Minimum volume of the magnetic shield

The first step in estimating the minimum volume is to estimate the minimum magnetic field necessary to deflect the solar wind at Mars’ orbit. To arrive at the minimum field required, we relate the ram pressure of the solar wind to the magnetic pressure at Mars. The magnetic field will deflect the solar wind when the magnetic pressure, B(r) ²/8π is of order the ram pressure of the solar wind, P _ram ~ ρv ²:

(1)$$\rho v^{2} \approx {B( r) ^{2}\over 8 \pi} .$$

In these expressions, ρ is the density of the solar wind, v is the wind speed and B is the magnitude of the magnetic field. Earth's magnetopause successfully deflects the solar wind; therefore, we derive the requirements on the strength of the Martian magnetic shield by scaling the conditions of Earth's magnetopause. A key parameter in the upcoming calculations is r _s, the standoff radius. The standoff radius is the distance from planet centre to the magnetopause, so a careful choice of this parameter affects the field intensity required. If we assume a dipole field for Earth's magnetic field, then the balance in pressures, equation (1), is

(2)$$P_{{\rm ram}, \oplus} \approx {B( r) ^{2}\over 8 \pi} = {B_\oplus^{2}\over 8\pi}\left({R_\oplus\over r_{\rm s}}\right)^{6} $$

where $R_\oplus$ is the Earth radius, r _s is the standoff radius of Earth's magnetopause and $B_\oplus$ is Earth's average magnetic field strength at the surface. The ram pressure of the solar wind as a function of distance, D, is P _ram ≈ ρv ². The mass loss rate from the Sun is $\dot {M} = \rho 4 \pi D^{2} v$. Assuming that $\dot {M}$ and v are constant at the distances of Earth and Mars, the ram pressure as a function of distance is

$$P_{\rm ram} \approx {\dot{M}v\over 4 \pi D^{2}} \Rightarrow P_{\rm ram, \ \male} = P_{\rm ram, \ \oplus} \left({D_\oplus\over D_{\male}} \right)^{2} ,\;$$

where $D_\oplus$ is the distance from the Sun to Earth, and $D_{\male}$ is the distance from the Sun to Mars. Now, having related the ram pressure to both the magnetic field and the distance from the Sun, we calculate the required strength of the Martian magnetic shield. Once again, we assume a dipole magnetic field and scale the strength of the magnetic field at the surface of Mars: $B_{\rm shield} = B_0 ( R_{\male}/r_{\rm s}^{\prime }) ^{3}$, where $R_{\male}$ is the radius of Mars and $r_{\rm s}^{\prime }$ is the standoff radius of the Martian magnetopause. This yields,

(3)$$B_0^{2} \left({R_{\male}\over r_{\rm s}^{\prime}}\right)^{6} = B_{\oplus}^{2} \left({R_{\oplus}\over r_{\rm s}} \right)^{6} \left({D_{\oplus}\over D_{\male}}\right)^{2}.$$

Roughly, $R_{\male} \approx R_{\oplus }/2$ and $D_{\male} \approx ( 3/2) D_{\oplus }$. Given these, the strength of the Martian magnetic shield is

(4)$$B_0 = {16\over 3} B_\oplus \left({r_{\rm s}^{\prime}\over r_{\rm s}} \right)^{3}.$$

