Analytical approximations for the self-field distribution of a superconducting tape between iron cores

The method of images is a standard method in electromagnetics to calculate the field without solving the Poisson's or Laplace's equations. It is based on the uniqueness theorem of the Poisson's equation [21]. We only give the conclusions, which will be used in this paper. More details about this method can be found in [22].

As shown in figure 3(a), the lower and upper half-planes are filled with media having a permeability of μ₀ and μ₁, and there is an arbitrary current density J in the medium μ₀. To calculate the magnetic field distribution in the medium μ₀, the method of images can be used. As shown in figure 3(b), the entire plane is now filled with medium μ₀. Compared with (a), the Poisson's equation in the lower half-plane is the same. To maintain the boundary conditions of B and H unchanged, an image current density J' is placed in the upper half-plane at the same distance to the boundary as the source current density J. And J' should satisfy

$$\begin{align}J^{^{\prime} } = \frac{{{\mu _1} - {\mu _0}}}{{{\mu _1} + {\mu _0}}}J.\end{align} \tag{ 6 }$$

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Now, with the same Poisson's equation and boundary conditions, the field distributions in the lower half-plane for the two cases in figure 3 are the same. The distribution of B or H in (b) can be easily calculated using the Biot-Savart law since it is the superposition of the fields generated by the image and source current density.

In the open-loop case, there are two media μ₀ and μ_1, and two parallel boundaries. According to the method of images, the HTS tape J should have two images to the two boundaries, J₁ and J₂, and

$$\begin{align}{J_1} = {J_2} = \frac{{{\mu _1} - {\mu _0}}}{{{\mu _1} + {\mu _0}}}J.\end{align} \tag{ 7 }$$

One should note that the image current density J₁ and J₂ should also have images J₃ and J₄. By repeating this process, a series of an infinite number of images is obtained as shown in figure 4. The values of these image current densities are

$$\begin{align}{J_{2n - 1}} = {J_{2n}} = \left(\frac{{{\mu _1} - {\mu _0}}}{{{\mu _1} + {\mu _0}}}\right)^nJ \qquad n = 1,{ }2,{ }3,{ } \ldots \ldots \end{align} \tag{ 8 }$$

Now the problem has been converted into calculating the self-field of a tape stack containing an infinite number of tapes. A Cartesian coordinate system is chosen as shown in figure 5. The tape width is 2a. The magnetic field H at point P (x, y) is the superposition of the field generated by the source current J and all the image current density J_n. According to the Ampere's law (or the Biot-savart law), the y component of the magnetic field strength H generated by the element (η) is

$$\begin{align}{h_y} = \frac{{(x - \xi ) \cdot J}}{{2\pi \left[ {{{(x - \xi )}^2} + {{(y - \eta )}^2}} \right]}}.\end{align} \tag{ 9 }$$

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Therefore, the y component of the field at P (x, y) generated by the source tape (from -a to a) is

$$\begin{align} {h_{y\_{\text{source}}}} = \int\limits_{ - a}^a {{h_y}} d\xi = - \frac{J}{{4\pi }}\ln \frac{{{{(x + a)}^2} + {y^2}}}{{{{(x - a)}^2} + {y^2}}} .\end{align} \tag{ 10 }$$

One should note that in the two equations above, J is the current line density, and

$$\begin{align}J = \frac{I}{{2a}}\end{align} \tag{ 11 }$$

where I is the DC current in the HTS tape. Using equations (8) and (10), the y component of the self-field of the infinite tape stack can be obtained by superposing all the currents

$$\begin{align}{H_y} = - \frac{I}{{8\pi a}}\mathop {\lim }\limits_{N \to \infty } \mathop {\sum }\limits_N^{n = - N} \left(\frac{{{\mu _1} - {\mu _0}}}{{{\mu _1} + {\mu _0}}}\right)^{\left| n \right|} \cdot \ln \frac{{{{(x + a)}^2} + {{(y - n \cdot g)}^2}}}{{{{(x - a)}^2} + {{(y - n \cdot g)}^2}}}\end{align} \tag{ 12 }$$

where N is an integer. One should note that equation (12) can only be used to calculate the field distribution in the air gap, but not in the iron core. In a similar way, the x component H_x can be calculated. However, since H_y is dominant in the air gap and the impact of the parallel field on the HTS tape is small [23], H_x is ignored in this paper.

