Time-Dependent Electromagnetic Fields

Abstract

The full Maxwell system of equations for time-dependent electromagnetic fields is written. The system is compatible with both conservation of charge and conservation of energy (Poynting’s theorem).

Appendices

Questions

• 1. Consider a region of space in which the e/m field is static. By using the Maxwell equations, show that there can be no time-dependent sources (charges or currents) within that region.

• 2. Show that, in a region of space where the magnetic field is time-dependent, the electric field cannot be the grad of some scalar function.

• 3. In a region of space the “displacement current” is everywhere equal to zero. Show that there can be no time-dependent charge distribution within that region.

• 4. The potential difference between the plates of a capacitor is V. A compass has been placed between the plates. Describe the expected behavior of the compass if (a) V is constant in time; (b) V is time varying.

• 5. By using the Maxwell equations, show that the total electric charge in Nature is constant in time. [Hint: Imagine that the entire Nature is surrounded by an imaginary closed surface S. What will be the total current coming “out” of S?].

• 6. Assume that by some experiment we determine the values of the e/m field $\left(\overrightarrow{E},\overrightarrow{B}\right)$ at all points of a region of space and at all times. Is it possible to determine the e/m potentials $\varPhi\;\left(\overrightarrow{r},t\right)$ and $\overrightarrow{A}\;\left(\overrightarrow{r},t\right)$ in that region uniquely? So, can Φ and $\overrightarrow{A}$ be regarded as measurable physical quantities with absolute significance, as the case is with $\overrightarrow{E}$ and $\overrightarrow{B}$?

• 7. (a) What is the physical significance of the Poynting vector? (b) A point charge q is firmly positioned at the center of a spherical surface S. What is the total e/m energy exiting S per unit time?

Problems

• 1. Show that in the interior of a “perfect” conductor (σ ≈ ∞) the electric field is zero and the magnetic field is time-independent. What can you say about the e/m field just outside the conductor?

• Solution: By Ohm’s law, $\overrightarrow{J}=\sigma\;\overrightarrow{E}\kern0.36em \Rightarrow \kern0.36em \overrightarrow{E}=\frac{\overrightarrow{J}}{\sigma}\simeq 0$ when σ ≈ ∞.

Then, by the Faraday-Henry law, $\overrightarrow{\nabla}\times \overrightarrow{E}=-\frac{\partial \kern0.1em \overrightarrow{B}}{\partial \kern0.2em t}=0\kern0.48em \Rightarrow \kern0.48em \overrightarrow{B}=\overrightarrow{B}\;\left(\overrightarrow{r}\right)$.

We regard the surface of the conductor as the interface of the conductor and its environment. From the boundary conditions (9.26) of Sect. 9.5 we know that the component ${\overrightarrow{E}}^{\parallel }$, parallel to the interface, is continuous, having the same value on both sides of the interface. Hence, given that ${{\overrightarrow{E}}^{\parallel}}_{in}=0$, it follows that ${{\overrightarrow{E}}^{\parallel}}_{out}=0$ also. That is, the electric field just outside the perfect conductor is normal to the surface of the conductor. On the other hand, the component ${\overrightarrow{B}}^{\perp }$, normal to the interface, is also continuous at the interface. So, given that $\partial \kern0.1em {{\overrightarrow{B}}^{\perp}}_{in}/\partial \kern0.2em t=0$, it follows that $\partial \kern0.1em {{\overrightarrow{B}}^{\perp}}_{out}/\partial \kern0.2em t=0$ also. That is, just outside the perfect conductor, the component of the magnetic field normal to the surface of the conductor is time-independent.

• 2. Show that a conductor cannot sustain net charge different from zero in its interior: any excessive charge is quickly transferred to the surface of the conductor. Thus, nonzero net charge may only exist on the surface of the conductor, while the substance is electrically neutral in its interior.

• Solution: Our basic equations are the following:

$$ {\mathrm{Ohm}}^{'}\mathrm{s}\ \mathrm{law}:\operatorname{}\overrightarrow{J}=\sigma \kern0.2em \overrightarrow{E} $$

(1)

$$ {\mathrm{Gauss}}^{'}\ \mathrm{law}:\operatorname{}\overrightarrow{\nabla}\cdot \overrightarrow{E}=\frac{\rho }{\varepsilon_0} $$

(2)

$$ \mathrm{Equation}\ \mathrm{of}\ \mathrm{continuity}:\operatorname{}\overrightarrow{\nabla}\cdot \overrightarrow{J}+\frac{\partial \rho }{\partial \kern0.2em t}=0 $$

(3)

The quantity ρ is the total density of free charge in the interior of the conductor, due to both the mobile electrons and the stationary positive ions of the metal. On the other hand, the current density $\overrightarrow{J}$ is only due to the mobile electrons. We have:

$$ (3)\kern0.36em \overset{(1)}{\Rightarrow}\kern0.36em 0=\overrightarrow{\nabla}\cdot \left(\sigma \overrightarrow{E}\right)+\frac{\partial \kern0.1em \rho }{\partial \kern0.2em t}=\sigma \kern0.22em \left(\overrightarrow{\nabla}\cdot \overrightarrow{E}\right)+\frac{\partial \kern0.1em \rho }{\partial \kern0.2em t}\kern0.24em \overset{(2)}{=}\kern0.24em \frac{\sigma }{\varepsilon_0}\;\rho +\frac{\partial \kern0.1em \rho }{\partial \kern0.2em t}\kern0.36em \Rightarrow \kern0.48em \frac{\partial \kern0.1em \rho }{\partial \kern0.2em t}=-\frac{\sigma }{\varepsilon_0}\kern0.22em \rho $$

Integrating the differential equation on the right with respect to t, we find:

$\rho\;(t)=\rho\;(0)\kern0.22em {e}^{-\left(\sigma /{\varepsilon}_0\right)\kern0.2em t}=\rho\;(0)\kern0.22em {e}^{-t/\tau }$ where we have put τ = ε₀/σ

We notice that τ → 0 as σ → ∞ (in general, τ is small for a good conductor). We also notice that ρ (t) → 0 as t increases. Thus, if at some moment there is net charge different from zero in the interior of the conductor, the excess of charge quickly flows toward the surface of the conductor so that the substance stays electrically neutral in the interior.

• 3. An inertial observer Ο records an electric field $\overrightarrow{E}$ and a magnetic field $\overrightarrow{B}$ in some region of space. Determine the electric field ${\overrightarrow{E}}^{\prime }$ recorded by another inertial observer Ο′ moving with velocity $\overrightarrow{\upsilon}$ with respect to Ο. As an application, assume that the observer Ο is inside a purely magnetic field (thus, $\overrightarrow{E}$=0). Show that the observer Ο′ will additionally record an electric field (${\overrightarrow{E}}^{\prime }$≠0). Assume that υ ≪ c or υ/c ≪ 1, so that the laws of non-relativistic Mechanics apply.

• Solution: We use the following trick: We consider a charge q that is momentarily at rest relative to Ο′, thus moves with velocity $\overrightarrow{\upsilon}$ relative to Ο. According to Ο, this charge is subject to a force $\overrightarrow{F}=q\;\left[\overrightarrow{E}+\left(\overrightarrow{\upsilon}\times \overrightarrow{B}\right)\right]$ by the e/m field. According to Ο′, however, the force on the (stationary) charge is ${\overrightarrow{F}}^{\prime }=q\kern0.1em {\overrightarrow{E}}^{\prime }$, due to the electric field alone. Now, since the relative velocity υ of the two observers is small, we can make the non-relativistic approximation $\overrightarrow{F}={\overrightarrow{F}}^{\prime }$. Thus, by eliminating q we find that

$$ {\overrightarrow{E}}^{\prime }=\overrightarrow{E}+\left(\overrightarrow{\upsilon}\times \overrightarrow{B}\right) $$

(1)

We emphasize that (1) is only approximately valid, for υ ≪ c, and requires correction in the context of Relativity. Now, in the special case where $\overrightarrow{E}$=0, relation (1) yields ${\overrightarrow{E}}^{\prime }=\overrightarrow{\upsilon}\times \overrightarrow{B}$. That is, whereas observer Ο records only a magnetic field, observer Ο′ records an electric field as well. We conclude that the separation of the e/m field into an electric and a magnetic component is not absolute but depends on the state of motion of the observer.

• 4. Show that if the e/m field $\left(\overrightarrow{E},\overrightarrow{B}\right)$ is static in a region R of space, this region cannot contain time-dependent sources $\left(\rho, \overrightarrow{J}\right)$. Is the converse true?

• Solution: Static e/m field ⇒ $\partial \kern0.1em \overrightarrow{E}/\partial \kern0.2em t=0,\kern0.36em \partial \kern0.1em \overrightarrow{B}/\partial \kern0.2em t=0,\kern0.36em \forall \overrightarrow{r}\in R$. We then have:

$$ {\displaystyle \begin{array}{l}\kern3.119999em \overrightarrow{\nabla}\cdot \overrightarrow{E}=\frac{\rho }{\varepsilon_0}\kern0.36em \Rightarrow \kern0.36em \frac{\partial \rho }{\partial \kern0.2em t}={\varepsilon}_0\;\frac{\partial }{\partial \kern0.2em t}\kern0.22em \left(\overrightarrow{\nabla}\cdot \overrightarrow{E}\right)={\varepsilon}_0\kern0.1em \overrightarrow{\nabla}\cdot \frac{\partial \kern0.1em \overrightarrow{E}}{\partial \kern0.2em t}=0,\\ {}\overrightarrow{\nabla}\times \overrightarrow{B}={\mu}_0\;\overrightarrow{J}+{\varepsilon}_0\;{\mu}_0\;\frac{\partial \kern0.1em \overrightarrow{E}}{\partial \kern0.2em t}={\mu}_0\;\overrightarrow{J}\kern0.36em \Rightarrow \kern0.36em \frac{\partial \kern0.1em \overrightarrow{J}}{\partial \kern0.2em t}=\frac{1}{\mu_0}\kern0.24em \frac{\partial }{\partial \kern0.2em t}\kern0.22em \left(\overrightarrow{\nabla}\times \overrightarrow{B}\right)=\frac{1}{\mu_0}\;\overrightarrow{\nabla}\times \frac{\partial \kern0.1em \overrightarrow{B}}{\partial \kern0.2em t}=0\end{array}} $$

The converse is not valid, in general. For example, the region R may even contain no sources at all but there may be some time-dependent sources outside that region. These sources will generate a time-dependent e/m field everywhere in space, including the region R.

• 5. A charge q moves with constant velocity $\overrightarrow{\upsilon}$ relative to an inertial observer. Show that, relative to this observer, the electric and the magnetic field produced by the charge are everywhere related by

$$ \overrightarrow{B}=\frac{1}{c^2}\kern0.22em \left(\overrightarrow{\upsilon}\times \overrightarrow{E}\right) $$

(1)

where $c=1/\sqrt{\varepsilon_0\kern0.1em {\mu}_0}$ is the speed of light in vacuum. [Although (1) is of general validity, consider for simplicity that the speed υ of the charge is much lower than the speed of light: υ ≪ c. This allows us to ignore the finite time needed for the observer to receive the information that the charge passes from a certain point of space at a certain time. In reality, this information – like any e/m signal, in general – travels at a finite speed c and does not reach the observer instantly (see Chap. 10).]

• Solution: We assume that q > 0. Let Α be the location of the observer (Fig. 9.5). Since υ ≪ c, we can consider that the electric field at Α is the same as the Coulomb field created by a stationary charge q a distance r from Α:

Fig. 9.5

Electric and magnetic field produced by a uniformly moving charge q

$$ \overrightarrow{E}=\frac{1}{4\pi \kern0.1em {\varepsilon}_0}\kern0.24em \frac{q}{r^2}\kern0.24em \hat{r}\kern1.08em \left(\upsilon <<c\right) $$

(2)

(Note that this is valid for a charge moving with constant velocity. The electric field of an accelerating charge is no longer radial relative to the charge.) The magnetic field lines are circular, centered on the axis of the velocity of q, each circle lying on a plane normal to that axis. The direction of traversing the circle is determined by the right-hand rule and by assuming that the moving positive charge corresponds to an electric current flowing in the direction of motion of the charge. Under these assumptions the Biot-Savart law (Sect. 7.1) is written:

$$ \overrightarrow{B}=\frac{\mu_0}{4\pi}\kern0.24em \frac{q\;\left(\overrightarrow{\upsilon}\times \hat{r}\right)}{r^2}\kern1.08em \left(\upsilon <<c\right) $$

(3)

By combining (2) and (3) and by taking into account that ε₀  μ₀ = 1/c² (see Appendix A) it is not hard to verify (1).

• 6. Compare the magnetic interaction between two electric charges with the electric interaction between them. Show that if the speeds of the charges, relative to an inertial observer, are small compared to the speed c of light, the magnetic force between the charges is negligible compared to the electric force between them, while in the limit of very high speeds the two forces become comparable to each other.

• Solution: We consider two charges q and q′ moving with corresponding velocities $\overrightarrow{\upsilon}$ and ${\overrightarrow{\upsilon}}^{\prime }$ relative to an inertial observer. We regard q′ as the source of an e/m field and q as a test charge within this field. We are interested in the force on q due to the e/m field produced by q′. Let $\left({\overrightarrow{E}}^{\prime },{\overrightarrow{B}}^{\prime}\right)$ be the value of this field at the location of q. The electric force on q is ${\overrightarrow{F}}_e=q\kern0.2em {\overrightarrow{E}}^{\prime }$ or, in magnitude, F_e = q  E^′, while the magnetic force on q is ${\overrightarrow{F}}_m=q\;\left(\overrightarrow{\upsilon}\times {\overrightarrow{B}}^{\prime}\right)$. Now, according to Problem 5,

$$ {\overrightarrow{B}}^{\prime }=\frac{1}{c^2}\kern0.22em \left({\overrightarrow{\upsilon}}^{\prime}\times {\overrightarrow{E}}^{\prime}\right) $$

Thus, ${\overrightarrow{F}}_m=\frac{q}{c^2}\kern0.24em \left[\overrightarrow{\upsilon}\times \left({\overrightarrow{\upsilon}}^{\prime}\times {\overrightarrow{E}}^{\prime}\right)\right]\kern0.6em \Rightarrow \kern0.48em {F}_m\approx \frac{q}{c^2}\kern0.22em \upsilon \kern0.2em {\upsilon}^{\prime}\kern0.1em {E}^{\prime }=\frac{\upsilon \kern0.2em {\upsilon}^{\prime }}{c^2}\;{F}_e$, and so, $\frac{F_m}{F_e}\approx \frac{\upsilon \kern0.2em {\upsilon}^{\prime }}{c^2}$.

We notice that F_m ≪ F_e when υ ≪ c and υ′ ≪ c, while F_m ~ F_e when υ ~ c and υ′ ~ c. This means that, while in the world of low energies (or low speeds) that we experience in our everyday lives the electric interaction between charged particles seems to be stronger than their magnetic interaction, in a world of high energies the two interactions become comparable to each other. This is natural since, after all, these interactions are the two “faces” of a single electromagnetic interaction!

• 7. Consider a circuit consisting of an ideal battery (i.e., one with no internal resistance) connected to an external resistor (Fig. 9.6). Show that the emf of the circuit in the direction of the current is equal to the voltage V of the battery. Also show that the emf in this case represents the work per unit charge done by the source (battery).

Fig. 9.6

An ideal battery connected to an external resistor. The circuit is positively oriented in the direction of the current

• Solution: We recall that, in general, the emf of a circuit C at time t is equal to the integral

$$ \mathrm{\mathcal{E}}={\oint}_C\;\overrightarrow{f}\cdot \overrightarrow{dl} $$

where $\overrightarrow{f}=\overrightarrow{F}/q$ is the force per unit charge at the location of the element $\overrightarrow{d\kern0em l}$ of the circuit, at time t. In essence, we assume that in every element $\overrightarrow{d\kern0em l}$ we have placed a test charge q (this could be, e.g., a free electron in the conducting part of the circuit). The force $\overrightarrow{F}$ on each q is then measured simultaneously for all charges at time t. Since here we are dealing with a static (time-independent) situation, however, we can treat the problem somewhat differently: The measurements of the forces $\overrightarrow{F}$ on the charges q need not be made at the same instant, given that nothing changes with time, anyway. So, instead of placing several charges q around the circuit and measuring the forces $\overrightarrow{F}$ on each of them at a particular instant, we imagine a single charge q making a complete tour around the loop C. We may assume, e.g., that the charge q is one of the (conventionally positive) free electrons taking part in the constant current Ι flowing in the circuit. We then measure the force $\overrightarrow{F}$ on q at each point of C.

We thus assume that q is a positive charge moving in the direction of the current Ι. We also assume that the direction of circulation of C is the same as the direction of the current (counterclockwise in the figure). During its motion, q is subject to two forces: (a) the force ${\overrightarrow{F}}_0$ by the source (battery) that carries q from the negative pole, a, to the positive pole, b, through the source; (b) the electrostatic force ${\overrightarrow{F}}_e=q\kern0.1em \overrightarrow{E}$ due to the electrostatic field $\overrightarrow{E}$ at each point of the circuit C (both inside and outside the source). The total force on q is

$$ \overrightarrow{F}={\overrightarrow{F}}_0+{\overrightarrow{F}}_e={\overrightarrow{F}}_0+q\kern0.1em \overrightarrow{E}\kern0.48em \Rightarrow \kern0.48em \overrightarrow{f}=\frac{\overrightarrow{F}}{q}=\frac{{\overrightarrow{F}}_0}{q}+\overrightarrow{E}\equiv {\overrightarrow{f}}_0+\overrightarrow{E} $$

Then,

$$ \mathrm{\mathcal{E}}={\oint}_C\;\overrightarrow{f}\cdot \overrightarrow{d\kern0em l}={\oint}_C\;{\overrightarrow{f}}_0\cdot \overrightarrow{d\kern0em l}+{\oint}_C\;\overrightarrow{E}\cdot \overrightarrow{d\kern0em l}={\oint}_C\;{\overrightarrow{f}}_0\cdot \overrightarrow{d\kern0em l} $$

(1)

since ${\oint}_C\;\overrightarrow{E}\cdot \overrightarrow{d\kern0em l}=0$ for an electrostatic field. However, the action of the source on q is limited to the region between the poles of the battery, that is, the section of the circuit from a to b. Hence, ${\overrightarrow{f}}_0=0$ outside the source, so that (1) reduces to

$$ \mathrm{\mathcal{E}}={\int}_a^b{\overrightarrow{f}}_0\cdot \overrightarrow{d\kern0em l} $$

(2)

Now, since the current Ι is constant, the charge q moves at constant speed along the circuit. This means that the total force on q in the direction of the path C is zero. In the interior of the resistor, the electrostatic force ${\overrightarrow{F}}_e=q\kern0.1em \overrightarrow{E}$ is counterbalanced by the force on q due to the collisions of the charge with the positive ions of the metal (this resistive force does not contribute to the emf and is not counted in its evaluation). In the interior of the (ideal) battery, however, where there is no resistance, the electrostatic force ${\overrightarrow{F}}_e$ must be counterbalanced by the opposing force ${\overrightarrow{F}}_0$ exerted by the source. Thus, in the section of the circuit between a and b,

$$ \overrightarrow{F}={\overrightarrow{F}}_0+{\overrightarrow{F}}_e=0\kern0.48em \Rightarrow \kern0.48em \overrightarrow{f}=\frac{\overrightarrow{F}}{q}={\overrightarrow{f}}_0+\overrightarrow{E}=0\kern0.48em \Rightarrow \kern0.48em {\overrightarrow{f}}_0=-\overrightarrow{E} $$

Equation (2) then takes the final form [cf. Eq. (5.17)]:

$$ \mathrm{\mathcal{E}}=-{\int}_a^b\overrightarrow{E}\cdot \overrightarrow{d\kern0em l}={V}_b-{V}_a=V $$

where V_a and V_b are the electrostatic potentials at a and b, respectively. This is, of course, what every student knows from elementary e/m courses!

The work done by the source on q upon transferring the charge from a to b is

$$ W={\int}_a^b{\overrightarrow{F}}_0\cdot \overrightarrow{d\kern0em l}=q\kern0.22em {\int}_a^b{\overrightarrow{f}}_0\cdot \overrightarrow{d\kern0em l}=q\;\mathrm{\mathcal{E}} $$

[where we have used (2)]. So, the work of the source per unit charge is⁴ W/q = ℰ. This work is converted into heat in the resistor, so that the source must again supply energy in order to carry the charges once more from a to b. This is similar to the torture of Sisyphus in ancient Greek mythology!

• 8. Examine Poynting’s theorem for the case of a metal wire of total resistance R, carrying a current Ι. Thus, assume that the power spent as Joule heat within the wire is transferred from the source (battery) to the wire by means of the e/m field surrounding the circuit, and show that this power is precisely equal to dU / dt = I ²R.

• Solution: We consider a long, cylindrical wire of radius ρ, carrying a constant current Ι (Fig. 9.7). We want to evaluate the Poynting vector $\overrightarrow{N}=\overrightarrow{E}\times \overrightarrow{H}=\left(1/{\mu}_0\right)\;\overrightarrow{E}\times \overrightarrow{B}$ on the surface of the wire and thus find the power entering the wire from outside through the wire’s surface. To be specific, we consider a section of the wire of length l and of total resistance R. Let ΔV be the potential difference between the ends of this section. The electric field inside the wire is uniform and has the direction of the current Ι, while its magnitude is E = ΔV / l (cf. Sect. 6.3). Since the component of the electric field parallel to the interface of two media is continuous upon crossing the interface (Sect. 9.5), we consider that, just outside the wire, the component of the electric field parallel to the surface of the wire is also in the direction of Ι and has magnitude E = ΔV / l. On the other hand, the current Ι produces a magnetic field whose direction just outside the wire is tangent to the circular cross-section of the wire and depends on the direction of the current, according to the right-hand rule (see figure). In accordance with Problem 7.3, the magnitude of the magnetic field is B = μ₀ Ι /2πρ, where ρ is the radius of the cross-section of the wire.

Fig. 9.7

Electromagnetic field and Poynting vector on the surface of a current-carrying wire

The Poynting vector $\overrightarrow{N}=\left(1/{\mu}_0\right)\;\overrightarrow{E}\times \overrightarrow{B}$ just outside the wire is normal to the surface of the wire and is directed toward the interior of the wire. That is, $\overrightarrow{N}$ is normal to the axis of the wire and directed toward that axis. The magnitude of $\overrightarrow{N}$ is

$$ N=\frac{1}{\mu_0}\mid \overrightarrow{E}\times \overrightarrow{B}\mid =\frac{1}{\mu_0}\;E\kern0.1em B=\frac{1}{\mu_0}\;\frac{\varDelta \kern0.3em V}{l}\;\frac{\mu_0\kern0.1em I}{2\pi \kern0.1em \rho }=\frac{I\kern0.1em \varDelta \kern0.3em V}{2\pi \kern0.1em \rho \kern0.2em l} $$

(1)

Let now $\overrightarrow{da}$ be a surface element of the wire. This element is normal to the wire’s surface and is directed outward. The energy per unit time coming out of the considered wire section is equal to ${\int}_S\overrightarrow{N}\cdot \overrightarrow{da}$, where S denotes the surface of this section. Therefore, the power entering the wire section is

$$ \frac{d\kern0.2em U}{d\kern0.1em t}=-{\int}_S\overrightarrow{N}\cdot \overrightarrow{da}={\int}_SN\kern0.1em da=N\;{\int}_S da=N\kern0.2em \left(2\pi \kern0.1em \rho \kern0.2em l\right) $$

(2)

(since Ν is constant on S). Substituting (1) into (2) and taking into account Ohm’s law (ΔV=IR), we finally get:

$$ \frac{d\kern0.2em U}{d\kern0.2em t}= I\varDelta \kern0.3em V={I}^2R $$

• 9. Prove equation (9.5) of Sect. 9.2,

$$ \mathrm{\mathcal{E}}=-\frac{d{\varPhi}_m}{d\kern0.1em t}, $$

for the case of a closed plane wire moving inside a static magnetic field $\overrightarrow{B}\kern0.2em \left(\overrightarrow{r}\right)$, where ℰ is the emf (9.4) along the wire at time t and where Φ_m is the magnetic flux passing through the wire at this time.

• Solution: Assume that, at time t, the wire describes a closed curve C that is the boundary of a plane surface S (Fig. 9.8). At time t′ = t + dt, the wire (which has moved in the meanwhile) describes another curve C′ that encloses a surface S′. Let $\overrightarrow{d\kern0.1em l}$ be an element of C in the direction of circulation of the curve, and let $\overrightarrow{\upsilon}$ be the velocity of this element relative to an inertial observer (the velocity of the elements of C may vary along the curve). The direction of the surface elements $\overrightarrow{da}$ and $\overrightarrow{d{a}^{\prime }}$ is consistent with the chosen direction of $\overrightarrow{d\kern0.1em l}$, according to the right-hand rule. The element of the side (“cylindrical”) surface S″ formed by the motion of C, is equal to

Fig. 9.8

A closed plane wire moving inside a static magnetic field

$$ \overrightarrow{d{a}^{{\prime\prime} }}=\overrightarrow{d\kern0.1em l}\times \left(\overrightarrow{\upsilon}\kern0.3em d\kern0.1em t\right)=\left(\overrightarrow{d\kern0.1em l}\times \overrightarrow{\upsilon}\kern0.1em \right)\kern0.2em d\kern0.1em t $$

Since the magnetic field is static, we can view the situation in a somewhat different way: Rather than assuming that the curve C moves within the time interval dt so that its points coincide with the points of the curve C′ at time t′, we consider two constant curves C and C′ at the same instant t. In the case of a static field $\overrightarrow{B}$, the magnetic flux through C′ at time t′ = t + dt (according to our original assumption of a moving curve) is the same as the flux through this same curve at time t, given that no change of the magnetic field occurs within the time interval dt. Now, we notice that the open surfaces S₁ = S and S₂ = S′ ∪ S″ share a common boundary, namely, the curve C. Since the magnetic field is solenoidal, the same magnetic flux Φ_m passes through S₁ and S₂ at time t. That is,

$$ {\int}_{S_1}\overrightarrow{B}\cdot \overrightarrow{da_1}={\int}_{S_2}\overrightarrow{B}\cdot \overrightarrow{da_2}\kern0.48em \Rightarrow \kern0.48em {\int}_S\overrightarrow{B}\cdot \overrightarrow{da}={\int}_{S\prime}\overrightarrow{B}\cdot \overrightarrow{d{a}^{\prime }}+{\int}_{S^{{\prime\prime} }}\overrightarrow{B}\cdot \overrightarrow{d{a}^{{\prime\prime} }} $$

But, returning to our initial assumption of a moving curve, we notice that

${\int}_S\overrightarrow{B}\cdot \overrightarrow{da}={\varPhi}_m\kern0.1em (t)=$magnetic flux through the wire at time t and

${\int}_{S\prime}\overrightarrow{B}\cdot \overrightarrow{d{a}^{\prime }}={\varPhi}_m\kern0.1em \left(t+d\kern0.1em t\right)=$ magnetic flux through the wire at time t + dt

Hence,

$$ {\displaystyle \begin{array}{l}\kern3.839998em {\varPhi}_m\kern0.1em (t)={\varPhi}_m\kern0.1em \left(t+d\kern0.1em t\right)+{\int}_{S^{{\prime\prime} }}\overrightarrow{B}\cdot \overrightarrow{d{a}^{{\prime\prime} }}\kern0.48em \Rightarrow \\ {}d{\varPhi}_m={\varPhi}_m\kern0.1em \left(t+d\kern0.1em t\right)-{\varPhi}_m\kern0.1em (t)=-{\int}_{S^{{\prime\prime} }}\overrightarrow{B}\cdot \overrightarrow{d{a}^{{\prime\prime} }}=-d\kern0.1em t\kern0.22em {\oint}_C\overrightarrow{B}\cdot \left(\overrightarrow{d\kern0.1em l}\times \overrightarrow{\upsilon}\right)\kern0.48em \Rightarrow \\ {}\kern2.52em -\frac{d{\varPhi}_m}{d\kern0.1em t}={\oint}_C\overrightarrow{B}\cdot \left(\overrightarrow{d\kern0.1em l}\times \overrightarrow{\upsilon}\right)={\oint}_C\left(\overrightarrow{\upsilon}\times \overrightarrow{B}\right)\cdot \overrightarrow{d\kern0.1em l}=\mathrm{\mathcal{E}}\end{array}} $$

in accordance with Eqs. (9.4) and (9.5).

• 10. (a) A conductor of capacitance C carries a total charge Q. Show that the electrostatic energy stored in the conductor is

$$ U=\frac{Q^2}{2\kern0.2em C} $$

• (b) Calculate the energy U for the case of a spherical conductor of radius R. By using this example, verify that the energy density of the electric field is equal to

$$ {u}_e=\frac{1}{2}\;{\varepsilon}_0\;{E}^2\kern0.48em \left(E=|\overrightarrow{E}|\;\right) $$

• Solution: (a) Let us follow the charging process of the conductor from zero initial net charge to a final net charge Q. We assume that charging is achieved by transferring small amounts of charge dq from infinity. Let q be the total charge already existing on the conductor at some instant, and let V be the electrostatic potential of the conductor at that instant, where we agree that V = 0 when there isn’t any net charge on the conductor. (We recall from Sect. 5.6 that nonzero net charge can only exist on the surface of the conductor. Moreover, the space occupied by a conductor is a space of constant potential. Therefore, the potential at every point of the conductor is V.) We have [Eq. (5.29)]:

$$ C=\frac{q}{V}\kern0.48em \Rightarrow \kern0.48em V=\frac{q}{C} $$

(1)

Now, suppose we transfer an additional charge dq to the conductor from infinity (we assume that dq has the same sign as q). This requires doing some work dW in order to overcome the repulsive force on dq due to the charge q already existing on the conductor. This work results in an increase of the electrostatic energy “stored” in the conductor and is the opposite of the work dW_e done on dq by the electrostatic field (we assume that dq moves with constant velocity, i.e., without acceleration, so that the force we exert on it is always equal and opposite to the force exerted on dq by the electrostatic field):

$$ d\kern0.2em W=-\kern0.3em d\kern0.2em {W}_e=- dq\;\left({V}_{\infty }-V\right)= dq\;\left(V-{V}_{\infty}\right) $$

where use has been made of Eq. (5.26). By making the assumption that V_∞ = 0 (consistently with the fact that no work is done on dq when the conductor is uncharged, i.e., when q = 0 and V = 0) we see that the increase dU of the electrostatic energy in the conductor is

$$ dU= dW=V\ dq $$

By using (1) and by integrating, we find the final value of the energy U in the conductor when the latter carries a total charge Q:

$$ d\kern0.2em U=\frac{q}{C}\; dq\kern0.48em \Rightarrow \kern0.48em {\int}_0^Ud\kern0.2em U=\frac{1}{C}\kern0.24em {\int}_0^Qq\kern0.3em d\kern0.1em q\kern0.48em \Rightarrow \kern0.48em U=\frac{Q^2}{2\kern0.2em C} $$

• (b) For a spherical conductor (see Chap. 5, Prob. 6) we have:

$$ V=\frac{1}{4\pi \kern0.1em {\varepsilon}_0}\kern0.24em \frac{Q}{R}\kern0.6em \Rightarrow \kern0.6em C=\frac{Q}{V}=4\pi \kern0.1em {\varepsilon}_0R $$

Hence,

$$ U=\frac{Q^2}{2\kern0.2em C}=\frac{Q^2}{8\pi \kern0.1em {\varepsilon}_0R} $$

Now, the magnitude of the electric field produced by the charged conductor is

$$ {\displaystyle \begin{array}{l}E=0,\kern0.48em r<R\\ {}\kern0.68em =\frac{1}{4\pi \kern0.1em {\varepsilon}_0}\kern0.24em \frac{Q}{r^2}\kern0.36em ,\kern0.48em r\ge R\end{array}} $$

We define the quantity ${u}_e=\frac{1}{2}\;{\varepsilon}_0\;{E}^2$, so that

$$ {\displaystyle \begin{array}{l}{u}_e=0,\kern0.48em r<R\\ {}\kern0.8em =\frac{Q^2}{32\kern0.2em {\pi}^2{\varepsilon}_0\kern0.1em {r}^4}\kern0.36em ,\kern0.48em r\ge R\end{array}} $$

We want to evaluate the integral ∫u_e d v over the entire space. It suffices, of course, to evaluate it at the exterior of the sphere (r ≥ R) since, inside the sphere, u_e = 0. The volume element is dv = 4πR²dr. Hence,

$$ \int {u}_e\;d\kern0.1em v={\int}_R^{\infty}\frac{Q^2}{32\kern0.2em {\pi}^2{\varepsilon}_0\kern0.1em {r}^4}\kern0.24em \left(4\pi {r}^2\kern0.1em d\kern0.1em r\right)=\frac{Q^2}{8\pi \kern0.1em {\varepsilon}_0}\kern0.24em {\int}_R^{\infty}\frac{d\kern0.1em r}{r^2}=\frac{Q^2}{8\pi \kern0.1em {\varepsilon}_0R}=U $$

The fact that the integral ∫u_e d v over all space equals the total electrostatic energy U suggests that the function ${u}_e=\frac{1}{2}\;{\varepsilon}_0\;{E}^2$ represents some sort of energy density. We can consider that the energy U of the conductor (equal to the work done in order to charge it) is stored in the electric field surrounding the conductor. According to this interpretation, the function u_e represents the energy density of the electric field.

• 11. A circuit having inductance L carries a current Ι(t). Show that the magnetic energy stored in the circuit at time t is given by

$$ {U}_m=\frac{1}{2}\;L\;{I}^2 $$

• Solution: Let R be the total resistance of the circuit and let V be the voltage of an ideal battery connected to the circuit and supplying the necessary power for the generation of the current I (Fig. 9.9). The moment we connect the battery to the circuit we apply to the latter an emf V, which tends to generate a current. The circuit then reacts, producing a self-induced emf V_L that tends to oppose the increase in the current. As soon as the current stabilizes to its final value V/R, however, V_L vanishes. At an arbitrary time t, the total emf of the circuit in the direction of the current Ι is the sum of the emf V, due to the battery alone (Prob. 7), and the emf V_L given by the Faraday-Henry law [Eq. (9.14)]:

Fig. 9.9

A circuit having total resistance R and inductance L

$$ \mathrm{\mathcal{E}}=V+{V}_L=V-L\kern0.22em \frac{d\kern0.1em I}{d\kern0.1em t} $$

By Ohm’s law (see [7]),

$$ \mathrm{\mathcal{E}}= IR\Rightarrow V-L\kern0.22em \frac{d\kern0.1em I}{d\kern0.1em t}=I\kern0.1em R\Rightarrow V=I\kern0.1em R+L\kern0.22em \frac{d\kern0.1em I}{d\kern0.1em t} $$

The total power supplied to the circuit by the battery at time t, when the current is Ι(t), is equal to

$$ P=I\kern0.2em V={I}^2\kern0.1em R+L\kern0.2em I\kern0.22em \frac{d\kern0.1em I}{d\kern0.1em t} $$

The term I ²R represents the power irreversibly lost as Joule heat in the resistor. This power must necessarily be supplied back by the battery in order to maintain the current against dissipative losses in the resistor. On the other hand, the term LIdI/dt represents the energy per unit time required to build up the current against the self-induced back emf V_L . This energy is retrievable and is given back to the source when the current decreases. It may be interpreted as the rate of change of the magnetic energy stored in the circuit or, alternatively, as energy per unit time required to establish the magnetic field surrounding the circuit. We thus have:

$$ {P}_m=\frac{d\kern0.2em {U}_m}{d\kern0.1em t}=L\kern0.2em I\kern0.22em \frac{d\kern0.1em I}{d\kern0.1em t}\ \mathrm{so}\ \mathrm{that}\ d\kern0.2em {U}_m=L\kern0.2em I\kern0.22em d\kern0.1em I\kern0.48em \Rightarrow \kern0.48em {\int}_0^{U_m}d\kern0.2em {U}_m=L\kern0.24em {\int}_0^II\kern0.22em d\kern0.1em I\kern0.48em \Rightarrow $$

$$ {U}_m=\frac{1}{2}\;L\;{I}^2 $$

• 12. A long solenoid of length l consists of Ν turns, each of area S. As can be proven⁵, the magnetic field in the interior of the solenoid is almost uniform and directed parallel to the axis of the solenoid, while its magnitude is

$B={\mu}_0\;\frac{N}{l}\;I={\mu}_0\kern0.2em n\;I$ where $n=\frac{N}{l}=$ number of turns per unit length

In the exterior of the solenoid the magnetic field is zero. (a) Find the inductance L of the solenoid. (b) Verify that the magnetic-energy density in the interior of the solenoid is ${u}_m=\frac{1}{2{\mu}_0}\;{B}^2$.

• Solution: (a) The magnetic flux through each turn is

$$ {\int}_S\overrightarrow{B}\cdot \overrightarrow{da}={\int}_SB\; da=B\;{\int}_S da=B\kern0.1em S $$

while the total flux through the Ν turns is Φ_m = NBS. The emf in the circuit due to the solenoid alone is, by the Faraday-Henry law,

$$ {\varepsilon}_m=-\frac{d\kern0.1em {\varPhi}_m}{d\kern0.1em t}=-N\kern0.1em S\kern0.22em \frac{d\kern0.1em B}{d\kern0.1em t}=-n\kern0.2em l\kern0.1em S\;\left({\mu}_0\kern0.1em n\;\frac{d\kern0.1em I}{d\kern0.1em t}\right)=-{\mu}_0\kern0.1em {n}^2\left(S\kern0.1em l\right)\;\frac{d\kern0.1em I}{d\kern0.1em t}=-{\mu}_0\kern0.1em {n}^2\kern0.2em V\frac{d\kern0.1em I}{d\kern0.1em t} $$

where V= Sl is the volume occupied by the solenoid. But, in general, ${\varepsilon}_m=-L\;\frac{d\kern0.1em I}{d\kern0.1em t}$ [Eq. (9.14)]. Hence, L = μ₀ n² (S l) = μ₀ n²  V.

• (b) According to Problem 11, the magnetic energy “stored” in the interior of the solenoid is

$$ {U}_m=\frac{1}{2}\;L\kern0.2em {I}^2=\frac{1}{2}\kern0.22em \left({\mu}_0\kern0.1em {n}^2\kern0.2em V\right)\;{\left(\frac{B}{\mu_0\kern0.1em n}\right)}^2=\frac{1}{2{\mu}_0}\;{B}^2\kern0.1em V $$

The energy density in the interior is, therefore, ${u}_m=\frac{U_m}{V}=\frac{1}{2{\mu}_0}\;{B}^2$.

(In the exterior we have B = 0 and u_m = 0.)

• 13. In Sect. 9.6 we showed that the Maxwell equations in differential form yield Eq. (9.31), which is the mathematical expression for conservation of charge. Show that the same equation can be obtained by using the integral form of the Maxwell equations [in particular, Eqs. (5.5) and (9.16)].

• Solution: Relation (5.5) (Gauss’ law), combined with (5.7), gives

$$ {\oint}_S\overrightarrow{E}\cdot \overrightarrow{da}=\frac{1}{\varepsilon_0}\kern0.24em {\int}_V\kern0.1em \rho \kern0.2em dv $$

(1)

while (9.16) (Ampère-Maxwell law), combined with (9.15), gives

$$ {\oint}_C\overrightarrow{B}\cdot \overrightarrow{dl}={\mu}_0\;{\int}_S\overrightarrow{J}\cdot \overrightarrow{da}+{\varepsilon}_0\kern0.1em {\mu}_0\;\frac{d}{dt}\kern0.24em {\int}_S\overrightarrow{E}\cdot \overrightarrow{da} $$

(2)

Note carefully that S in (1) is a closed surface, whereas in (2) the surface S is open (see Fig. 9.10). We may, however, close the latter surface by letting the closed curve C (which constitutes the boundary of S) shrink to a point, taking care that the surface element $\overrightarrow{da}$ point toward the exterior of the ensuing closed surface S (Fig. 9.11).

Fig. 9.10

An open surface S bordered by a closed curve C

Fig. 9.11

The surface S of Fig. 9.10 becomes closed when its boundary C shrinks to a point

Now, as C shrinks to a point, the closed line integral on the left-hand side of (2) vanishes, so that this relation takes on the new form (after eliminating μ₀):

$$ {\oint}_S\overrightarrow{J}\cdot \overrightarrow{da}+{\varepsilon}_0\kern0.22em \frac{d}{dt}\kern0.24em {\oint}_S\overrightarrow{E}\cdot \overrightarrow{da}\kern0.34em =0 $$

(3)

By assuming that the closed surfaces S in (1) and (3) coincide, and by substituting (1) into (3), we find:

$$ \frac{d}{dt}\kern0.24em {\int}_V\kern0.1em \rho \kern0.2em dv=-{\oint}_S\overrightarrow{J}\cdot \overrightarrow{da}, $$

which is precisely Eq. (9.31).