The Conducting Core

Abstract

Here, we consider what happens when the current in a unit sphere is turned on suddenly while a smaller perfectly conducting sphere of radius 𝜖 is centered inside. As in the case without the central core, two waves will leave the current source, one heading outward away from the sphere and the other heading inward toward the center. If the core were not present, the incoming wave would turn around when it reached the center and then move back out through the sphere to form the trailing edge of an outgoing bubble of thickness 2R (where R = 1 for the unit sphere).

Here, we consider what happens when the current in a unit sphere is turned on suddenly while a smaller perfectly conducting sphere of radius 𝜖 is centered inside. As in the case without the central core, two waves will leave the current source, one heading outward away from the sphere and the other heading inward toward the center. If the core were not present, the incoming wave would turn around when it reached the center and then move back out through the sphere to form the trailing edge of an outgoing bubble of thickness 2R (where R = 1 for the unit sphere).

The infinite conductivity of the central core will prevent the magnetic field from penetrating it and cause the ingoing wave to turn around prematurely. Thus, the separation between the outgoing waves would be smaller by the amount, 2𝜖, and a static, dipole-like field would be left behind, rooted, not in the unit sphere, but in the portion of the unit sphere external to the conducting core. Intuitively, this remnant field would be equivalent to the static field of two uniformly magnetized spheres having the same central field strength, but opposite polarities, similar to the field that is shown in the left panel of Fig. 14.4. Now, let us see what a calculation actually gives.

Three waves will contribute to the field. Outside the unit sphere, there will be an outgoing wave associated with the leading edge of the disturbance: $f_{L}^{out}(r,r_{L})=f_{o}(r,r_{L})$, where f_o is the external f-value given by Eq. (2.11a). Inside the unit sphere, there will be an ingoing wave associated with the following edge of the disturbance: $f_{F}^{in}(r,r_{L})=f_{i}(r,r_{L})$, where f_i(r, r_L) is the internal f-value given by Eq. (2.11b). Also, inside the unit sphere, there will be an outgoing wave, $f_{F}^{out}(r,r_{L})$, produced by the reflection of the ingoing wave at the conducting core. Our first objective will be to determine this outgoing wave.

The outgoing wave, $f_{F}^{out}(r,r_{L})$, must satisfy three conditions. First, it must be constructed from outgoing solutions to the wave equation (15.22). As we have seen in Eq. (15.4), these solutions are generally of the form $(1-kr/{\epsilon }) \exp \{k(r-r_{L})/{\epsilon }\}$, where k is a constant. Second, the outgoing wave must combine with the ingoing wave to make the radial component of magnetic field vanish at the surface of the core, so that B_r(𝜖, r_L) = 0 for all r_L. This is equivalent to

$$\displaystyle \begin{aligned} f_{F}^{in}({\epsilon},r_{L})+f_{F}^{out}({\epsilon},r_{L})= f_{i}({\epsilon},r_{L})+f_{F}^{out}({\epsilon},r_{L})=0. {} \end{aligned} $$

(16.1)

Third, the outgoing wave must have a “leading edge” at r = r_L − 2(1 − 𝜖), given by

$$\displaystyle \begin{aligned} f_{F}^{out}(r_{L}+2{\epsilon}-2,r_{L})=0,{} \end{aligned} $$

(16.2)

for all r_L ≥ 2 − 𝜖.

We can use these conditions to determine $f_{F}^{out}(r,r_{L})$ as follows. We begin with the particular combination of outgoing waves:

$$\displaystyle \begin{aligned} f_{F}^{out}(r,r_{L})= \int_{0}^{\infty}A(k)dk(1-kr/{\epsilon})e^{k(r-r_{L}-2{\epsilon}+2)/{\epsilon}}+ c_{1}(1-r/{\epsilon})e^{(r-r_{L}-2{\epsilon}+2)/{\epsilon}},{} \end{aligned} $$

(16.3)

where c₁ is a constant to be determined. In this equation, the continuum spectrum, A(k), allows us to satisfy the boundary condition (16.1), and the “line” at k = 1 allows us to satisfy the leading-edge condition (16.2). Thus, when we put r = 𝜖 in Eq. (16.3), the c₁-dependent term vanishes and the resulting expression for $f_{F}^{out}({\epsilon },r_{L})$ may be equated to − f_i(𝜖, r_L), according to Eq. (16.1). The result is

$$\displaystyle \begin{aligned} \int_{0}^{\infty}A(k)(1-k)e^{-sk}dk=-f_{i}\{{\epsilon},(s-1){\epsilon}+2\},{} \end{aligned} $$