Equation (12) shows that H_y is a 2D infinite series function dependent on several parameters, I, a, μ, μ₀, and g. It is straightforward to use software (e.g. MATLAB) to calculate the exact value of this solution by assigning N a rather great value (e.g. 10 000). However, it is possible to further simplify this solution if more assumptions are taken.

In practice, the permeability of the iron μ₁ is usually much greater than that of the vacuum μ₀, which means equation (8) can be approximated as

$$\begin{align}{J_{2n - 1}} = {J_{2n}} \approx J \qquad n = 1,{\text{ }}2,{\text{ }}3,{\text{ }} \ldots \ldots\end{align} \tag{ 13 }$$

And for an infinite tape stack, the thickness is infinite, which is much greater than the width of the stack, 2a. Therefore, as shown in figure 6, the self-field of such a stack can be approximated as the self-field of a line current with an infinite length in the y-direction and zero-width in the x-direction. The field distribution of such an infinite line current can be easily calculated using Ampere's law. The magnetic flux density on the left and right half-planes are in the opposite directions and of the same value, which is

$$\begin{align}{B_{\max }} = \frac{{{\mu _0}I}}{{2g}}.\end{align} \tag{ 14 }$$

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Equation (14) is much simpler than (12), but it can be only used to calculate the field distribution when |x| ⩾ a since the width of the tape stack is ignored. The problem we are solving is to calculate the field on the surface of the tape, which means that equation (14) is useful to obtain the flux density on the left and right edges of the tape where |x| = a. When |x| < a, the flux density should be smaller than that on the edges, so we name the flux density B_max in equation (14). With this maximum value known, the field distribution from -a to a can be fully obtained if we know the field changing trend when |x| < a.

For the infinite stack in figure 5, we have the Ampere's law

$$\begin{align}\frac{{d{B_y}}}{{dx}} - \frac{{d{B_x}}}{{dy}} = {\mu _0}{J_{\text{s}}}\end{align} \tag{ 15 }$$

where J_s is the average current density of the stack and

$$\begin{align}{J_{\text{s}}} = \frac{I}{{2ga}}.\end{align} \tag{ 16 }$$

Here we have taken a further assumption that the current density is also homogeneous along the y-direction, which is sensible because the gap length g is usually much smaller than the tape width 2a. Now, the model resembles the 'infinite slab' in the Bean model [24]. When B is homogeneous along the y-direction in this infinite tape stack, equation (15) becomes

$$\begin{align}\frac{{d{B_y}}}{{dx}} = {\mu _0}{J_{\text{s}}}.\end{align} \tag{ 17 }$$

This means that B_y changes linearly along the x-direction with a constant slope. Given that it is obvious that B_y is 0 at x = 0, the perpendicular field distribution on the tape surface can be obtained as

$$\begin{align}{B_y} = - \frac{{{\mu _0}I}}{{2ga}}x\end{align} \tag{ 18 }$$

where |x| < a. This result agrees with equation (14) at |x| = a.

In the closed-loop case, the iron is connected at infinity, which means there is a second magnetic circuit in addition to the one through the air gap. As shown in figure 7, the two magnetic circuits are parallel and thus independent of each other. Therefore, the magnetic field in the air gap should be the superposition of the field calculated from these two magnetic circuits.

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Since magnetic circuit 1 is the same as the open-loop case, the analytical solution to the open-loop case can be used directly. According to the simplified solution, the perpendicular magnetic flux density on the tape surface should be a straight line along the x-direction as shown in figure 8. The maximum value B₁ can be known from equation (18) as

$$\begin{align}{B_1} = \frac{{{\mu _0}I}}{{2g}}.\end{align} \tag{ 19 }$$

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Superposed by magnetic circuit 2, the field of the closed-loop case should be of the same profile as the open-loop case, but with a DC offset B₂, which can be calculated from magnetic circuit 2. If we introduce an unknown k to describe the length ratio of the tape where the magnetic flux density is positive, an equation can be obtained from the similar triangles in figure 8.

