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Chapter 7: Problem 18

Two expressions for the energy (general) *** The task of Problem \(2.24\) was to demonstrate that two different expressions for the electrostatic energy, \(\int\left(\epsilon_{0} E^{2} / 2\right) d v\) and \(\int(\rho \phi / 2) d v\), are equivalent (as they must be, if they are both valid). The latter expression can quickly be converted to \(C \phi^{2} / 2\) in the case of oppositely charged conductors in a capacitor (see Exercise \(3.65\) ). The task of this problem is to demonstrate the analogous relation for the magnetic energy, that is, to show that if a circuit (of finite extent) with self-inductance \(L\) contains current \(I\), then \(\int\left(B^{2} / 2 \mu_{0}\right) d v\) equals \(L I^{2} / 2 .\) This is a bit trickier than the electrostatic case, so here are some hints: (1) a useful vector identity is \(\nabla \cdot(\mathbf{A} \times \mathbf{B})=\mathbf{B} \cdot(\nabla \times \mathbf{A})-\mathbf{A} \cdot(\nabla \times \mathbf{B}),(2)\) the vector potential and magnetic field satisfy \(\nabla \times \mathbf{A}=\mathbf{B}\), (3) \(\nabla \times \mathbf{B}=\mu_{0} \mathbf{J}\), (4) \(\Phi=\int \mathbf{A} \cdot d l\) from Eq. (7.52), and (5) \(L\) is defined by \(\Phi=L I\).

Short Answer

Expert verified
Using vector calculus identities and integral operations, it is shown that the two expressions for magnetic energy, \(\int\left(B^{2} / 2 \mu_{0}\right) dv\) and \(LI^{2} / 2\), are indeed equivalent.

Step by step solution

01

Express the Magnetic Field in Terms of the Vector Potential

The relationship between the magnetic field B and the vector potential A is given by \(\nabla \times \mathbf{A} = \mathbf{B}\). Now, substitute this expression into the formula for the magnetic energy and simplify.
02

Use the Vector Identity and Simplify

Apply the vector identity \(\nabla \cdot(\mathbf{A} \times \mathbf{B}) = \mathbf{B} \cdot(\nabla \times \mathbf{A}) - \mathbf{A} \cdot(\nabla \times \mathbf{B})\) to simplify the expression obtained from the previous step.
03

Express the Curl of B in Terms of the Current Density

Since \(\nabla \times \mathbf{B} = \mu_{0} \mathbf{J}\), substitute this relationship into the obtained expression and simplify.
04

Integrate the Magnetic Flux

By definition, magnetic flux is given by \(\Phi = \int \mathbf{A} \cdot dl\) (Eq. (7.52)). Integrate this over the entire volume.
05

Use the Definition of the Inductance and Conclude the Proof

The self-inductance L is defined by the relationship \(\Phi = LI\). Substitute this into the expression for magnetic energy derived in the previous steps, thereby getting \(LI^{2} / 2\), which proves the equivalence of the two energy formulations.