(16.4)

where s = (r_L + 𝜖 − 2)∕𝜖. Recognizing that the left term is the Laplace transform of A(k)(1 − k) with respect to the variable s, and inverting this transform, we obtain A(k)(1 − k) as a series of delta functions and their derivatives:

$$\displaystyle \begin{aligned} A(k)(1-k)=-\frac{3}{2}{\epsilon}^{2}~{\delta}'(k)+ \frac{3}{4}{\epsilon}^{2}(1-{\epsilon})~{\delta}''(k)+\frac{1}{4}{\epsilon}^{3} ~{\delta}'''(k).{} \end{aligned} $$

(16.5)

Equation (16.2) tells us that if we set r = r_L + 2𝜖 − 2 in Eq. (16.3), the result must vanish for all r_L ≥ 2 − 𝜖. This can be true only if the constant c₁ is given by

$$\displaystyle \begin{aligned} c_{1}=-\int_{0}^{\infty}A(k)kdk=-\int_{0}^{\infty}A(k)dk= -3{\epsilon}^{2}(1-{\epsilon})={\epsilon}^{3}+f_{i}\{0,2(1-{\epsilon})\}.{} \end{aligned} $$

(16.6)

Substituting these values of A(k) and c₁ into Eq. (16.3), we finally obtain the expression for $f_{F}^{out}(r,r_{L})$:

$$\displaystyle \begin{aligned} f_{F}^{out}(r,r_{L})=r^{3}-{\epsilon}^{3}-f_{i}(r,r_{L})+3{\epsilon} (1-{\epsilon})(r-{\epsilon})e^{- \{ r_{L}-2(1-{\epsilon})-r \} /{\epsilon}}.{} \end{aligned} $$

(16.7)

After the following edge has passed out through the unit sphere and the bubble has moved outward, the residual f-values inside and outside the unit sphere are given by $f_{i}(r,r_{L})+f_{F}^{out}(r,r_{L})$ and $f_{o}(r,r_{L})+f_{F}^{out}(r,r_{L})$, respectively. We can express these values as

$$\displaystyle \begin{aligned} &f(r,r_{L})\\&\quad = \left\{ \begin{array}{llll} r^{3}-{\epsilon}^{3}+3{\epsilon}(1-{\epsilon})(r-{\epsilon}) e^{-\{r_{L}-2(1-{\epsilon})-r \}/{\epsilon}},&{\epsilon}~{\leq}~r~{\leq}~1 \\ {}1-{\epsilon}^{3}+3{\epsilon}(1-{\epsilon})(r-{\epsilon}) e^{-\{r_{L}-2(1-{\epsilon})-r \}/{\epsilon}},&1~{\leq}~r~{\leq}~r_{L}-2(1-{\epsilon}). \end{array} \right.{} \end{aligned} $$

(16.8)

The f-value is still f_o(r, r_L) between the two outgoing waves, but now this region extends only 2(1 − 𝜖) units behind the leading edge compared to 2 units behind the leading edge of the bubble produced in the absence of the conducting core. The difference, 2𝜖, corresponds to the length of the path that the ingoing wave did not take. Note that the first part of Eq. (16.8) is valid behind the trailing edge as soon as that wave reflects off the conducting core and heads outward, but the second part of the equation does not apply until the trailing edge crosses the surface at r = 1.

Referring to Eq. (16.8), we see that the fields left behind the bubble are not static when a conducting core is present inside the unit sphere. However, the time-varying terms decay exponentially and become greatly reduced when the field points are more than a few units of 𝜖 behind the following edge of the bubble. In that case, f(r, r_L) → r³ − 𝜖³ inside the unit sphere, and f(r, r_L) → 1 − 𝜖³ outside. As we guessed initially, these f-values represent the field produced by two uniformly magnetized spheres of equal strength, but opposite polarity, one of radius 𝜖, and the other of radius 1. This field is identical to the static field whose field lines are plotted in the left panel of Fig. 14.4. In both cases, the field vanishes inside the inner sphere and is a weakened dipole outside the outer sphere. In the transition region between the two spheres, the small dipole contribution causes the uniform field lines to bend smoothly around the inner sphere.

