability of free space
The permeability of free space, denoted as \( \mu_0 \), is a fundamental physical constant essential in the study of electromagnetism. Its value is approximately \(4\pi \times 10^{-7} \, \text{T} \cdot \text{m/A} \). This constant describes how a magnetic field propagates through a vacuum, which is considered 'free space'.

In practical applications, \( \mu_0 \) is used to determine how a magnetic field behaves in various environments. It forms a baseline measure and allows us to calculate the magnetic field produced by devices such as solenoids, coils, and even permanent magnets.

Understanding \( \mu_0 \) helps identify how different materials can alter or enhance the properties of a magnetic field. When materials with greater permeability than free space are used (such as iron), they increase the efficiency of a magnetic circuit, enabling innovations in technology, like MRI machines and high-efficiency transformers.
turns per unit length
Turns per unit length, often denoted as \( n \), is a key parameter in the functioning of a solenoid. It refers to the number of wire loops or turns packed into a specific length of the solenoid. This measure is critical because it directly influences the strength of the magnetic field generated by the solenoid.

The formula to determine the turns per unit length is:\[ n = \frac{N}{L} \]where \( N \) is the total number of turns, and \( L \) is the solenoid's length. The greater the \( n \), the stronger and more concentrated the magnetic field within the solenoid.

This parameter is often adjusted in the design of electromagnets to meet specific requirements. For instance, dense coil windings will produce a stronger magnetic field, beneficial in applications requiring intense magnetic forces, like electric motors and magnetic locks. It highlights the significant role of engineering decisions in optimizing electromagnetic devices for their designated functions.

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Most popular questions from this chapter

At a certain place, Earth's magnetic field has magnitude \(B=0.590\) gauss and is inclined downward at an angle of \(70.0^{\circ}\) to the horizontal. A flat horizontal circular coil of wire with a radius of \(10.0 \mathrm{~cm}\) has 1000 turns and a total resistance of \(85.0 \Omega\). It is connected in series to a meter with \(140 \Omega\) resistance. The coil is flipped through a half- revolution about a diameter, so that it is again horizontal. How much charge flows through the meter during the flip?

A certain elastic conducting material is stretched into a circular loop of \(12.0 \mathrm{~cm}\) radius. It is placed with its plane perpendicular to a uniform \(0.800 \mathrm{~T}\) magnetic field. When released, the radius of the loop starts to shrink at an instantaneous rate of \(75.0 \mathrm{~cm} / \mathrm{s}\). What emf is induced in the loop at that instant?

At time \(t=0\), a \(45 \mathrm{~V}\) potential difference is suddenly applied to the leads of a coil with inductance \(L=50 \mathrm{mH}\) and resistance \(R=180 \Omega\). At what rate is the current through the coil increasing at \(t=1.2 \mathrm{~ms} ?\)

Two long, parallel copper wires of diameter \(2.5 \mathrm{~mm}\) carry currents of \(10 \mathrm{~A}\) 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 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?

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