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Key Concepts

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

Self-Inductance
Self-inductance is an important electrical property that explains how a change in current influences the voltage in a circuit. It occurs in circuits that have inductors, which are often coils of wire.
  • When the current in an inductor changes, it creates a varying magnetic field.
  • This magnetic field, in turn, induces a voltage in the conductor itself 鈥� this is self-inductance.
  • The formula for self-inductance is given by the equation \( \Phi = L I \), where \( \Phi \) is the magnetic flux, \( L \) is the inductance, and \( I \) is the current.
The unit of self-inductance is the henry (H). Inductance depends on the geometry of the circuit, specifically how the wire is coiled, and the material's magnetic properties surrounding it. This property is key in the formulation of magnetic energy, as it provides a way to quantify how energy is stored in the magnetic field of a circuit.
Vector Potential
In magnetism, the vector potential \( \mathbf{A} \) is a fundamental concept that helps describe and analyze magnetic fields. Although \( \mathbf{A} \) itself does not have a direct physical meaning like the magnetic field \( \mathbf{B} \), it is a very useful tool in computations.
  • The vector potential is related to the magnetic field through the equation \( abla \times \mathbf{A} = \mathbf{B} \).
  • This means that the magnetic field is the curl of the vector potential.
Because of this relationship, vector potential is critical when evaluating problems associated with magnetic energy. By manipulating \( \mathbf{A} \), you can derive various characteristics of \( \mathbf{B} \), and therefore facilitate the calculations related to the magnetic properties of a system. Understanding vector potential can simplify the analysis of electromagnetic systems, especially when calculating magnetic flux through a given region.
Magnetic Field
The magnetic field \( \mathbf{B} \) is a vector field that describes the magnetic influence on moving electric charges, magnetic materials, and on varying electric fields. It is a crucial element in understanding magnetic energy and is directly evaluated when solving problems related to inductance.
  • The magnetic field is defined by lines of force represented by vectors, which show the direction and magnitude of the force.
  • It is commonly represented in Tesla (T) in the SI unit system.
  • The relation \( abla \times \mathbf{A} = \mathbf{B} \) links the magnetic field to the vector potential; this indicates that \( \mathbf{B} \) is calculated as the curl of the vector potential.
Understanding the characteristics and behaviors of magnetic fields allows for better analysis of how energy is stored and transferred in electronic systems. This is especially relevant in crafting efficient devices or solving complex electromagnetic problems.
Current Density
Current density \( \mathbf{J} \) is a measure of the electric current (the flow of electric charge) per unit area of cross-section. It is important in the context of magnetic energy because it helps in understanding the distribution of currents in a conductor, which in turn affects the magnetic field and energy.
  • Measured in amperes per square meter (A/m\(^2\)), current density provides a detailed picture of how electric currents flow through different materials.
  • The Amp猫re's circuital law connects current density to magnetic fields, specifically through the equation \( abla \times \mathbf{B} = \mu_{0} \mathbf{J} \), where \( \mu_{0} \) is the permeability of free space.
  • This relationship is vital for integrating magnetic fields to determine stored energy.
By analyzing current density, one can deduce factors such as energy loss due to resistance or the effectiveness of a conductor in transporting energy. Current density is a foundational concept that supports deeper investigations into electromagnetism.

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Most popular questions from this chapter

Critical frequency of a dynamo *** A dynamo like the one in Exercise \(7.47\) has a certain critical speed \(\omega_{0}\). If the disk revolves with an angular velocity less than \(\omega_{0}\), nothng happens. Only when that speed is attained is the induced \(\mathcal{E}\) arge enough to make the current large enough to make the mag- netic field large enough to induce an \(\mathcal{E}\) of that magnitude. The critial speed can depend only on the size and shape of the conductors, ic dimension expressing the size of the dynamo, such as the radius f the disk in our example. a) Show by a dimensional argument that \(\omega_{0}\) must be given by a relation of this form: \(\omega_{0}=K / \mu_{0} \sigma d^{2}\), where \(K\) is some dimensionless numerical factor that depends only on the arrangement and relative size of the various parts of the dynamo. (b) Demonstrate this result again by using physical reasoning that relates the various quantities in the problem \((R, \mathcal{E}, E, I, B,\), etc.). You can ignore all numerical factors in your calculations and absorb them into the constant \(K\). Additional comments: for a dynamo of modest size made owever, with \(d\) measured in hundreds of kilometers rather than neters, the critical speed is very much smaller. The earth's magnetic field is almost certainly produced by a nonferromagnetic dynamo involving motions in the fluid metallic core. That fluid happens to be molten iron, but it is not even slightly ferromagnetic because it is too hot. (That will be explained in Chapter 11.) We don't know how the conducting fluid moves, or what configuration of electric currents and magnetic fields its motion generates in the core. The magnetic field we observe at the earth's surface is the external field of the dynamo in the core. The direction of the earth's field a million years ago is preserved in the magnetization of rocks that solidified at that time. That magnetic record shows that the field has reversed its direction nearly 200 times in the last 100 million years. Although a reversal cannot have been instantaneous (see Exercise 7.46), it was a relatively sudden event on the geological time scale. The immense value of paleomagnetism as an indelible record of our planet's history is well explained in Chapter 18 of Press and Siever (1978).

