IPhO 1994 · Theory problem 2

Superconducting Magnet

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Super conducting magnets are widely used in laboratories. The most common form of super conducting magnets is a solenoid made of super conducting wire. The wonderful thing about a superconducting magnet is that it produces high magnetic fields without any energy dissipation due to Joule heating, since the electrical resistance of the superconducting wire becomes zero when the magnet is immersed in liquid helium at a temperature of 4.2 K. Usually, the magnet is provided with a specially designed superconducting switch, as shown in Fig. 1. The resistance \(r\) of the switch can be controlled: either \(r=0\) in the superconducting state, or \(r=r_n\) in the normal state. When the persistent mode, with a current circulating through the magnet and superconducting switch indefinitely. The persistent mode allows a steady magnetic field to be maintained for long periods with the external source cut off.

The details of the superconducting switch are not given in Fig. 1. It is usually a small length of superconducting wire wrapped with a heater wire and suitably thermally insulated from the liquid helium bath. On being heated, the temperature of the superconducting wire increases and it reverts to the resistive normal state. The typical value of \(r_n\) is a few ohms. Here we assume it to be \(5\,\Omega\). The inductance of a superconducting magnet depends on its size; assume it be \(10\,\mathrm H\) for the magnet in Fig. 1. The total current \(I\) can be changed by adjusting the resistance \(R\).

This problem will be graded by the plots only!

The arrows denote the positive direction of \(I\), \(I_1\) and \(I_2\).

Figure 1

Fig. 1

1) If the total current \(I\) and the resistance \(r\) of the superconducting switch are controlled to vary with time in the way shown in Figs, 2a and 2b respectively, and assuming the currents \(I_1\) and \(I_2\) flowing through the magnet and the switch respectively are equal at the beginning (Fig. 2c and Fig. 2d), how do they vary with time from \(t_1\) to \(t_4\)? Plot your answer in Fig. 2c and Fig. 2d

Figure 2

Fig.2a

Figure 3

2b

Figure 4

2c

Figure 5

2d

2) Suppose the power switch \(K\) is turned on at time \(t=0\) when \(r=0\), \(I_1=0\) and \(R=7.5\,\Omega\), and the total current \(I\) is \(0.5\,\mathrm A\). With \(K\) kept closed, the resistance \(r\) of the superconducting switch is varied in he way shown in Fig. 3b. Plot the corresponding time dependences of \(I\), \(I_1\) and \(I_2\) in Figs. 3a, 3c and 3d respectively.

Figure 6

Fig. 3a

Figure 7

3b

Figure 8

3c

Figure 9

3d

3) Only small currents, less than \(0.5\,\mathrm A\), are allowed to flow through the superconducting switch when it is in the normal state, with larger currents the switch will be burnt out. Suppose the superconducting magnet is operated in a persistent mode, i. e. \(I=0\), and \(I_1=i_1\)(e. g. \(20\,\mathrm A\)), \(I_2=-i_1\), as shown in Fig. 4, from \(t=0\) to \(t=3\mathrm{min}\). If the experiment is to be stopped by reducing the current through the magnet to zero, how would you do it? This has to be done in several operation steps. Plot the corresponding changes of \(I\), \(r\), \(I_1\) and \(I_2\) in Fig. 4

Figure 10

Fig. 4a

Figure 11

4b

Figure 12

4c

Figure 13

4d

4) Suppose the magnet is operated in a persistent mode with a persistent current of \(20\,\mathrm A\) (\(t=0\) to \(t=3\mathrm{min}\). See Fig. 5). How would you change it to a persistent mode with a current of 30a? plot your answer in Fig. 5.

Figure 14

Fig. 5a

Figure 15

5b

Figure 16

5c

Figure 17

5d

Official English text of the problem.

Answers

  1. Throughout the prescribed sequence, \[ \boxed{I_1=\frac{I_0}{2},\qquad I_2=I-\frac{I_0}{2}.} \] Thus \(I_2\) falls linearly from \(+I_0/2\) at \(t_1\) to \(-I_0/2\) at \(t_2\), rises linearly to zero as \(I\) reaches its plateau, and remains zero through \(t_3\) and \(t_4\).

  2. The source emf is \(3.75\,\mathrm V\). Initially, \[ (I,I_1,I_2)=(0.50,0,0.50)\,\mathrm A. \] At \(1\,\mathrm{min}\), \(I\) and \(I_2\) jump to \(0.30\,\mathrm A\), while \(I_1\) remains zero. They then approach \((0.50,0.50,0)\,\mathrm A\) exponentially with time constant \[ \boxed{\tau_R=\frac{10}{3}\,\mathrm s\approx3.33\,\mathrm s.} \] After \(2\,\mathrm{min}\), \(I=0.50\,\mathrm A\), \(I_1\) is constant, and \(I_2\approx0\). Exact expressions appear below.