Once again, $r_{\rm s}^{\prime }$ is the standoff radius of Mars’ magnetosphere, a free parameter, and r _s is the standoff radius for Earth's magnetopause, $r_{\rm s} = 6 R_\oplus$. Note that the minimum field strength depends on a choice of the Martian magnetopause, r _s′. To calculate the minimum requirements for the magnetic field, we choose a relatively small standoff radius for the Martian magnetopause. The absolute minimum would be $r_{\rm s}^{\prime } = R_{\male}$. However, such a small standoff radius would barely protect the Martian atmosphere, so we choose a slightly larger radius. To provide a little buffer, we choose $r_{\rm s}^{\prime } = 2R_{\male}$, which is $r_{\rm s}^{\prime } \approx R_{\oplus }$, since $2R_{\male} \approx R_{\oplus }$. This choice also simplifies the ratio of standoff radii, $r_{\rm s}^{\prime }/r_{\rm s} \approx 1/6$. Given these choices, the minimum field strength is $B_0 \approx B_\oplus /40$, which corresponds to a magnetic field strength of 1.25 μT at the Martian surface. This simple order-of-magnitude estimate for the necessary magnetic field strength is consistent with the results of magneto-hydro simulations (Green et al. Reference Green, Hollingsworth, Brain, Airapetian, Pulkkinen, Dong and Bamford2017). Next, we calculate the mass of the magnetic shield given B ₀ and B _c. In general, the magnetic field is a function of distance, B(r). At the edge of the superconducting magnet, the magnetic field can be no larger than B _c, yet the magnetic must be have a characteristic strength larger than B ₀ to deflect the solar wind. The design of the Martian magnetic shield is fundamentally a solenoid with N loops, each with a loop radius of a; the bundle of wires has a radius d. In the following derivation, we show that the constraints on the magnetic field impose constraints on the dimensions for the solenoid. For an illustration of the solenoid and magnetic field strength, see Fig. 1. Since the goal is to compress planetary scale currents and magnetic fields into as small a volume as possible, this solenoid will need to be made with superconducting wires.Using Biot–Savart's law, the magnetic field in the plane of the solenoid a distance r from the centre is

(5)$$\eqalign{ B( r) & = - {\mu_0 N I_{\rm c}\over 4\pi a} \int_0^{2\pi} {\left(( {r}/{a}) \cos\theta - 1 \right){\rm d}\theta\over \left(1 + ( {r}/{a}) ^{2} - ( {2r}/{a}) \cos\theta \right)^{3/2}}\cr & = - {\mu_0 NI_{\rm c}\over 4\pi a}f\left({r\over a}\right). }$$

The prefactor sets the overall scale of the magnetic field, and the dimensionless integral gives the radial dependence. In general, the integral, f(r/a) does not have a closed form, so we explore limiting cases to show that this equation reduces to well-known fields. These limiting cases are also useful in providing an analytic constraint equation for the magnetic shield. There are three limiting cases that are analytic and are useful in the design of the magnetic shield. First, at the centre of solenoid, r = 0, the magnetic field in equation 5) is

(6)$$\vert B( 0) \vert = {\mu_0 N I_{\rm c}\over 2 a} .$$

Second, in the far-field limit, r ≫ a, B(r) approaches the dipole approximation

(7)$$\lim_{r \gg a} \vert B( r) \vert \approx B( 0) \left({a\over r} \right)^{3} .$$

Third, for a solenoid with a large loop radius a (bundle radius d is much smaller than a) the magnetic field just outside the wire bundle approaches the limit for an long straight wire,

(8)$$\lim_{d \ll a} \vert B( a + d) \vert \approx {\mu_0 N I_{\rm c}\over 2 \pi d} .$$

Equations (6)–(8) represent three useful limiting cases that we will use later in the derivation. Before we use the approximate magnetic field expressions, we first use the exact expression, equation (5), to find algebraic expressions for the magnetic field constraints. For the first condition, the magnetic field at the surface of the wire bundle should be less than the critical magnetic field, B(a + d) < B _c. In this case, the exact equation, equation (5), becomes

(9)$$\vert B( a + d) \vert = {\mu_0 N I_{\rm c}\over 4 \pi a} f\left(1 + {d\over a}\right) = B_{\rm c} .$$

For the second condition, the magnetic field at a standoff radius of $r_{\rm s} = 2R_{\male}$ is the minimum (B ₀) required to deflect the solar wind, $B( 2 R_{\male}) = B_0$. In this case, the exact equation, equation (5), becomes

(10)$$\vert B( 2R_{\male}) \vert = {\mu_0 N I_{\rm c}\over 4 \pi a} f\left({2 R_{\male}\over a}\right) = B_0 .$$

The approximate analytic equivalents of the these two constraint equations are as follows. The equivalent of equation (9) is

(11)$$\vert B( a + d) \vert \approx {\mu_0 N I_{\rm c}\over 2 \pi d} \approx B_{\rm c} ,\; $$

and the equivalent of equation (10) is

(12)$$\vert B( 2R_{\male}) \vert \approx {\mu_0 N I_{\rm c}\over 2 a} \left({a\over 2 R_{\male}} \right)^{3} \approx B_0 .$$

Together, the exact constraint equations, equations (9) and (10), provide a joint constraint on the loop radius, a, and the wire bundle radius, d. Solving both equations for μ₀ N I _c/(4πa) and then equating them gives the following constraint

(13)$${\,f \left({2 R_{\male}}/{a}\right)\over f \left(1 + {d}/{a} \right)} = {B_0\over B_{\rm c}}.$$

Fig. 1.

