Magnetobaric Battery

Magnetobaric Battery, or formally known as Superconducting Toroidal Loops Energy Storage, is an energy storage system that stores energy as magnetic pressure inside a looped superconducting coil.

Introduction

Superconductor is a material with the ability to conduct electrical currents with no resistance. It means any current sent through a superconducting wire will not dissipate into heat like any normal wire. Given that any electric current produce magnetic field lines perpendicular to the direction of the current, a wire could be coiled into a solenoid to concentrate the magnetic field lines into an electromagnet with two poles.

If both ends of the solenoid are connected together, a torus will form. Inside such torus, all magnetic field lines are contained within the interior. As more currents are pushed through the toroidal solenoid, the magnetic fields will push back harder. The pressure will force the toroidal solenoid to expand outward, as if a balloon the shape of a doughnut is filled with air.

The energy is contained as magnetic pressure inside the toroidal solenoid. Therefore one of the primary limit of such energy storage is how much pressure can the toroidal wall support before the structural integrity is compromised. This is why this type of energy storage is called a magnetobaric battery, a battery that stores energy as magnetic pressure.

If the solenoid is comprised of a superconducting material, for as long as the electrical circuit is closed, the electric current will continue to flow, and therefore energy is conserved. Such magnetobaric battery can last practically forever, for as long as the superconductivity is maintained. If the circuit is broken, or if the superconductivity is lost, energy will begin to dissipate out of the system.

To extract the energy out of the loop, the loop is disconnected by introducing another circuit where the electric current is converted into work. By doing so, the magnetic pressure within the loop will be relieved by the amount of energy extracted from it. Close the circuit again, and the battery will preserve the remaining magnetic pressure. Extract the energy out of the loop until the very last pascal, and the pressure will be completely gone.

Magnetic Pressure

In fluid dynamic, pressure is considered a measure of energy per unit volume, or formally known as energy density. Moreover, pressure as a unit of force per unit area, is dimensionally equivalent to energy density, as demonstrated below:

$$ P = \frac{F}{A} = \frac{F\times d}{A\times d} = \frac{W}{V} $$

Energy confined within a magnetobaric battery manifests as magnetic pressure pushing against the interior of the toroidal cavity. In practical units, magnetic pressure is described as:

$$P_B[Bar] = \left(\frac{B[T]}{0.501}\right)^2$$

Since the unit of the equation is in Bar, the conversion to SI unit is:

$$1 \text{ Bar} = 100 \text{ kPa} = 10^5 \text{ Pa}$$

Therefore, in SI unit, magnetic pressure is described as:

$$ P_B[Pa] = \left(\frac{B[T]}{0.501}\right)^2 \times 10^5 \text{ Pa} $$ $$ P_B[Pa] = B[T]^2 \times 398,404,787 \text{ Pa} $$

Reversing the process to obtain magnetic field strength $B[T]$, it becomes:

$$ B[T] = \sqrt{\frac{0.501^2 \times P_B[Pa]}{10^5\text{ Pa}}} $$ $$ B[T] = \sqrt{\frac{P_B[Pa]}{398,404.787\text{ Pa}}} $$ $$ B[T] \approx \sqrt{\frac{P_B[Pa]}{4\times10^5\text{ Pa}}} $$

Toroidal Solenoid

A magnetobaric battery’s primary structure is a torus shell, where its external volume can be described as:

$$V_{ext} = (\pi a^2)(2\pi b)$$ $$V_{ext} = 2\pi^2ba^2$$

Where:

• $V$ is the torus’ external volume.
• $a$ is the torus’ minor radius, and
• $b$ is the torus’ major radius

Magnetic pressure of the runninc current inside the solenoid would produce an outward push, toward the wall of the solenoid. If the wire is not strong enough, the wire would break, dissipating the energy within. Therefore, to reinforce the solenoid, it is supported by a stronger material, typically it is supported by a toroidal diamondoid shell. The support structure will then be the one pushing back against the magnetic pressure, and therefore maintain the structural integrity of the toroidal cell. The maximum energy density allowed in the torus is determined solely by the strength of the support structure.