  3. With \(r=0\), first raise \(I\) from zero to \(20\,\mathrm A\), leaving \(I_1=20\,\mathrm A\). Then set \(r=5\,\Omega\), reduce \(I\) slowly to zero, and open \(K\). The remaining magnet current decays exponentially with time constant \(2.0\,\mathrm s\). The explicit three-minute ramp below keeps \[ \boxed{|I_2|<0.223\,\mathrm A<0.5\,\mathrm A} \] throughout the normal state.

  4. First match \(I=I_1=20\,\mathrm A\) with \(r=0\). Set \(r=5\,\Omega\) and increase \(I_1\) to \(30\,\mathrm A\), using \(I_2=1/9\,\mathrm A\) during a three-minute ramp. Then set \(I=30\,\mathrm A\), cool the switch to \(r=0\), and reduce \(I\) to zero. The final persistent state is \[ \boxed{I=0,\qquad I_1=30\,\mathrm A,\qquad I_2=-30\,\mathrm A,\qquad r=0.} \]

Solution

We treat the magnet as an ideal inductor with constant inductance \(L=10\,\mathrm H\). We neglect lead resistance, stray inductance and switching time, as implicit in the problem.

With the indicated current directions, Kirchhoff’s current law gives \[ I=I_1+I_2. \] The magnet and superconducting switch have the same terminal voltage: \[ V=L\frac{dI_1}{dt}=rI_2. \] Therefore the central equation is \[ \boxed{L\frac{dI_1}{dt}=r(I-I_1).} \]

When \(r=0\), the magnet current is constant. Changes in the source current then pass entirely through the superconducting switch: \[ \frac{dI_1}{dt}=0,\qquad I_2=I-I_1. \]

When \(r=r_n\) and the total current is prescribed, the equation becomes \[ \tau\frac{dI_1}{dt}+I_1=I, \qquad \boxed{\tau=\frac{L}{r_n}=2.0\,\mathrm s.} \] The inductor current remains continuous at switching instants; \(I\) and \(I_2\) may jump.

1) Prescribed total current and switch resistance

Initially the two branch currents are equal, so \[ I_1=I_2=\frac{I_0}{2}. \]

Until \(t_3\), the switch is superconducting. Hence \[ I_1=\frac{I_0}{2}, \qquad I_2=I-\frac{I_0}{2}. \]

By \(t_3\), the imposed current has reached \(I_0/2\). Thus \(I_2=0\) when the switch becomes resistive. During \(t_3<t<t_4\), \[ L\frac{dI_1}{dt}=r_n\left(\frac{I_0}{2}-I_1\right)=0. \] The currents consequently remain unchanged. Returning the switch to \(r=0\) at \(t_4\) also causes no change.

The requested plots are:

  • Fig. 2c: a horizontal line at \(I_1=I_0/2\) throughout.
  • Fig. 2d: a horizontal line at \(I_0/2\) until \(t_1\), followed by a straight decline to \(-I_0/2\) at \(t_2\). It crosses zero halfway between \(t_1\) and \(t_2\). It then rises linearly to zero when the prescribed \(I\) reaches \(I_0/2\), and stays at zero through \(t_3\), \(t_4\), and afterward.

The negative portion of Fig. 2d must be drawn below the original zero axis.

2) Fixed source resistance

Let the source emf be \(\mathcal E\). Initially \(r=0\), so the parallel combination has zero voltage: \[ \mathcal E=RI=(7.5\,\Omega)(0.50\,\mathrm A) =\boxed{3.75\,\mathrm V}. \]

For \(0\le t<60\,\mathrm s\), the magnet current stays at its initial value zero: \[ \boxed{I=0.50\,\mathrm A,\qquad I_1=0,\qquad I_2=0.50\,\mathrm A.} \]

At \(t=60\,\mathrm s\), the switch becomes resistive. Since \(I_1\) cannot jump, its immediate value remains zero, giving \[ I(60^+)=I_2(60^+) =\frac{\mathcal E}{R+r_n} =\frac{3.75}{12.5} =\boxed{0.30\,\mathrm A}. \]