Magnetic field strength as a function of distance from the centre of the solenoid. The solid curve represents the numerical integration of equation (5). We also show three standard approximations for solenoids. The central magnetic field, B(0), is roughly equal to the core magnetic field for a current loop. The magnetic field at the surface of the wire is roughly that of a straight wire. The dashed line is the far-field dipole approximation.

In summary, this constraint defines the magnetic shield dimensions required to achieve the magnetic field ratio, B ₀/B _c. It is a two parameter constraint that depends upon a and d. For a given loop radius, a, one may find the wire bundle radius, d, that satisfies the constraint. Again, due to equation (5) not having a closed form for intermediate values of r, we used numerical methods to both compute the magnetic field (Fig. 1) and the wire bundle radius, d (Fig. 2). Using the approximations for the magnetic field near the wire (equation (11)) and far from the wire equation (12), we derive an analytic expression for equation (13):

(14)$${B_0\over B_{\rm c}} = {\pi d a^{2}\over 8R_{\male}^{3}}$$

The material volume is dependent on the dimensions of the loop bundle as well as the loop radius,

(15)$$V_{\rm superconductor} = 2\pi^{2} d^{2} a$$

where we calculate d using the constraint equation, equation (13). Using the analytic constraint, equation (14), we can express the volume in terms of the magnetic field constraints and the loop radius,

(16)$$V_{\rm superconductor} = 128 \left({B_0\over B_{\rm c}} \right)^{2} {R_{\male}^{6}\over a^{3}} .$$

This equation shows that the physics of superconductors and solenoids apply strong constraints on the design of the magneto-shield. For one, the superconducting volume goes down as the critical magnetic field squared. The strongest dependence is on the loop radius. The material required for solenoid goes down as the loop radius cubed. Figure 2 shows both d and V as a function of a. The solid line shows the numerical result of solving equation (13) for d; the dashed line shows the result when using the analytic result, equation (14). As a reminder, constraints on the magnetic field are that $B_0 = B_\oplus /40$, and B _c = 200 T. The critical field value is of order the critical magnetic field for bismuth strontium calcium copper oxide (BSCCO), which was discovered to have one of the largest B _c for a superconducting wire (Golovashkin et al. Reference Golovashkin, Ivanenko, Kudasov, Mitsen, Pavlovsky, Platonov and Tatsenko1991). Another unique quality about BSCCO is that it is able to produce round wires, ideal for this type of shielding construction. Based on this result, if we wrapped a wire around Mars’ equator (operating near the critical field of ~200 T), the material volume would be roughly 2.25 × 10⁵ m³. Moreover, a smaller loop radius a requires a much larger wire thickness and hence volume of superconductor. For example, a loop radius of a ~ 10 km, would require a wire radius of d ~ 10 km, giving a volume of ~ 10¹² m³. An illustration of the different geometries we consider for the magnetic shield is shown in Fig. 3.

Fig. 2.

Physical constraints on the solenoid dimensions: wire radius (d) and superconducting material volume as a function of loop radius (a). The dashed purple line shows the analytic result, equation (7), and the solid yellow curve is the numerical result, equation (13). These dimensions are calculated by requiring the field strength to be $B( 2 R_{\male}) = B_\oplus / 40$ and below a critical superconducting field of 200 T at the wire. The left axis shows the superconducting bundle radius (d) as a function of the solenoid radius (a); the right axis shows the material volume (2π² d ² a) as a function of a. The required volume goes down for a wider loop. Therefore, the most optimum solution is to wrap the magnetic shield around Mars.

Fig. 3.