The supporting structure, or the backend, would be in the shape of a torus as well, covering an inner solenoid torus. The thin superconducting films comprising the solenoid covers the interior wall of such toroidal shell, so the majority of the wall’s thickness is the backend. Therefore the battery can be modeled with just considering the thickness of the toroidal wall, and the tensile strength of the backend. Trivially, thicker and stronger backend equals more energy stored within the solenoid coil.

Therefore the shape can be modeled as a torus with interior cavity the shape of a torus with smaller minor radius. If the torus has a minor radius $a$, and a major radius $b$, the cavity would share the same major radius $b$, but the cavity minor radius ($c$) is smaller than $a$ ($c \lt a$). The value of $c$ can approach $a$, however since the wall’s thickness is nonzero and is always positive, $c$ is always less than $a$.

The torus wall’s thickness ($t$) can then be defined as:

$$t = a - c$$

The backend is a wall with thickness $t$ covering an interior cavity, having the minor radius $a$ and major radius $b$, comprising the exterior of the cell. The inner cavity is empty, usually a vacuum, shares the same major radius $b$, but the minor radius is $c$. Therefore:

$$V_{cavity} = (\pi c^2)(2\pi b)$$ $$V_{cavity} = 2\pi^2bc^2$$

Then the majority material volume of the cell (assuming the solenoid is very thin or of a negligible fraction of $t$) is the external volume $V_{ext}$ subtracted by its inner cavity $V_{cavity}$:

$$\sum V_{torus} = V_{ext} - V_{cavity}$$ $$\sum V_{torus} = (\pi a^2)(2\pi b) - (\pi c^2)(2\pi b)$$

Typically, to maximize battery cell volume, we want to maximize the minor radius to extinguish the empty center of the torus. The maximum value of the minor radius $a$ is where $a = b$ since $a \leq b$. Therefore, the equation can be simplified to:

$$\sum V_{torus} = (\pi a^2)(2 \pi a) - (\pi c^2)(2 \pi a)$$ $$\sum V_{torus} = 2a\pi^2 (a^2 - c^2)$$ $$\sum V_{torus} = 2\pi^2b(a - c)(a + c)$$

Then, the body of a torus is nothing much different to a cylindrical pipe with diameter $D$ and thickness $t$. The pressure in the interior of the torus will push away from the major radius, therefore exerting outward pressure to the solenoid interior. Since the body of the torus is a tube, the pressure will then induce tensile stress circumferentially along the length of the tube, that is the circumference of the major radius.

In engineering, this circumferential tensile stress is known as Tangential Stress ($\sigma_t$), that is the forces acting are the total pressure caused by the internal pressure and the total tension of the wall ($t$). It is described as follows:

$$\sigma_t = \frac{PD}{2t} = \frac{Pr}{t}$$

Where:

• $P$ is pressure,
• $D$ is diameter,
• $r$ is radius.

Pertaining to the examined toroidal cell, it contains a concentric toroidal cavity internally. It has been known that: $t = a - c$, and the radius of the tube is the minor radius of the torus ($r = c$), therefore:

$$ \sigma_t = \frac{Pc}{(a-c)} $$

To find the maximum energy density ($\rho_{energy}$) is to find the maximum internal pressure ($P_{max}$) the structure can withstand. The limit is determined by the material’s maximum tensile strength ($\sigma_{backend}$), that is when the maximum tangential stress is equal to the tensile strength ($\sigma_t = \sigma_{backend}$). The $P$ can then be calculated as:

$$\sigma_t \times (a-c) = P c$$ $$P = \frac{\sigma_t \times (a-c)}{c}$$

Substituting $\sigma_{backend}$ for $\sigma_t$ therefore indicate maximum energy density as:

$$\rho_{energy} = \frac{\rho_{backend} \cdot (a-c)}{c}$$

However it only consider energy density inside the cavity. To obtain averaged energy density of the entire torus, which is the amount of energy stored per external volume of the torus itself:

$$\rho_{energy[torus]} = \frac{E_{max}}{V_{ext}}$$ $$\rho_{energy[torus]} = \frac{2\pi^2 \sigma_{backend} b c (a-c)}{2\pi^2ba^2}$$ $$\rho_{energy[torus]} = \frac{\sigma_{backend} c (a-c)}{a^2}$$ This formula gives the energy density averaged over the entire volume of the torus. It’s a valuable metric for understanding how efficiently the torus’s overall volume is being used to store energy.

Specific energy for the torus ($S_E$) is defined by its maximum energy capacity ($E_{max}$) and the mass of the torus ($M_{torus}$).

$$ S_E = \frac{E_{max}}{M_{torus}} $$

Meanwhile maximum energy capacity is obtained as the product of the energy density ($\rho_{energy}$), and the volume of the cavity ($V_{cavity}$).

$$ E_{max} = \rho_{energy} \cdot V_{cavity} $$

By substituting $\rho_{energy}$ and $V_{cavity}$ into the equation:

$$E_{max} = \left( \frac{\sigma_{backend} \cdot (a-c)}{c} \right) \cdot (2\pi^2bc^2)$$

Then simplifying the expression yields:

$$E_{max} = \sigma_{backend} \cdot (a-c) \cdot (2\pi^2bc)$$ $$E_{max} = 2\pi^2 \sigma_{backend} b c (a-c)$$

The torus mass ($M_{torus}$) is the product of the mass of the torus ($\sum V_{torus}$) and the density of the backend material ($\rho_{backend}$).

$$ M_{torus} = \sum V_{torus} \cdot \rho_{backend} $$ Substituting for $\sum V_{torus}$ and$\rho_{backend}$ followed by a simplification of terms yields:

$$M_{torus} = \left( 2\pi^2b(a - c)(a + c) \right) \cdot \rho_{backend}$$ $$M_{torus} = 2\pi^2 \rho_{backend} b (a - c)(a + c)$$

Therefore, substituting the derived values,

$$ S_E = \frac{\rho_{energy} \cdot V_{cavity}}{\rho_{backend} \cdot \sum V_{torus}} $$ $$S_E = \frac{2\pi^2 \sigma_{backend} b c (a-c)}{2\pi^2 \rho_{backend} b (a - c)(a + c)}$$ $$S_E = \frac{\sigma_{backend} \cdot c}{\rho_{backend} \cdot (a + c)}$$

At last, the averaged density of the torus can be obtained by:

$$ \rho_{torus[avg]} = \frac{M_{torus}}{V_{ext}} $$ $$\rho_{torus[avg]} = \frac{2\pi^2 \rho_{backend} b (a - c)(a + c)}{2\pi^2ba^2}$$ $$\rho_{torus[avg]} = \frac{\rho_{backend} (a - c)(a + c)}{a^2}$$ $$\rho_{torus[avg]} = \rho_{backend} \frac{a^2 - c^2}{a^2}$$

Battery Cell

In the previous sections, we’ve explored the case of a single torus. However in practice, a number of torus is arranged into a hexagonal matrix grid, and then the grid is stacked atop one another. This section explores the stacking of magnetobaric battery tori into layered hexagonal grids, with the plane of the hexagon grid is parallel to the plane of the major radius. Therefore each torus can be considered to occupy a hexagonal prism, where the height is the minor diameter, and the entire major diameter is contained within the hexagonal plane.

So to calculate the volume of a hexagonal prism, first we need to know the hexagonal surface area, and multiply it by the height that is equal to the minor diameter of the torus, or the minor radius multiplied by two ($2a$).