Pulling a square frame \(*\) A square wire frame with side length \(\ell\) has total resistance \(R\). It is being pulled with speed \(v\) out of a region where there is a uniform \(\mathbf{B}\) field pointing out of the page (the shaded area in Fig. 7.28). Consider the moment when the left corner is a distance \(x\) inside the shaded area. (a) What force do you need to apply to the square so that it moves with constant speed \(v\) ? (b) Verify that the work you do from \(x=x_{0}\) (which you can assume is less than \(\ell / \sqrt{2}\) ) down to \(x=0\) equals the energy dissipated in the resistor.

Current in a bottle \(*\) An ocean current flows at a speed of 2 knots (approximately \(1 \mathrm{~m} / \mathrm{s}\) ) in a region where the vertical component of the earth's magnetic field is \(0.35\) gauss. The conductivity of seawater in that region is 4 (ohm-m) \(^{-1}\). On the assumption that there is no other horizontal component of \(\mathbf{E}\) than the motional term \(\mathbf{v} \times \mathbf{B}\), find the density \(J\) of the horizontal electric current. If you were to carry a bottle of seawater through the earth's field at this speed, would such a current be flowing in it?

Magnetic energy in the galaxy * A magnetic field exists in most of the interstellar space in our galaxy. There is evidence that its strength in most regions is between \(10^{-6}\) and \(10^{-5}\) gauss. Adopting \(3 \cdot 10^{-6}\) gauss as a typical value, find, in order of magnitude, the total energy stored in the magnetic field of the galaxy. For this purpose you may assume the galaxy is a disk roughly \(10^{21} \mathrm{~m}\) in diameter and \(10^{19} \mathrm{~m}\) thick. To see whether the magnetic energy amounts to much, on that scale, you might consider the fact that all the stars in the galaxy are radiating about \(10^{37}\) joules/second. How many years of starlight is the magnetic energy worth?

Decay time for current in the earth \(* *\) Magnetic fields inside good conductors cannot change quickly. We found that current in a simple inductive circuit decays exponentially with characteristic time \(L / R\); see Eq. (7.71). In a large conducting body such as the metallic core of the earth, the "circuit" is not easy to identify. Nevertheless, we can find the order of magnitude of the decay time, and what it depends on, by making some reasonable approximations. Consider a solid doughnut of square cross section, as shown in Fig. 7.42, made of material with conductivity \(\sigma\). A current \(I\) flows around it. Of course, \(I\) is spread out in some manner over the cross section, but we shall assume the resistance is that of a wire of area \(a^{2}\) and length \(\pi a\), that is, \(R \approx \pi / a \sigma .\) For the field \(B\) we adopt the field at the center of a ring with current \(I\) and radius \(a / 2\). For the stored energy \(U\), a reasonable estimate would be \(B^{2} / 2 \mu_{0}\) times the volume of the doughnut. Since \(d U / d t=-I^{2} R\), the decay time of the energy \(U\) will be \(\tau \approx U / I^{2} R\). Show that, except for some numerical factor depending on our various approximations, \(\tau \approx \mu_{0} a^{2} \sigma .\) The radius of the earth's core is \(3000 \mathrm{~km}\), and its conductivity is believed to be \(10^{6}(\mathrm{ohm}-\mathrm{m})^{-1}\), roughly one-tenth that of iron at room temperature. Evaluate \(\tau\) in centuries.
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