Figure 16.1 compares the evolution of the field lines for a sudden turn-on without a conducting core (top panels) and the evolution with conducting cores of radii 𝜖 = 1/3, 1/2, and 2/3 shown in the second, third, and fourth rows, respectively. The dotted circles mark the positions of the leading and trailing waves at the times r_L = 2.6, 3.6, and 4.6. In each case, the spatial scale is varied to follow the leading edge of the outgoing wave. For progressively larger conducting cores of radius, 𝜖, the separation between the leading and trailing waves shows the expected decrease as 2(1 − 𝜖). At r_L = 2.6 and 𝜖 = 0, the ingoing wave has turned around and is moving outward through the sphere at the location r = 0.6. For 𝜖 ≥ 1∕3, the wave is already outside the sphere trailing behind the leading wave. As expected for 𝜖 > 0, the field lines inside the sphere stream smoothly around the conducting core and have no components normal to its surface.

Fig. 16.1

Field lines for sudden turn-ons with conducting cores of radius 𝜖 = 0, 1/3, 1/2, 2/3 (top to bottom). The wave fronts and conducting cores are indicated by dashed circles and blue disks, respectively

These maps reveal two additional features of the field when a conducting core is present. First, an X-type neutral point (line in three dimensions) forms well behind the reflected wave front, and marks the end of the region of detached flux (i.e., the back end of the bubble). As a result, this region of detached flux falls behind the two waves and becomes increasingly larger with time. Second, between the two wave fronts, the O-type neutral point, familiar to us when 𝜖 = 0, lies progressively closer to the following boundary at r = r_L − 2(1 − 𝜖) as 𝜖 increases from 0, and the circulating contours become significantly flattened for values of 𝜖 > 1∕2.

For the case of 𝜖 = 0, we have already seen in Chap. 3 that the O-type neutral point moves toward the center of the bubble and asymptotically approaches the location r = r_L − 1. This asymptotic location will fall behind the trailing wave at r = r_L − 2(1 − 𝜖) when 𝜖 ≥ 1∕2. In this case, we might expect the neutral point to disappear, changing to a discontinuity analogous to the one that lies at the trailing edge of the bubble when 𝜖 = 0.

It is interesting to examine these discontinuities in detail. We begin with 𝜖 = 0 and r = 1. Dropping the common factor of $\sin {\theta }$, we find that B_θ jumps from − 1 (corresponding to the uniform field, B₀, in the sphere) to + 1/2 for a total discontinuity of + 3/2 (corresponding to the strength of the source current on the sphere). Similar jumps of + 3/2 are also present at r = 1 when 𝜖 > 0. The jump at r = r_L − 2(1 − 𝜖) is ΔB_θ∕B₀ = +3∕(4r) (corresponding to ΔE_ϕ∕(cB₀) = −3∕(4r)), and the jump at r = r_L is ΔB_θ∕B₀ = −3∕(4r) (corresponding to ΔE_ϕ∕(cB₀) = +3∕(4r)). As mentioned in Chap. 3, in the absence of currents, the jumps of B_θ and E_ϕ are related by ΔE_ϕ = −c ΔB_θ.

In the presence of a conducting core, there is also a discontinuity at the surface of the core where r = 𝜖. Using the f-value given in Eq. (16.8) with 𝜖 ≤ r ≤ 1, it is easy to show that when r = 𝜖, the jump is

$$\displaystyle \begin{aligned} \frac{{\Delta}B_{\theta}}{B_{0}}&=\frac{1}{2}r^{-3} {\Delta}(f-rf')= -\frac{1}{2}r^{-2} {\Delta}(f')\\&= -\frac{3}{2} \left \{1+\Bigg(\frac{1-{\epsilon}}{{\epsilon}}\Bigg)~ e^{-\{r_{L}-(2-{\epsilon})\}/{\epsilon}}\right \}\!,{} \end{aligned} $$

(16.9)

again dropping the common factor of $\sin {\theta }$. Here, we see that the jump of B_θ∕B₀ approaches − 3∕2 as r_L→∞. If we restore the $\sin {\theta }$-dependence of B_θ and then integrate B_θ over the surface of the conducting core of radius 𝜖, then we obtain the total induced current on that surface as a function of time:

$$\displaystyle \begin{aligned} \begin{array}{rcl} i(r_{L})= \left\{ \begin{array}{ll} 0,&r_{L}~{\leq}~2(1-{\epsilon}), \\ \\ -3(B_{0}R/{\mu}_{0}) \left [ {\epsilon}+(1-{\epsilon})~e^{- \left \{ r_{L}-(2-{\epsilon}) \right \}/{\epsilon}} \right ],&r_{L}>2(1-{\epsilon}). \end{array} \right.{} \end{array} \end{aligned} $$

(16.10)

From this equation, we see that the induced current begins with the value − 3B₀R∕μ₀ and approaches the asymptotic value of − (3B₀R∕μ₀)𝜖 on an exponential time scale of 𝜖R∕c. Thus, after a few time scales, the induced current is similar to the current on the outer sphere, except that the induced current is of opposite sign and is smaller because it scales as the size of the conducting core. Also, note that as 𝜖 → 1, the induced current loses its time dependence, and maintains a static value of − (3B₀R∕μ₀)𝜖.