$$\begin{align}\frac{{{B_1} - {B_2}}}{{k \cdot 2a}} = \frac{{{B_1}}}{a}\end{align} \tag{ 20 }$$

which can be simplified as

$$\begin{align}\frac{{{B_2}}}{{{B_1}}} = 1 - 2k.\end{align} \tag{ 21 }$$

One should note that the magnetomotive force (MMF) of magnetic circuit 2 is kI rather than I. According to the Ohm's law in magnetic circuits, we have

$$\begin{align}{B_2} = \frac{{kI}}{{\left(\frac{{{l_i}}}{{{\mu _1}}} + \frac{g}{{{\mu _0}}} + {R_{\text{m}}}\right)}}\end{align} \tag{ 22 }$$

where R_m is the magnetic resistance at infinity. In normal cases, μ₁ is much greater than μ₀. If we assume at infinity there is an air gap with a length of g' (which is not shown in figure 7), the equation can be approximated as

$$\begin{align}{B_2} = \frac{{kI}}{{\left(\frac{g}{{{\mu _0}}} + \frac{{g^{^{\prime} }}}{{{\mu _0}}}\right)}} \approx \frac{{{\mu _0}kI}}{{g + g^{^{\prime} }}}.\end{align} \tag{ 23 }$$

Combing equations (19) and (23), we have

$$\begin{align}\frac{{{B_2}}}{{{B_1}}} = 2k\frac{g}{{g + g^{^{\prime} }}}.\end{align} \tag{ 24 }$$

Solving simultaneous equations (21) and (24), B₂ and k can be calculated as

$$\begin{align}k = \frac{{g + g^{^{\prime} }}}{{2(2g + g^{^{\prime} })}}\end{align} \tag{ 25 }$$

$$\begin{align}{B_2} = \frac{{{\mu _0}I}}{{2(2g + g^{^{\prime} })}}.\end{align} \tag{ 26 }$$

Now we can conclude that the field distribution along the tape width direction for closed-loop switches is a straight line from B₁–B₂ to -B₁–B₂, with B₁ and B₂ both known.

The two cases in figure 1(a) are different in terms of the self-field distribution discussed in this paper. For the open-loop case, the iron core and the tape along the tape width direction are symmetric, which fits the model of the open-loop case in figure 2; for the closed-loop case, the distribution is asymmetric, because there is an additional magnetic circuit in the iron core allowing the magnetic flux to form a closed loop around the tape, which fits the model of the closed-loop case in figure 2. The parameters of the switches are shown in table 1. When the charging process is completed, there is no current in the copper coils of the magnetic switches and a DC current has been injected in the tape as part of the HTS magnet. With the analytical approximations, the perpendicular field distribution on the tape surface can be calculated. With the calculation result and the I_c-perpendicular field curve obtained from the HTS tape manufacturer, the I_c reduction can be determined.

To verify the analytical solution, a full-scale 3D FEM model is built in COMSOL Multiphysics. This model uses the T–A formulation [25] to introduce the non-linear E–J relationship (27) into equations (1)–(5). In the T–A formulation, the current vector potential T is used to calculate the current density J , and the magnetic vector potential A is used to calculate the magnetic flux density B . The E–J relationship is

$$\begin{align}{\boldsymbol{E}} = \rho {\boldsymbol{J}} = {E_0}\frac{{\boldsymbol{J}}}{{{J_c}(B)}}\left(\frac{{|{\boldsymbol{J}}|}}{{{J_c}(B)}}\right)^{n - 1}\end{align} \tag{ 27 }$$

where E₀ = 1 × 10⁻⁴ V m⁻¹. ρ is the resistivity of the superconductor. The value of n can be found in table 1. J_c(B) is the perpendicular field-dependent critical current density and is obtained from AMSC. It should be noted that the AMSC HTS tapes used in this paper is undoped and the critical current is maximumly reduced when the external field is perpendicular to the ab plane of the tape. The parallel field dependency is ignored in this paper [23]. In this model, the thickness of the HTS layer is ignored [25]. Rather than the constant permeability, the B–H curve of the iron core is used for an accurate result. The HTS tape length in the model is 10 cm, which makes the quench voltage criterion 10 μV. The voltage is calculated in the model by the integral

$$\begin{align}V\_{\text{tape}} = \frac{{\mathop {\iint }\limits_S {E_z}dS}}{{w\_{\text{tape}}}}\end{align} \tag{ 28 }$$

where S is the total area of the tape, and w_tape is the HTS layer width (10 mm). The tape length is in the z-direction. The air gap length g is set as 0.2 mm, the same as the tape thickness, which means there is no extra space between the tape and the iron core.

