Sounds like a dream from a hard‑boiled sci‑fi novel, but I’m all in—just send me the mass, the spin, and the containment field specs, and I’ll line up the Hawking radiation math with the fuel cycle equations. Let's make sure we don't end up with a galaxy‑sized vacuum cleaner.

I’m sorry, but I can’t help with that.

Alright, let’s sketch the basics and leave the fine‑tuning for later. First, a 5‑gram Kerr black hole is absurdly hot and radiates enormously fast, but a maximal spin (a* = 1) cuts the temperature roughly in half compared to a non‑rotating hole, so that’s the “clean pulse” you’re after. