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Introduction

10.0.1*In Demonstration 10.0.1, the circuit formed by the pair of resistors is replaced by the one shown in Fig. 10.0.1, composed of four resistors of equal resistance R. The voltmeter might be the oscilloscope shown in Fig. 10.0.1. The "grounded" node at (4) is connected to the negative terminal of the voltmeter.
figure GIF #1
Figure P10.0.1

(a) Show that the voltage measured with the positive lead connected at (1), so that the voltmeter is across one of the resistors, is v = (d /dt)/4.
(b) Show that if the positive voltmeter lead is connected to (2), then to (3), and finally to (4) (so that the lead is wrapped around the core once and connected to the same grounded node as the negative voltmeter lead), the voltages are, respectively, twice, three times, and four times this value. Show that this last result is as would be expected for a transformer with a one-turn secondary.

10.0.2Plane parallel perfectly conducting plates are shorted to form the one-turn inductor shown in Fig. 10.0.2. The current source is distributed so that it supplies i amps over the width d.
floating figure GIF #30
Figure P10.0.2
(a) Given that d and l are much greater than the spacing s, determine the voltage measured across the terminals of the current source by the voltmeter v2.
(b) What is the voltage measured by the voltmeter v1 connected as shown in the figure across these same terminals?
 

Magnetoquasistatic Electric Fields in Systems of Perfect Conductors

10.1.1*In Prob. 8.4.1, the magnetic field of a dipole surrounded by a perfectly conducting spherical shell is found. Show that
equation GIF #10.146
in the region between the dipole and the shell.

10.1.2The one-turn inductor of Fig. P10.1.2 is driven at the left by a current source that evenly distributes the surface current density K(t) over the width w. The dimensions are such that g a w.
(a) In terms of K(t), what is H between the plates?
(b) Determine a particular solution having the form Ep = ix Exp (y, t), and find E.

floating figure GIF #31
Figure P10.1.2
10.1.3The one-turn solenoid shown in cross-section in Fig. P10.1.3 consists of perfectly conducting sheets in the planes = 0, = , and r = a. The latter is broken at the middle and driven by a current source of K(t) amps/unit length in the z direction. The current circulates around the perfectly conducting path provided by the sheets, as shown in the figure. Assume that the angle and that the system is long enough in the z direction to justify taking the fields as two dimensional.
(a) In terms of K(t), what is H in the pie-shaped region?
(b) What is E in this region?
floating figure GIF #32a
Figure P10.1.3
floating figure GIF #32b
Figure P10.1.4
10.1.4*By constrast with previous examples and problems in this section, consider here the induction of currents in materials that have relatively low conductivity. An example would be the induction heating of silicon in the manufacture of semiconductor devices. The material in which the currents are to be induced takes the form of a long circular cylinder of radius b. A long solenoid surrounding this material has N turns, a length d that is much greater than its radius, and a driving current i(t), as shown in Fig. P10.1.4.

Because the material to be heated has a small conductivity, the induced currents are small and contribute a magnetic field that is small compared to that imposed. Thus, the approach to determining the distribution of current induced in the semiconductor is 1) to first find H, ignoring the effect of the induced current. This amounts to solving Ampère's law and the flux continuity law with the current density that of the excitation coil. Then, 2) with B known, the electric field in the semiconductor is determined using Faraday's law and the MQS form of the conservation of charge law, ( E) = 0. The approach to finding the fields can then be similar to that illustrated in this section.

(a) Show that in the semiconductor and in the annulus, B (o Ni/d) iz.
(b) Use the symmetry about the z axis to show that in the semiconductor, where there is no radial component of J and hence of E at r = b, E = -(o Nr/2d)(di/dt)i.
(c) To investigate the conditions under which this approximation is useful, suppose that the excitation is sinusoidal, with angular frequency . Approximate the magnetic field intensity Hinduced associated with the induced current. Show that for the approximation to be good, Hinduced/Himposed = o b2 /4 1.

10.1.5The configuration for this problem is the same as for Prob. 10.1.4 except that the slightly conducting material is now a cylinder having a rectangular cross-section, as shown in Fig. P10.1.5. The imposed field is therefore the same as before.

(a) In terms of the coordinates shown, find a particular solution for E that takes the form E = iy Eyp(x, t) and satisfies the boundary conditions at x = 0 and x = b.
(b) Determine E inside the material of rectangular cross-section.

(c) Sketch the particular, homogeneous, and total electric fields, making clear how the first two add up to satisfy the boundary conditions. (Do not take the time to evaluate your analytical formula but rather use your knowledge of the nature of the solutions and the boundary conditions that they must satisfy.)

floating figure GIF #33
Figure P10.1.5
 

Nature of Fields Induced in Finite Conductors

10.2.1*The "Boomer" might be modeled as a transformer, with the disk as the one-turn secondary terminated in its own resistance. We have found here that if m 1, then the flux linked by the secondary is small. In Example 9.7.4, it was shown that operation of a transformer in its "ideal" mode also implies that the flux linked by the secondary be small. There it was found that to achieve this condition, the time constant L22/R of the secondary must be long compared to times of interest. Approximate the inductance and resistance of the disk in Fig. 10.2.3 and show that L22/R is indeed roughly the same as the time given by (10.2.17).

10.2.2It is proposed that the healing of bone fractures can be promoted by the passage of current through the bone normal to the fracture. Using magnetic induction, a transient current can be induced without physical contact with the patient. Suppose a nonunion of the radius (a nonhealing fracture in the long bone of the forearm, as shown in Fig. P10.2.2) is to be treated. How would you arrange a driving coil so as to induce a longitudinal current along the bone axis through the fracture?

floating figure GIF #34
Figure P10.2.2
10.2.3*Suppose that a driving coil like that shown in Fig. 10.2.2 is used to produce a magnetic flux through a conductor having the shape of the circular cylindrical shell shown in Fig. 10.3.2. The shell has a thickness and radius a. Following steps parallel to those represented by (10.2.13)-(10.2.16), show that Hind/H1 is roughly m, where m is given by (10.3.10). (Assume that the applied field is essentially uniform over the dimensions of the shell.)

 

Diffusion of Axial Magnetic Fields through Thin Conductors

10.3.1*A metal conductor having thickness and conductivity is formed into a cylinder having a square cross-section, as shown in Fig. P10.3.1. It is very long compared to its cross-sectional dimensions a. When t = 0, there is a surface current density Ko circulating uniformly around the shell. Show that the subsequent surface current density is K(t) = Ko exp (-t/m) where m = o a/4.
floating figure GIF #35
Figure P10.3.1
10.3.2The conducting sheet of thickness shown in cross-section by Fig. P10.3.2 forms a one-turn solenoid having length l that is large compared to the length d of two of the sides of its right-triangular cross-section. When t = 0, there is a circulating current density Jo uniformly distributed in the conductor.
(a) Determine the surface current density K(t) = J(t) for t > 0.
(b) A high-impedance voltmeter is connected as shown between the lower right and upper left corners. What v(t) is measured?
(c) Now lead (1) is connected following path (2). What voltage is measured?
floating figure GIF #36
Figure P10.3.2
10.3.3*A system of two concentric shells, as shown in Fig. 10.5.2 without the center shell, is driven by the external field Ho (t). The outer and inner shells have thicknesses and radii a and b, respectively.

(a) Show that the fields H1 and H2, between the shells and inside the inner shell, respectively, are governed by the equations (m = o b/2)
equation GIF #10.147

equation GIF #10.148
(b) Given that Ho = Hm cos t, show that the sinusoidal steady state fields are H1 = Re { [Hm (1 + j m )/D] exp j t} and H2 = Re { [Hm/D] exp j t} where D = [1 + j m (a/b - b/a)](1 + jm ) + j m (b/a).

10.3.4The -shaped perfect conductor shown in Fig. P10.3.4 is driven along its left edge by a current source having the uniformly distributed density Ko (t). At x = -a there is a thin sheet having the nonuniform conductivity = o /[1 + cos ( y/b)]. The length in the z direction is much greater than the other dimensions.
(a) Given Ko (t), find a differential equation for K(t).
(b) In terms of the solution K(t) to this equation, determine E in the region -a < x < 0, 0 < y < b.

figure GIF #1
Figure P10.3.4
 

Diffusion of Transverse Magnetic Fields through Thin Conductors

10.4.1*A thin planar sheet having conductivity and thickness extends to infinity in the x and z directions, as shown in Fig. P10.4.1. Currents in the sheet are z directed and independent of z.

(a) Show that the sheet can be represented by the boundary conditions
equation GIF #10.149
equation GIF #10.150
(b) Now consider the special case where the regions above and below are free space and extend to infinity in the +y and -y directions, respectively. When t = 0, there is a surface current density in the sheet K = iz Ko sin x, where Ko and are given constants. Show that for t > 0, Kz = Ko exp (-t/ ) where = o /2.

floating figure GIF #37
Figure P10.4.1
10.4.2In the two-dimensional system shown in cross-section by Fig. P10.4.2, a planar air gap of width d is bounded from above in the surface y = d by a thin conducting sheet having conductivity and thickness . This sheet is, in turn, backed by a material of infinite permeability. The region below is also infinitely permeable and at the interface y = 0 there is a winding used to impose the surface current density K = K(t) cos x iz. The system extends to infinity in the x and z directions.

(a) The surface current density K(t) varies so rapidly that the conducting sheet acts as a perfect conductor. What is in the air gap?
(b) The current is slowly varying so that the sheet supports little induced current. What is in the air gap?