The hexagonal area $A_{hexagon}$ correlates to the length of its sides $S$, whose relationship is determined by the following expression:

$$ A_{\text{hexagon}} = \frac{3\sqrt{3}}{2} \times S^2 $$

The hexagon must be able to cover the entire width of the torus major circle. The diameter of the major circle equals double the sum of the major radius and minor radius ($D = 2(a+b)$) and the external radius $r$ of a torus on the plane of its major radius is equal to the sum of major radius and minor radius ($r = a + b$). A hexagon can be constructed out of six equilateral triangles, and as each equilateral triangles is comprised of three sides of equal lengths, the length of the equilateral triangle’s side is equal to the length of the hexagon’s side. Given that the external radius of the torus intersects the interior side of the hexagonal prism, the radius is equal to the height of the equilateral triangle’s height ($r = h$, consequently, $h = a + b$).

The height of a triangle is the shortest length that can be drawn from an angle to a triangle to the side opposite the angle. The resultant length is then perpendicular to the side opposite the angle. For an equilateral triangle, the resultant length will intersect the side opposite the angle right at the middle, therefore dividing the side into two. Then the right angle triangle produced in this procedure with the hypotenuse $S$.

Since the hypotenuse $S$ is not known, and at this point only $h$ is known and the angle opposite of $h$ is known ($\theta = 60^\circ$), using trigonometry to find the value of $S$ is trivial:

$$ h = S \cdot \sin 60^\circ $$ $$ S = \frac{h}{\sin 60^\circ} $$

Since $\sin 60^\circ = \frac{1}{2}\sqrt{3}$, then:

$$ S = \frac{2h}{\sqrt{3}} $$

For the case $a = b$ (maximum torus volume for a given major radius), the height of the hexagonal prism would also be $h$. Since volume of a prism can be calculated by area multiplied to height, and a hexagonal prism is the shape of a single magnetobaric battery cell, we get:

$$ V_{\text{cell}} = h \times A_{\text{hexagon}} $$ $$ V_{\text{cell}} = h \times \frac{3\sqrt{3}}{2} \times S^2 $$ $$ V_{\text{cell}} = h \times \frac{3\sqrt{3}}{2} \times \left(\frac{2h}{\sqrt{3}}\right)^2 $$ $$ V_{\text{cell}} = h \times \frac{3\sqrt{3}}{2} \times \frac{4h^2}{3} $$ $$ V_{\text{cell}} = \frac{12\sqrt{3} \times h^3}{6} $$ $$ V_{\text{cell}} = 2\sqrt{3} \times h^3 $$

Since $h = a + b$, substituting the relationship into the volume equation:

$$ V_{\text{cell}} = 2\sqrt{3} \times (a + b)^3 $$

Then the difference between $V_{\text {cell}}$ and $V_{\text{ext}}$ can be filled with support substrates responsible for regulating and transferring energy flow to and from each torus in the stack. Therefore to get the volume of the support structure $V_{support}$:

$$ V_{\text{support}} = V_{\text{cell}} - V_{\text{ext}} $$

However, one of the most interesting property is the ratio between $V_{ext}$ and $V_{cell}$.

$$ \frac{V_{ext}}{V_{cell}} = \frac{(\pi a^2)(2\pi b)}{2\sqrt{3}\times (a+b)^3} $$ $$ \frac{V_{ext}}{V_{cell}} = \frac{2\pi^2 a^2 b}{2\sqrt{3}\times (a+b)^3} $$ $$ \frac{V_{ext}}{V_{cell}} = \frac{\pi^2 a^2 b}{\sqrt{3}(a+b)^3} $$

For the case where $a=b$,

$$ \frac{V_{ext}}{V_{cell}} = \frac{\pi^2 a^2 b}{\sqrt{3}(a+b)^3} $$ $$ \frac{V_{ext}}{V_{cell}} = \frac{\pi^2 a^3}{\sqrt{3}(2a)^3} $$ $$ \frac{V_{ext}}{V_{cell}} = \frac{\pi^2 a^3}{\sqrt{3}(8a^3)} $$