According to the simulation results, when the DC currents in the tape are 160 A for the open-loop case and 150 A for the closed-loop case, V_tape reaches the quench voltage criterion criterion of 10 μV. If we input the corresponding parameters into the analytical solution (12), the distribution of B_y long the x-direction can be obtained. For the open-loop case, the comparison of the analytical and simulation results is shown in figure 9. When calculating equation (12), the integer N must be a certain value instead of infinity. Therefore, the results of different N are given in the figure. As N becomes larger, the curve is closer to the simulation result. The fact that curves for N= 1001 and N= 10 001 are almost overlapped indicates that the tapes from the 1000th away in the infinite tape stack hardly influences the field distribution around the source tape. The analytical solution shows the field curve is linear along the x-direction, which agrees with the simplified solution (18). However, the simulation result shows that the curve is not completely linear. This difference is caused by the assumption that the current density is homogeneous along the x-direction, which does not completely represent the real case. Besides this, the analytical results and the simulation show good consistency. As shown in figure 9. The maximum value of B_y is around 0.5 T. This value can also be calculated from the simplified solution (18) as 0.503 T.

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For the closed-loop case, there is an identical air gap in the left part of the iron core, which means g' = g in equations (25) and (26). Then we have

$$\begin{align}\frac{{{B_2}}}{{{B_1}}} = k = \frac{1}{3}\end{align} \tag{ 29 }$$

where B₁ = 0.47 T, which can be obtained from the open-loop case when the current is 150 A. The solution is a straight line from B₁-B₂ (0.31 T) to -B₁-B₂ (0.63 T). The comparison of the analytical and simulation results is shown in figure 10. The tape length ratio k can be seen as 1/3 from the figure, and again, a good consistency can be observed.

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Now, with the analytical approximations for both cases in figure 1(a), we can determine the critical current reduction of the HTS tape based on the given conditions. If the air gap length is fixed, for any given current value I, the perpendicular field distribution along the tape width can be calculated. Since the field is nearly linearly distributed, the average field can be easily obtained. Then, with this average perpendicular field, a corresponding critical current can be found from the J_c(B) curve provided by the tape manufacturer. Therefore, the critical current of the tape is dependent on the current flowing inside it and can be seen as a function of the current I_c (I).

Figure 11 shows the function I_c (I) for both cases in figure 1(a). To find the degraded critical current of the tape, we need to find the point where

$$\begin{align}{I_{\text{c}}}(I) = I\end{align} \tag{ 30 }$$

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which is also shown in figure 11. For the open-loop case, the degraded critical current is 159 A, which means within this point, the critical current is larger than the current and the tape is in the 'safe region'; beyond this point, the critical current would be lower than the current, and the tape quenches. This result also agrees with the simulation results above, which indicate a current 160 A when the voltage reaches the quench criterion of 1 × 10⁻⁴ V m⁻¹.

For the closed-loop case, the critical current is 153 A, slightly smaller than that of the open-loop case. The small difference seems unreasonable because the DC offset B₂ in the open-loop case is 1/3 of B₂. But if we calculate the average absolute value of the perpendicular magnetic flux density for the two cases, we have

$$\begin{align}\left\{\begin{aligned} {B_{{\text{avg}}\text{-}{\text{I}}}} &= \frac{{{B_1}}}{2} \\ {B_{{\text{avg}}\text{-}{\text{II}}}} &= \left(1 + k - 2{k^2}\right)\frac{{{B_1}}}{2} \\ \end{aligned} \right.\end{align} \tag{ 31 }$$

where k = 1/3. So

$$\begin{align}\frac{{{B_{{\text{avg-II}}}}}}{{{B_{{\text{avg-I}}}}}} = \frac{{10}}{9} = 1.\dot 1\end{align} \tag{ 32 }$$

which indicates that the impact of k is quite small on the average perpendicular field to the tape. This justifies the small critical current difference between the two cases.