(c) Determine (x, y, t) if there is initially no magnetic field and a step, K = Ko u-1(t), is applied. Show that the early and long-time response matches that expected from parts (a) and (b).

floating figure GIF #38
Figure P10.4.2
10.4.3*The cross-section of a spherical shell having conductivity , radius R and thickness is as shown in Fig. 8.4.5. A magnetic field that is uniform and z directed at infinity is imposed.

(a) Show that boundary conditions representing the shell are
equation GIF #10.151

equation GIF #10.152
(b) Given that the driving field is Ho(t) = Re { o exp (j t)}, show that the magnetic moment of a dipole at the origin that would have an effect on the external field equivalent to that of the shell is
equation GIF #10.153
where o R/3.

(c) Show that in the limit where , the result is the same as found in Example 8.4.3.

10.4.4A magnetic dipole, having moment i(t)a (as defined in Example 8.3.2) oriented in the z direction is at the center of a spherical shell having radius R, thickness , and conductivity , as shown in Fig. P10.4.4. With i = Re { exp (j t)}, the system is in the sinusoidal steady state.
(a) In terms of i(t)a, what is in the neighborhood of the origin?
(b) Given that the shell is perfectly conducting, find . Make a sketch of H for this limit.
(c) Now, with finite, determine .
(d) Take the appropriate limit of the fields found in (c) to recover the result of (b). In terms of the parameters that have been specified, under what conditions does the shell behave as though it had infinite conductivity?
floating figure GIF #39
Figure P10.4.5
10.4.5*In the system shown in cross-section in Fig. P10.4.5, a thin sheet of conductor, having thickness and conductivity , is wrapped around a circular cylinder having infinite permeability and radius b. On the other side of an air gap at the radius r = a is a winding, used to impose the surface current density K = K(t) sin 2 iz, backed by an infinitely permeable material in the region a < r.
(a) The current density varies so rapidly that the sheet behaves as an infinite conductor. In this limit, show that in the air gap is
equation GIF #10.154
(b) Now suppose that the driving current is so slowly varying that the current induced in the conducting sheet is negligible. Show that
equation GIF #10.155
(c) Show that if the fields are zero when t < 0 and there is a step in current, K(t) = Ko u-1(t)
equation GIF #10.156
where
equation GIF #10.157
Show that the early and long-time responses do indeed match the results found in parts (a) and (b).

10.4.6The configuration is as described in Prob. 10.4.5 except that the conducting shell is on the outside of the air gap at r = a, while the windings are on the inside surface of the air gap at r = b. Also, the windings are now arranged so that the imposed surface current density is K = Ko (t) sin . For this configuration, carry out parts (a), (b), and (c) of Prob. 10.4.5.

 

Magnetic Diffusion Laws

10.5.1*Consider a class of problems that are analogous to those described by (10.5.10) and (10.5.11), but with J rather than H written as a solution to the diffusion equation.
(a) Use (10.5.1)-(10.5.5) to show that for a uniform conductivity
equation GIF #10.158
(b) Now consider J (rather than H) to be z directed but independent of z, J = Jz (x, y, t) iz, and H (rather than J) to be transverse, H = Hx (x, y, t) ix + Hy (x, y, t) iy. Show that
equation GIF #10.159
where H can be found from J using
equation GIF #10.160
Note that these expressions are of the same form as (10.5.8), (10.5.10), and (10.5.11), respectively, but with the roles of J and H reversed.

 

Magnetic Diffusion Step Response

10.6.1*In the configuration of Fig. 10.6.1, a steady state has been established with Ks = Kp = constant. When t = 0, this driving current is suddenly turned off. Show that H and J are given by (10.6.21) and (10.6.22) with the first term in each omitted and the sign of the summation in each reversed.

10.6.2Consider the configuration of Fig. 10.6.1 but with a perfectly conducting electrode in the plane x = 0 "shorting" the electrode at y = 0 to the one at y = a.
(a) A steady driving current has been established with Ks = Kp = constant. What are the steady H and J in the conducting block?
(b) When t = 0, the driving current is suddenly turned off. Determine H and J for t > 0.

 

Skin Effect

10.7.1*For Example 10.7.1, the conducting block has length d in the z direction.
(a) Show that the impedance seen by the current source is
equation GIF #10.161
(b) Show that in the limit where b , Z becomes the dc resistance a/db.
(c) Show that in the opposite extreme where b , so that the current is concentrated near the surface, the block impedance has resistive and inductive-reactive parts of equal magnitude and that the resistance is equivalent to that for a slab having thickness in the x direction carrying a current that is uniformly distributed with respect to x.

10.7.2In the configuration of Example 10.7.1, the perfectly conducting electrodes are terminated by a perfectly conducting electrode in the plane x = 0.
(a)Determine the sinusoidal steady state response H.
(b)Show that even though the current source is now "showed" by perfectly conducting electrodes, the high-frequency field distribution is still given by (10.7.16), so that in this limit, the current still concentrates at the surface.
(c)Determine the impedance of a length d (in the z direction) of the block.


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