During the normal-state interval, Kirchhoff’s voltage law gives \[ \mathcal E=R(I_1+I_2)+r_nI_2, \] and therefore \[ I_2=\frac{\mathcal E-RI_1}{R+r_n}. \] Combining this with the inductor equation, \[ L\frac{dI_1}{dt} =\frac{r_nR}{R+r_n}\left(\frac{\mathcal E}{R}-I_1\right). \] Thus \[ \boxed{\tau_R=\frac{L(R+r_n)}{Rr_n} =\frac{10(7.5+5)}{7.5(5)}\,\mathrm s =\frac{10}{3}\,\mathrm s.} \]

For \(60\le t<120\,\mathrm s\), define \[ x=\frac{t-60\,\mathrm s}{\tau_R}. \] The currents are \[ \boxed{ \begin{aligned} I_1(t)&=(0.50\,\mathrm A)(1-e^{-x}),\\ I_2(t)&=(0.30\,\mathrm A)e^{-x},\\ I(t)&=0.50\,\mathrm A-(0.20\,\mathrm A)e^{-x}. \end{aligned}} \]

At \(t=120\,\mathrm s\), \(x=18\), and the switch returns to \(r=0\). The source current becomes \(\mathcal E/R=0.50\,\mathrm A\), while the magnet current freezes at its existing value. Hence, for all \(t\ge120\,\mathrm s\), \[ \boxed{ \begin{aligned} I&=0.50\,\mathrm A,\\ I_1&=(0.50\,\mathrm A)(1-e^{-18}),\\ I_2&=(0.50\,\mathrm A)e^{-18} \approx7.6\,\mathrm{nA}. \end{aligned}} \]

The plot features are:

  • Fig. 3a: \(I=0.50\,\mathrm A\) until one minute; a vertical drop to \(0.30\,\mathrm A\); then an exponential, concave-down rise toward \(0.50\,\mathrm A\).
  • Fig. 3c: \(I_1=0\) until one minute; then a continuous, concave-down exponential rise toward \(0.50\,\mathrm A\). It becomes exactly constant at two minutes.
  • Fig. 3d: \(I_2=0.50\,\mathrm A\) until one minute; a vertical drop to \(0.30\,\mathrm A\); then exponential decay toward zero.

At two minutes, \(I\) and \(I_2\) have upward jumps of only about \(3.0\,\mathrm{nA}\), invisible on the supplied scale. All visible transients occupy only a small fraction of a minute.

3) Reducing the persistent current to zero

The current restriction applies to the magnitude of the normal-state switch current: \[ |I_2|<0.5\,\mathrm A. \] Since \[ I_2=\frac{L}{r_n}\frac{dI_1}{dt}, \] it requires \[ \boxed{\left|\frac{dI_1}{dt}\right| <\frac{r_n}{L}(0.5\,\mathrm A) =0.25\,\mathrm{A\,s^{-1}}.} \]

Heating the switch immediately in the initial persistent state would put \(20\,\mathrm A\) through it. We must first transfer its current to the external source while it is still superconducting.

Here is one definite schedule for the \(20\,\mathrm A\) example.

From 0 to 3 min: retain the given persistent state: \[ I=0,\qquad r=0,\qquad I_1=20\,\mathrm A,\qquad I_2=-20\,\mathrm A. \]

From 3 to 6 min: keep \(r=0\), close \(K\) with sufficiently large \(R\), and raise \(I\) linearly from zero to \(20\,\mathrm A\). Then \[ I_1=20\,\mathrm A,\qquad I_2=I-20\,\mathrm A. \] Thus \(I_2\) rises linearly from \(-20\,\mathrm A\) to zero. At six minutes the source supplies the whole magnet current.

From 6 to 9 min: set \(r=r_n=5\,\Omega\) and reduce the total current linearly from \(20\,\mathrm A\) to zero.

Let \[ u=t-360\,\mathrm s,\qquad T=180\,\mathrm s,\qquad a=\frac{20\,\mathrm A}{T}=\frac19\,\mathrm{A\,s^{-1}}. \] During this interval, \[ I=20\,\mathrm A-au. \] Solving \(\tau\dot I_1+I_1=I\), with \(I_1(0)=20\,\mathrm A\), gives \[ \boxed{ \begin{aligned} I_1&=20\,\mathrm A-au+a\tau(1-e^{-u/\tau}),\\ I_2&=-a\tau(1-e^{-u/\tau}), \end{aligned} \qquad \tau=2.0\,\mathrm s.} \]

Consequently, \[ |I_2|\le a\tau=\frac29\,\mathrm A \approx0.222\,\mathrm A<0.5\,\mathrm A. \] The magnet current follows slightly above the declining source current.