Martian magnetic shield designs. Panel (a) shows the current interaction between the Martian atmosphere and the solar wind. A lack of magnetic field allows the solar wind to ablate away the Martian atmosphere, making Mars inhospitable to life. The challenge of building a Martian magnetic shield is to build an electromagnet that can produce planetary scale magnetic fields with the least amount of superconducting material; however, the critical magnetic field of superconductors limits the size of the electromagnet. The mass of the electromagnet scales as $B_{\rm c}^{-2}a^{-3}$, where B _c is the superconducting critical magnetic field, and a is the loop radius. Panel (b) shows the most compact design possible: a solenoid with a wire radius and loop radius of order 10 km. Counter-intuitively, this ‘compact’ design would require the most superconducting material and would require mining 10% of Mars to extract the rare superconducting material. This is clearly an impractical design. Panel (c) shows a more practical design: the electromagnet is a loop of wire around the Martian equator. The wire radius is b = 5 cm and requires mining only 0.1% of Olympus Mons for the rare superconducting material.

Power requirements to keep the conducting wire cool

We now visit the power needed to maintain the superconductor at a temperature of order 100 K. The optimal electromagnet design is a superconducting wire around Mars that has a wire diameter of 5 cm. Insulating the superconducting wire will require building an insulating tube around the wire. No insulator is perfect and some heat will leak into the system. As a result, there will need to be active cooling systems placed along the wire, and these cooling systems will require some power to maintain the low temperatures. Typically the power required is of order the power of heat flowing through the insulator. For the purposes of this order-of-magnitude calculation, we imagine that the insulator is a 1 m diameter tube that envelopes the entire loop. The tube has a near vacuum. Except for the loop radius being the size of Mars, the other dimensions (tube diameter) and design are similar to standard dewars (vacuum flasks). The power of heat flow through an insulator is given by the Fourier equation for unidirectional conduction

(17)$${} \dot{Q} = {A\over L}\int_{T_0}^{T} k( T') {\rm d}T' = {A\over L}\bar{k}\Delta T$$

where A is the cross-sectional area, L is the length of the heat conduction path, $\dot {Q}$ is the heat flow rate, T is temperature, k is the thermal conductivity of the material and $\bar {k} = \int _{T_0}^{T} k( T') {\rm d}T'/\Delta T$ is the mean thermal conductivity of the material. Vacuum-insulated panels can achieve thermal conductivity as low as 0.004 W m⁻¹ K⁻¹ (Véjelis et al. Reference Véjelis, Gailius, Véjeliene and Vaitkus2010). Using this value, a temperature difference ΔT = (T _ambient − T _c) ~ 173 K, a cross-sectional area of 10⁷ m², and a conduction path of 1 m, we obtain a vacuum dewar heat flow rate of 1 MW. This is the average power capacity of a wind turbine. Hence, the primary challenge in constructing the electromagnet is not in cooling it.

Can the device fit in a rocket? Critical magnetic field required to fit the electromagnet on today's rockets

To further illustrate the extreme constraints for building the Martian magneto-shield, we calculate what the required B _c would be if one built the superconducting wire on Earth, shipped it in a rocket and placed it a the L1 Lagrange point. This strategy requires fitting the wire inside the rocket fairing. Once completed, the SpaceX Starship will have the largest payload capacity. Starship has a 9 m diameter; for ease of calculation we will just assume that today's current rockets could fit a coil of superconductor that has a volume of 10³ m³. If one kept the superconductor compact with a ~ d ~ 10 m, then equation (14) indicates that the critical magnetic field would have to be 10¹¹ T or 10¹⁵ G. This is a magnetar field, and the energy density of such a field far exceeds the bulk modulus of any terrestrial solid. Clearly, this solution is not feasible. Another alternative is to transport the spool to Mars, unwind it to make a loop around Mars. With this configuration, the wire radius would have to be d ~ 7 mm, and the critical magnetic field would have to be B _c ~ 1400 T. This is a factor of ten larger than current B _c limits. Fitting the superconducting material in one of today's rockets is not feasible either. These estimates show that it is not possible to build the electromagnet on Earth and transport it to Mars. Rather, the electromagnet must be constructed on or near Mars using local material.