As $a \neq 0$,

$$ \frac{V_{ext}}{V_{cell}} = \frac{\pi^2}{8\sqrt{3}} $$

By numerically approximating $\pi^2 \approx 9.8698$ and $\sqrt{3} \approx 1.73205$,

$$ \frac{V_{ext}}{V_{cell}} \approx 0.7122 $$

It means about 71.22% of the cell volume is the torus volume. Meanwhile the remaining 28.78% is the support structure of the cell. Therefore to get the averaged density of the cell $\rho_{agg}$, we have to consider the ratio:

$$ \rho_{agg} = 0.7122\rho_{torus[avg]} + 0.2878\rho_{support} $$

However it is not unreasonable to assume $\rho_{torus[avg]} \approx \rho_{support}$ for simplicity, therefore:

$$ \rho_{agg} \approx \rho_{torus[avg]} $$

However, the aggregated specific energy $S_{E[agg]}$, would use $\rho_{agg}$ instead of $\rho_{backend}$:

$$ S_{E[agg]} = \frac{E_{max}}{M_{cell}} $$

As $E_{max}$ remained the same, only the mass of the cell ($M_{cell}$) needs to be defined:

$$M_{cell} = M_{torus} + M_{support}$$

While $M_{torus}$ has been derived, $M_{support}$ hasn’t, so it can be derived as follows:

$$M_{support} = V_{support} \cdot \rho_{support}$$ $$M_{support} = (V_{cell} - V_{ext}) \cdot \rho_{support}$$ Then the mass of the cell ($M_{cell}$) can be defined: $$M_{cell} \approx (\sum V_{torus} \cdot \rho_{backend}) + \left( V_{support} \cdot \rho_{backend} \frac{a^2 - c^2}{a^2} \right)$$ $$M_{cell} \approx \rho_{backend} \left( \sum V_{torus} + V_{support} \frac{a^2 - c^2}{a^2} \right)$$ $$M_{cell} \approx \rho_{backend} \left( 2\pi^2b(a^2 - c^2) + (V_{cell} - V_{ext}) \frac{a^2 - c^2}{a^2} \right)$$ $$M_{cell} \approx \rho_{backend} \frac{a^2 - c^2}{a^2} \left( \frac{a^2 \cdot 2\pi^2b(a^2 - c^2)}{a^2 - c^2} + (V_{cell} - V_{ext}) \right)$$ $$M_{cell} \approx \rho_{backend} \frac{a^2 - c^2}{a^2} (2\pi^2ba^2 + V_{cell} - V_{ext})$$ $$M_{cell} \approx \rho_{backend} \frac{a^2 - c^2}{a^2} (V_{cell})$$

As $M_{cell}$ is derived already, the focus can be shifted back to $S_{E[agg]}$ that by substituting $E_{max}$ and $M_{cell}$ yields:

$$S_{E[agg]} = \frac{E_{max}}{M_{cell}}$$ $$S_{E[agg]} \approx \frac{E_{max}}{\rho_{backend} \frac{a^2 - c^2}{a^2} V_{cell}}$$ $$S_{E[agg]} \approx \frac{2\pi^2 \sigma_{backend} b c (a-c)}{\rho_{backend} \frac{a^2 - c^2}{a^2} (2\sqrt{3}(a+b)^3)}$$ $$S_{E[agg]} \approx \frac{2\pi^2 \sigma_{backend} b c (a-c) a^2}{\rho_{backend} (a-c)(a+c) (2\sqrt{3}(a+b)^3)}$$

Finally, since 2 cancels and likewise for $(a-c)$, it can further be simplified as: $$S_{E[agg]} \approx \frac{\pi^2 \sigma_{backend} a^2 b c}{\sqrt{3} \rho_{backend} (a+c) (a+b)^3}$$

The aggregated energy density ($\rho_{energy[agg]}$) is maximum stored energy per cell ($E_{max}$) divided by a cell’s volume ($V_{cell}$).