At 9 min and afterward: the source current has reached zero. Open \(K\), leaving the switch normal. The remaining magnet current is \[ D=\frac29(1-e^{-90})\,\mathrm A\approx0.222\,\mathrm A. \] It circulates through the magnet and normal switch and decays as \[ \boxed{ I=0,\qquad I_1=D e^{-(t-540\,\mathrm s)/\tau},\qquad I_2=-D e^{-(t-540\,\mathrm s)/\tau}. } \] After another \(10\,\mathrm s\), \(I_1\approx1.5\,\mathrm{mA}\), and it continues toward zero.

The complete Fig. 4 plots are therefore:

Plot Required shape
4a: \(I\) Zero until 3 min; linear rise to \(20\,\mathrm A\) at 6 min; linear fall to zero at 9 min; zero afterward.
4b: \(r\) Zero until 6 min; step to \(5\,\Omega\); remain normal during the final discharge.
4c: \(I_1\) \(20\,\mathrm A\) until 6 min; smoothly enters an almost linear decline, reaching approximately \(0.222\,\mathrm A\) at 9 min; then decays exponentially to zero.
4d: \(I_2\) \(-20\,\mathrm A\) until 3 min; linear rise to zero at 6 min; becomes slightly negative, approaching \(-0.222\,\mathrm A\); after 9 min returns exponentially toward zero from below.

As in any passive \(RL\) discharge, exact zero is approached asymptotically. On the minute-scale plot, the final tail is indistinguishable from zero. Returning the switch to \(r=0\) before that tail has decayed would trap the remaining current.

For a different initial \(i_1\), the same procedure applies with a sufficiently slow ramp to respect the \(0.25\,\mathrm{A\,s^{-1}}\) bound.

4) Changing the persistent current from \(20\,\mathrm A\) to \(30\,\mathrm A\)

Again, first match the source current to the existing magnet current before heating the switch.

We can choose an exactly linear magnet-current ramp lasting three minutes: \[ \alpha=\frac{30-20}{180}\,\mathrm{A\,s^{-1}} =\frac1{18}\,\mathrm{A\,s^{-1}}. \] The required switch current during this ramp is \[ \boxed{\delta=\frac{L}{r_n}\alpha =\frac19\,\mathrm A\approx0.111\,\mathrm A,} \] well below the normal-state limit.

To produce this ramp exactly, prescribe \[ I=I_1+\delta. \] For \(u=t-360\,\mathrm s\), \(0<u<180\,\mathrm s\), \[ \boxed{ I_1=20\,\mathrm A+\alpha u,\qquad I_2=\delta,\qquad I=20\,\mathrm A+\alpha u+\delta. } \] Indeed, \(r_nI_2=r_n\delta=L\alpha=L\,dI_1/dt\).

A complete schedule, specifying all four plots, is:

Time (min) \(I\) (A), Fig. 5a \(r\) (\(\Omega\)), Fig. 5b \(I_1\) (A), Fig. 5c \(I_2\) (A), Fig. 5d
0–3 \(0\) \(0\) \(20\) \(-20\)
3–6 Linear \(0\to20\) \(0\) \(20\) Linear \(-20\to0\)
6–9 Linear \(20+\delta\to30+\delta\) \(5\) Linear \(20\to30\) \(\delta=1/9\)
9–12 \(30\) \(5\) \(30\) \(0\)
12–15 Linear \(30\to0\) \(0\) \(30\) Linear \(0\to-30\)
After 15 \(0\) \(0\) \(30\) \(-30\)

Here the numerical value of \(\delta\) in the table is in amperes.

The switching details matter:

  • At 6 min: \(I=I_1=20\,\mathrm A\), so \(I_2=0\). Make the switch normal, then increase \(I\) by the small amount \(\delta\). This starts the linear magnet-current ramp without any jump in \(I_1\).
  • At 9 min: \(I_1\) has reached exactly \(30\,\mathrm A\). Reduce \(I\) from \(30+\delta\) to \(30\,\mathrm A\). Then \(I_2=0\), so the magnet current remains constant.
  • At 12 min: return the switch to \(r=0\). Reduce \(I\) to zero over the next three minutes. The magnet current stays fixed at \(30\,\mathrm A\), while \(I_2\) becomes increasingly negative.
  • At 15 min: open \(K\). The magnet and superconducting switch carry equal and opposite currents of magnitude \(30\,\mathrm A\).

Thus Fig. 5a includes small \(0.111\,\mathrm A\) steps at 6 and 9 min. Fig. 5d has a small positive rectangular pulse during 6–9 min. The large negative switch currents occur only while \(r=0\), so they do not violate the stated normal-state restriction.

The solution was checked against the official solution. Where this solution differs from the official solution, the official solution contains an error or a misprint.