Bulk rock mass required to mine superconducting material

Using the solar abundances shown in Fig. 4 (Cameron Reference Cameron1973), we estimate the amount of bulk rock required to mine the elemental ingredients of the superconductor. For BSCCO superconductors, the rarest element is bismuth (Bi), so its abundance would dictate the bulk rock required. Asteroids are often dominated by carbonaceous materials, so the relative ratio of rock to Bi is roughly given by the abundance ratio of C to Bi. The larger planets are predominantly silicates, of which silicon is the most abundant element. Therefore, the ratio of Si to Bi gives a rough scaling for the ratio of rock to Bi. For the optimum magneto-shield design (loop around Mars’ equator), the superconductor volume is 2 × 10⁵ m³. Assuming an average rock density of 3 g cm⁻³, the superconducting mass is 6 × 10¹¹ g. The ratio of mass in Bi to the bulk rock mass is given by the equivalent abundance ratios:

(18)$${M_{\rm Bi}\over M_{\rm C/Si}} = {X_{\rm Bi}\over X_{\rm C/Si}}$$

Here, M _Bi and M _C/Si are the Bi ore masses and silicate/carbon masses, respectively. Given these ratios, the mass of bulk rock as a function of the loop and bundle radii is:

(19)$$M( a,\; d) = 2\pi^{2} \rho a d^{2} {X_{\rm C/Si}\over X_{\rm Bi}} .$$

Fig. 4.

Relative abundance of chemical elements in the Solar System (Cameron Reference Cameron1973). The x-axis represents atomic number and the y-axis represents abundance of elements for every 10⁶ atoms of Si. The relevant elements for this paper are Si, a primary constituent in silicate bodies such as Earth, and C, a primary constituent of most asteroids. The rare elements Bi and Y represent two important elements in the construction of high T _c superconductors (e.g. the magnetic shield). Figure credit: public domain.

Figure 5 shows the required superconductor mass and bulk rock mass as a function of the loop radius, a. In addition, to Bi, we also calculate the bulk rock required if the limiting element for the superconductor is yttrium or carbon. Yttrium is the limiting element for yttrium barium copper oxide (YBCO), which is a ceramic superconductor that has a high critical temperature (~93 K) and magnetic field (Wu et al. Reference Wu, Ashburn, Torng, Hor, Meng, Gao, Huang, Wang and Chu1987) but is usually constructed in an anisotropic tape form as opposed to a round wire (e.g. Kametani et al. Reference Kametani, Jiang, Matras, Abraimov, Hellstrom and Larbalestier2015). Carbon is included for the case of carbon nanotubes; the benefit to carbon nanotubes is that carbon is many orders of magnitude more abundant than most superconducting materials, but the drawback is that carbon nanotubes are difficult to construct long wires and they have a low critical temperature ($\sim \!5\, \rm \, K$) and magnetic field ($\sim \! 4\, \rm \, T$). Figure 5 clearly indicates that the most feasible magneto-shield is one with a loop radius of order the size of the planet. If the loop radius is of order ~10 km and one fabricates the superconductor using Bi, then the amount of mined bulk rock is 10% of Mars. Even if the magneto-shield is composed of carbon nanotubes and a more common element (C), the amount of bulk rock is equivalent to the mass of a very large ~10 km asteroid. On the other hand, if the loop radius is a ~ 10³ km, then the amount of bulk rock required for Bi is 0.1% of Olympus MonsFootnote ¹, the largest mountain on Mars. In the case of carbon nanotubes, the bulk rock is only 10⁻⁶ of Olympus Mons. Therefore, building a loop around Mars is the most feasible solution when constructing a Martian magnetosphere.

Fig. 5.

Constraints on the superconductor mass as a function of electromagnet loop radius, a. The larger loop radius requires the least amount of superconducting mass. The three dots indicate three potential solutions. The annotations for each dot show estimates for the amount of bulk rock that is required to mine key elements for superconductors. For the smallest loop radius, a ~ 10 km, the construction project would need ~ 10¹⁹ g of Bi, and given Solar System abundances, this amount of Bi would require mining 7 × 10²⁵ g of bulk rock. This amount of bulk rock is equivalent to ~10% of Mars. Alternatively, a larger loop radius, the radius of Mars, would require ~ 10¹² g of Bi and ~ 5 × 10¹⁸ g of bulk rock, or 0.1% of Olympus Mons. The most reasonable solution is a magnetic shield wrapped around Mars.