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Chapter 7: Problem 25

It has been estimated that the magnetic field strength at the surface of a neutron star, or pulsar, may be as high as \(10^{12}\) gauss. What is the energy density in such a field? Express it, using the massenergy equivalence, in grams per \(\mathrm{cm}^{3}\).

Short Answer

Expert verified
The energy density is approximately 4.42 g/cm鲁.

Step by step solution

01

Understand the Energy Density Formula

The energy density of a magnetic field is given by the formula \( u = \frac{B^2}{8\pi} \), where \( B \) is the magnetic field strength in gauss, and \( u \) is the energy density in \( \text{erg/cm}^3 \). {In this problem, we know \( B = 10^{12} \) gauss.
02

Calculate Energy Density in erg/cm鲁

Substitute \( B = 10^{12} \) gauss into the formula: \[ u = \frac{(10^{12})^2}{8\pi} = \frac{10^{24}}{8\pi} \]Carry out the division to find \( u \) in \( \text{erg/cm}^3 \).
03

Convert Energy Density to Mass Density

Use mass-energy equivalence, \( E = mc^2 \), to convert energy density to mass density:\[ \rho = \frac{u}{c^2} \] where \( c \) is the speed of light, approximately \( 3 \times 10^{10} \text{ cm/s} \).Using the calculated \( u \), find \( \rho \) in \( \text{g/cm}^3 \).\[ \rho = \frac{10^{24}/(8\pi)}{(3 \times 10^{10})^2} = \frac{10^{24}}{8\pi \times 9 \times 10^{20}} \]
04

Simplify Final Expression

Simplify the expression:\[ \rho = \frac{10^{24}}{72\pi \times 10^{20}} \approx \frac{10^{24}}{226.2 \times 10^{20}} \]\[ \rho \approx \frac{10^{24}}{2.262 \times 10^{22}} \approx 4.42 \text{ g/cm}^3 \]

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Key Concepts

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

Energy Density
Energy density is a fundamental concept in understanding how much energy is stored in a given volume of space. In the context of magnetic fields, energy density can be thought of as how much energy is present per cubic centimeter due to the magnetic field. This is important when studying objects like neutron stars, which can have extremely powerful magnetic fields.
Formula for magnetic energy density:
  • For a magnetic field of strength \( B \) in gauss, the energy density \( u \) can be calculated using: \[ u = \frac{B^2}{8\pi} \]
  • In simple terms, this formula provides the energy stored per cubic centimeter, expressed in \( \text{erg/cm}^3 \).
Understanding the energy density of neutron stars gives insight into the immense forces at play in such extreme environments.
Mass-Energy Equivalence
The mass-energy equivalence principle is a cornerstone of Einstein's theories, encapsulated in the famous equation \( E = mc^2 \). This equation highlights the relationship between mass \( m \) and energy \( E \), showing that they are interchangeable.
  • The constant \( c \) represents the speed of light, \( 3 \times 10^{10} \text{ cm/s} \), which is extremely large, meaning even a small amount of mass can convert into a huge amount of energy.
  • In the exercise given, we use this principle to convert the energy density of a magnetic field into a mass density, expressed in grams per cubic centimeter \( \text{g/cm}^3 \).
This transformation is crucial for relating the immense energy stored in neutron stars to their gravitational impacts and dynamics.
Neutron Star
Neutron stars are among the densest objects in the universe, formed from the remnants of massive stars that have exploded in supernovae. They possess extreme properties:
  • A small size of about 10 km (6 miles) in radius but with masses up to twice that of the Sun.
  • Despite their small size, their magnetic fields can reach up to \( 10^{12} \) gauss, much stronger than anything we can produce on Earth.
  • The strong magnetic field is a result of their rapid rotation and the collapse of the star's core, concentrating their magnetic field lines.
These magnetic fields contribute to the neutron star's incredible energy density, making them fascinating objects of study for astrophysicists.

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Most popular questions from this chapter

A taut wire passes through the gap of a small magnet, where the field strength is 5000 gauss. The length of wire within the gap is \(1.8 \mathrm{~cm} .\) Calculate the amplitude of the induced alternating voltage when the wire is vibrating at its fundamental frequency of \(2000 \mathrm{~Hz}\) with an amplitude of \(0.03 \mathrm{~cm}\).

In the central region of a solenoid which is connected to a radiofrequency power source, the magnetic field oscillates at \(2.5 \times\) \(10^{6}\) cycles per sec with an amplitude of 4 gauss. What is the amplitude of the oscillating electric field, in statvolts \(/ \mathrm{cm}\), at a point \(3 \mathrm{~cm}\) from the axis? (This point lies within the region where the magnetic field is nearly uniform.)

A circular coil of wire, with \(N\) turns of radius \(a\), is located in the field of an electromagnet. The magnetic field is perpendicular to the coil, and its strength has the constant value \(B_{0}\) over that area. The coil is connected by a pair of twisted leads to an external resistance. The totai resistance of this closed circuit, including that of the coil itself, is \(R\). Suppose the electromagnet is turned off, its field dropping more or less rapidly to zero. The induced electromotive force causes current to flow around the circuit. Derive a formula for the total charge \(Q=\int I d t\) which passes through the resistor, and explain why it does not depend on the rapidity with which the field drops to zero.

In this question the term dynamo will be used for a generator which works in the following way. By some external agency-the shaft of a steam turbine, for instance- a conductor is driven through a magnetic field, inducing an electromotive force in a circuit of which that conductor is part. The source of the magnetic field is the current which is caused to flow in that circuit by that electromotive force. An electrical engineer would call it a self- excited dc generator. One of the simplest dynamos conceivable is sketched below. It has only two essential parts. One part is a solid metal disk and axle which can be driven in rotation. The other is a two-turn "coil" which is stationary but is connected by sliding contacts, or "brushes," to the axle and to the rim of the revolving disk. One of the two devices pictured is, at least potentially, a dynamo. The other is not. Which is the dynamo? Note that the answer to this question cannot depend on any convention about handedness or current directions. An intelligent extraterrestrial being inspecting the sketches could give the answer, provided only that it knows about arrows! What do you think determines the direction of the current in such a dynamo? What will determine the magnitude of the current?

What is the maximum electromotive force induced in a coil of 4000 turns, average radius \(12 \mathrm{~cm}\), rotating at 30 revolutions per sec in the earth's magnetic field where the field intensity is \(0.5\) gauss?
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