When they exist, the neutral points are determined by setting B_θ(r, π∕2, r_L) = 0, which is equivalent to f − rf′ = 0. The O-type neutral point is obtained by substituting f = f_o, which gives the result that we obtained in Chap. 3 in the absence of the conducting core (i.e., r → r_L − 1 as r_L → ∞). However, as we mentioned above, this solution is valid only when 𝜖 < 1∕2. There is no O-type neutral point when 𝜖 ≥ 1∕2, and the detached field lines circulate around the discontinuity at r = r_L − 2(1 − 𝜖).

The X-type neutral point is obtained by substituting f from Eq. (16.8) with 1 ≤ r ≤ r_L − 2(1 − 𝜖), which gives

$$\displaystyle \begin{aligned} f-rf'=1-{\epsilon}^{3}-3(1-{\epsilon})\Bigg(\frac{r^{3}+{\epsilon}^{3}}{r+{\epsilon}}\Bigg) e^{-\{r_{L}-2(1-{\epsilon})-r \}/{\epsilon}}=0,{} \end{aligned} $$

(16.11)

which can be solved for r_L as a function of r:

$$\displaystyle \begin{aligned} r_{L}=2(1-{\epsilon})+r+{\epsilon}~ ln \left \{ 3\left (\frac{1-{\epsilon}}{1-{\epsilon}^{3}} \right ) \left (\frac{r^{3}+{\epsilon}^{3}}{r+{\epsilon}} \right) \right \}{} \end{aligned} $$

(16.12)

For 𝜖 ≪ 1 and r ≫ 1, this equation has the approximate solution:

$$\displaystyle \begin{aligned} r_{L}-2(1-{\epsilon})-r~{\approx}~{\epsilon}~ln(3r^{2}),{} \end{aligned} $$

(16.13)

which means that

$$\displaystyle \begin{aligned} r~{\approx}~r_{L}-2(1-{\epsilon})-{\epsilon}~ln(3r^{2}),{} \end{aligned} $$

(16.14)

and the neutral point lies a distance 𝜖ln(3r²) behind the trailing wave. Thus, as time increases, the neutral point falls increasingly farther behind the trailing wave, while still moving outward from the sphere, and the thickness of the bubble increases accordingly.

Next, we consider the flux budget for this sudden turn-on. The amount of flux that is ultimately contained in the external dipole field is Φ_e∕ Φ₀ = f_o(r_L, r_L)∕r_L − f(1, r_L), where Φ₀ = πR²B₀ with R = 1, and f(1, r_L) is given by Eq. (16.8). The first term vanishes because it refers to f_o at the leading edge of the disturbance. The second term is obtained by substituting the expression for r_L given in Eq. (16.12) into Eq. (16.8) and letting r_L → ∞. The result is Φ_e∕ Φ₀ = −(1 − 𝜖³), where the minus sign indicates that B_θ is positive. Equivalently, the corresponding amount of flux trapped in the sphere is Φ_t∕ Φ₀ = 1 − 𝜖³.

Next, we consider the radiated flux. If 𝜖 ≤ 1∕2, then the amount of detached flux is Φ_d∕ Φ₀ = f_o(r_L − 1, r_L)∕(r_L − 1) − f(r, r_L)∕r, where f(r, r_L) is given by Eq. (16.8) with 1 ≤ r ≤ r_L − 2(1 − 𝜖), and r is related to r_L by Eq. (16.12). As we found in Chap. 3, the first term approaches 3/4 as r_L → ∞. The second term vanishes when r → ∞. Consequently, for 𝜖 ≤ 1∕2, the amount of radiated flux is Φ_r∕ Φ₀ = 3∕4, just as we found in the absence of a conducting core.

However, if 𝜖 > 1∕2, there is no O-type neutral point and the flux circulates around the location r = r_L − 2(1 − 𝜖). In this case, the amount of detached flux is Φ_d∕ Φ₀ = f[r_L − 2(1 − 𝜖), r_L]∕[r_L − 2(1 − 𝜖)] − f(r, r_L)∕r. As before, the second term vanishes and the first term approaches 3𝜖(1 − 𝜖). Thus, the radiated flux Φ_r∕ Φ₀ = 3𝜖(1 − 𝜖) for 𝜖 > 1∕2, which matches the value of 3/4 at 𝜖 = 1∕2, and falls continuously to 0 as 𝜖 increases to 1. With a little algebra, we can summarize this result as:

$$\displaystyle \begin{aligned} \begin{array}{rcl} \frac{{\Phi}_{r}}{{\Phi}_{0}}= \left\{ \begin{array}{ll} 3/4,&0{\leq}~ {\epsilon}~{\leq}~1/2 \\ \\ (3/4)-3({\epsilon}-1/2)^{2},&1/2<{\epsilon}~{\leq}~1. \end{array} \right.{} \end{array} \end{aligned} $$

(16.15)

The conducting core does not affect the radiated flux unless its radius, 𝜖, exceeds 1/2, in which case the amount of radiated flux is smaller than it is without the core.