$$ \rho_{energy[agg]} = \frac{E_{max}}{V_{cell}} $$ $$\rho_{energy[agg]} = \frac{2\pi^2 \sigma_{backend} b c (a-c)}{2\sqrt{3} (a+b)^3}$$ $$\rho_{energy[agg]} = \frac{\pi^2 \sigma_{backend} b c (a-c)}{\sqrt{3} (a+b)^3}$$

Numerical Analysis

Numerical analysis can be performed by using a carbon-nanotube based torus with assumed material density of $\rho_{backend} = 2,000 \text{ kgm}^{-3}$ and tensile strength of $\sigma_{backend} = 1.00 \times 10^{11} \text{ Pa}$ or 100 GPa. For the size of the torus, the value of $a = b = 1 \text{ m}$ is selected for simplicity. Then we vary the ratio of $c$ and $a$ from 0.001 to 0.999 (very close to zero to very close to one) to get the optimal ratio of $c$ and $a$ in real world application. To simplify calculations, the value $a = b$ is used.

The aggregated material density $\rho_{agg}$ is in $\text{kg/m}^3$, specific energy $S_E$ is in $J/kg$ then the energy density $\rho_{energy}$ is in $J/m^3$, and finally the maximum stored energy $E_{max}$ is in $J$, all in SI units.

$c/a$	$\rho_{agg}$	$\rho_{energy}$	$S_E$	$E_{max}$	$\rho_{energy[torus]}$	$\rho_{energy[agg]}$	$S_{E[agg]}$
0.001	2000	9.99E+13	5.00E+04	1.97E+09	9.99E+07	7.11E+07	3.56E+04
0.250	1875	3.00E+11	1.00E+07	3.70E+11	1.88E+10	1.34E+10	7.12E+06
0.500	1500	1.00E+11	1.67E+07	4.93E+11	2.50E+10	1.78E+10	1.19E+07
0.750	875	3.33E+10	2.14E+07	3.70E+11	1.88E+10	1.34E+10	1.53E+07
0.999	4.0	1.00E+08	2.50E+07	1.97E+09	9.99E+07	7.11E+07	1.78E+07

The following equations are used for this numerical analysis, along with their associated conclusions:

1. Energy Density (cavity): The maximum energy stored per unit volume of the internal toroidal cavity. This is equivalent to the maximum pressure the walls can withstand and is highest when the walls are thickest. $$\rho_{energy} = \frac{\sigma_{backend} \cdot (a-c)}{c}$$

2. Specific Energy (torus only): The energy-to-mass ratio for an individual torus, excluding the support structure. It is highest when the walls are thinnest (c/a approaches 1). $$S_E = \frac{\sigma_{backend} \cdot c}{\rho_{backend} \cdot (a + c)}$$

3. Maximum Stored Energy: The absolute maximum amount of energy a single torus can store. It represents the trade-off between having a large cavity volume and maintaining high pressure. $$E_{max} = 2\pi^2 \sigma_{backend} b c (a-c)$$

4. Average Energy Density (torus only): The energy density averaged over the entire external volume of a single torus (its “footprint”). This metric peaks at c/a = 0.5. $$\rho_{energy[torus]} = \frac{\sigma_{backend} c (a-c)}{a^2}$$

5. Aggregated Energy Density (full cell): The practical energy density for a complete battery cell, accounting for the volume of both the torus and its hexagonal support substrate. This metric also peaks at c/a = 0.5. $$\rho_{energy[agg]} = \frac{\pi^2 \sigma_{backend} b c (a-c)}{\sqrt{3} (a+b)^3}$$

6. Aggregated Specific Energy (full cell): The practical energy-to-mass ratio for a complete battery cell, accounting for the mass of both the torus and the support structure. It is also highest when the walls are thinnest (c/a approaches 1). $$S_{E[agg]} \approx \frac{\pi^2 \sigma_{backend} a^2 b c}{\sqrt{3} \rho_{backend} (a+c) (a+b)^3}$$