To verify the results above, we measured the critical currents of the taps in the two switches in figure 1(a). The distance between the two voltage taps is 10 cm. The results are shown in figure 12. The critical current for the open-loop case is measured as 136 A, and that for the closed-loop case is slightly smaller. The self-field critical current of the tape is measured as 248 A, and that for a half iron core is slightly smaller. This means if the magnetic circuit is not closed, it could barely influence the critical current of the tape. The experimental results (136 A) show there is an error of around 17% for the analytical solution (159 A) and the 3D FEM model (160 A). This error suggests that the practical current reduction is even worse than that seen in the analytical approximations. This is sensible because, except for the experimental errors and the assumptions we take to simplify the problem, the analytical approximations only consider the electromagnetic aspect of the problem. There are other factors that impact the HTS tape performance, e.g. mechanical factors. For instance, according to the analytical and simulation results, when the current is around 160 A in the tape, the maximum magnetic flux density on the tape surface is around 0.5 T. There is a large attractive force between the two iron cores, which is applied to the tape in the middle. The impact of this force on the tape performance is not considered in the analytical approximations and the simulation model. Therefore, the current reduction suggested by the analytical approximations in this paper is only a lower limit, while the practical reduction is usually worse.

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The magnetic switches are supposed to generate voltage high enough to charge the magnet when used in a flux pump. Previous studies [17] show that the voltage is proportional to the exposed length of the tape where the AC field is applied, which is the contact length of the iron core and the tape in figure 1(a). From the simulation results obtained from the 3D model, figure 13, we can also see the current density distribution is rather homogeneous along the tape length direction, which means the voltage generated by the tape per unit length should be equal. The current density ratio J/J_c(B) of the tape contacting the iron core is higher than 1 because this part of the tape needs to generate a voltage high enough to reach the quench criterion of 10 µV for the entire tape length (10 cm) while the rest of the tape generates almost zero voltage with a ratio below 0.8.

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However, since the analytical approximations are based on a 2D infinite long model, it seems impossible to investigate the impact of this contact length on the critical current reduction. Therefore, the contact lengths are changed in the 3D FEM model of the closed-loop case, and the results are shown in and table 2.

From the table we can tell the critical current is not sensitive to the change in the contact length. The original length of the iron core is 35 mm, and with the length reduced by half twice, the critical current only increases by 2 A and 1 A. The reason for this lies in the non-linear E–J characteristic of the HTS tape (27). When the length is reduced from 35 mm to 17.5 mm, if the current is maintained at 150 A, the voltage would decrease from 10 µV to 5 µV. To compensate for this 5 µV difference, the current must increase. However, the tape is already working at the steep region of the E–J power curve, which means even if the current increases slightly, the voltage will increase sharply. Therefore, as shown in table 2, the 2 A current will generate this 5 µV to make the total voltage back to the criterion of 10 µV. Thanks to this insensitivity of the critical current reduction to the contact length, the 2D analytical approximations are still able to give a good estimation for magnetic switches with different contact lengths.

According to the magnetic circuit law, if the MMF of the copper coils of a magnetic switch is fixed, the magnetic flux density in the air gap is approximately inversely proportional to the air gap length. Therefore, a small g means that the effective AC field applied to the tape can be increased, which can be used to generate a high voltage to charge the magnet. However, the analytical approximations also show that a small air gap length can cause a large critical current reduction. Using the method shown in figure 11, we have calculated the critical currents for different air gap lengths, and the results are shown in figure 14. The reduction for a g smaller than 0.5 mm can be over 30%, which must be considered in the design process of the PCSs or the flux pumps. When the air gap length is larger than 1 mm, the reduction is only less than 10%. Hence, we recommend the air gap length in a magnetic switch should be larger than 1 mm to mitigate the reduction, given that it is very unlikely the critical current of a magnet can reach 90% of the tape critical current.

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