Next, we consider the energy budget. To evaluate the total energy provided by the current source, we return to the volume integral given by Eq. (4.6), but with the limits of r_L-integration changed from (1, 3) to (r_L1, r_L2). This integral gives the total energy provided by the source between the times r_L1 and r_L2:

$$\displaystyle \begin{aligned} \frac{\mathcal{E}_{tot}}{\mathcal{E}_{0}}=3 \left \{f(1,r_{L2})-f(1,r_{L1}) \right \},{} \end{aligned} $$

(16.16)

where $\mathcal {E}_{0}=(4{\pi }R^{3}/3)(B_{0}^{2}/2{\mu }_{0})$ is the field energy in a uniformly magnetized sphere of radius, R. For our source at r = 1 with the conducting core at r = 𝜖, we set r_L1 = 1 and recognize that f(1, 1) = f_o(1, 1) = 0 so that the second term in Eq. (16.16) vanishes. Then, we let r_L2 → ∞ and evaluate f(1, r_L2) from Eq. (16.8) to obtain

$$\displaystyle \begin{aligned} \mathcal{E}_{tot}=3(1-{\epsilon}^{3})\mathcal{E}_{0}= 3 \left \{ \frac{4}{3}{\pi}R^{3}-\frac{4}{3}{\pi}({\epsilon}R)^{3} \right \} \Bigg(\frac{B_{0}^{2}}{2{\mu}_{0}}\Bigg),{} \end{aligned} $$

(16.17)

where R = 1 for the unit sphere. Thus, the total energy provided by the source is reduced by three times the energy that would have been in the volume now occupied by the conducting core. This reduction is equal to the energy difference, ${\Delta }\mathcal {E}=M_{i}B_{0}=3\mathcal {E}_{i}$, obtained from Eq. (14.25) when B_i = −B_o, for the field of anti-parallel dipole moments shown in the left panel of Fig. 14.4.

But how much of this energy is stored in the field and how much is radiated away? Returning this time to Eq. (4.3), we set r_L1 = r and let r_L2 → ∞ to obtain the Poynting energy that ultimately crosses the surface at radius r. For the f-value given by Eq. (16.8) with 1 ≤ r ≤ r_L − 2(1 − 𝜖), we obtain

$$\displaystyle \begin{aligned} \frac{\mathcal{E}}{\mathcal{E}_{0}}= \left \{\frac{3}{2}+\frac{1}{2r^{3}} \right \}(1-{\epsilon}^{3}),{} \end{aligned} $$

(16.18)

which approaches $\mathcal {E}_{rad}/\mathcal {E}_{0}=(3/2)(1-{\epsilon }^{3})$ as r → ∞. Thus, like the case without a conducting core, half the energy is radiated and therefore half is stored in the static magnetic field that is ultimately left behind. A separate evaluation of the field energy gives (3∕2)(1 − 𝜖³), of which (1 − 𝜖³){3∕2 − (1∕2)(1 − 𝜖³)} lies inside the sphere of unit radius, and (1∕2)(1 − 𝜖³)² lies outside. For 𝜖 ≪ 1, this corresponds to 1 − (1∕2)𝜖³ inside and (1∕2) − 𝜖³ outside. Thus, of the total field energy deficit, (3∕2)𝜖³, caused by the conducting core, approximately 2/3 is removed from outside the sphere and 1/3 from inside the sphere. Of course, no field energy lies inside the conducting core.

All of these energies approach 0 as 𝜖 → 1, indicating that little energy is transferred to either the field or the radiation when only a small space is left between the current source and the central conducting core. Also, unlike the radiated flux, the radiated energy is diminished by a finite amount even for cores with 𝜖 < 1∕2. However, this is a small effect, varying as 𝜖³, and when 𝜖 = 0, we regain all of the energies previously obtained without the core: 1 unit stored inside the sphere, 1/2 stored outside, and 3/2 radiated away. Of course, the radiated energy would be smaller if the current were turned on gradually.