It is found that at $c/a$ ratio of 0.5 the maximum aggregated energy density ($\rho_{energy[torus]}$) is reached, the same for maximum stored energy ($E_{max}$) and aggregated energy density ($\rho_{energy[agg]}$). Meanwhile specific energy of both the torus only and the full cell ($S_E$ and $S_{E[agg]}$) peaked as $c/a$ approaches 1. Therefore for applications requiring maximum aggregated energy density, the most ideal condition is where $c/a \approx 0.5$ or where the minor radius of the inner cavity is half the minor radius of the torus. Meanwhile for applications requiring maximum mass saving per unit energy, the condition should approach $c/a \approx 1$ where the thinner the toroidal wall, the higher the mass savings per unit energy. It also means that $c/a$ is a very useful metric on designing a magnetobaric battery, other than the volume, density, and material strength.

Real World Application

While it is easy to build a macroscopic magnetobaric battery, an array of microscopic magnetobaric batteries can be of any reasonably small size, and would function essentially the same. It is more practical to have microscopic ones stacked into a three dimensional matrix of any arbitrary shape. Therefore the properties of such battery cells can be aggregated.

For nanometer-scaled individual magnetobaric battery cells, it can macroscopically be filled into any arbitrary volume, thereby freeing the form factor in actual application, and ensures smooth and continuous consistency on macroscopic level. Since the macroscopic property is approximately isomorphic, important properties of such batterinium substrate reduces to just energy density and specific energy. Therefore for the case where support structure density and averaged torus density is equal ($\rho_{torus[avg]} \approx \rho_{support}$),

For identical constituent materials, aggregated metrics for magnetobaric battery cells such as $\rho_{energy[agg]}$ and $S_{E[agg]}$ are also identical for all practical scales. It means $\rho_{energy[agg]}$ and $S_{E[agg]}$ are important metrics for bulk magnetobaric battery arrays, as long as it is stacked as suggested in this paper. Therefore the battery arrays can fill any macroscopic volume, while the amount of energy it stores dependent only on the two metrics, knowing the battery array volume or mass. In the case of a carbon-nanotube based magnetobaric battery cells, the property is as follows:

Metrics	Values
$c/a$	$0.5$
$\rho_{agg}$	$1,500 \text{ kg/m}^3$
$\rho_{energy[agg]}$	$1.78 \times 10^{10} \text{ J/m}^3$
$S_{E[agg]}$	$1.19 \times 10^7 \text{ J/kg}$

Safety Factor

However one would not just fill up the magnetobaric battery to full power, as it risks tearing the backend wall. As energy is contained within as magnetic pressure, punctured backend wall would potentially damage the superconducting lining. At a sufficiently high energy density, that would lead into electric discharges in form of plasma arcs. The generation of plasma arcs would atomize the area around the puncture, which will generate more arc, until the energy that was within the loop’s structure is depleted or the entire loop is disintegrated, whichever one comes first. Essentially a sufficiently charged and energy dense magnetobaric battery can also function as a bomb.

In real life, the full energy capacity would not be used. Depending on the risk level involved, a magnetobaric battery is typically filled up only to about 0.5 to 0.8 times of its full capacity (safety margin of 2 to ~1.2), therefore leaving enough wiggle room to avoid disintegration. Even with safety factors included, magnetobaric battery still yields as an attractive energy storage reaching the limit of how much energy can be stored mechanically at 5.95MJ/kg to 9.52MJ/kg.

For comparison, a typical information age lithium-ion battery has a specific energy of 0.46-0.72MJ/kg, while specific energy of hydrogen at 141.6MJ/kg, gasoline at 46.4MJ/kg, and lithium at 43MJ/kg.¹ Therefore magnetobaric battery is still around ten times more energy dense than lithium-ion battery, but still far below specific energy of hydrogen, gasoline, and lithium.

External links

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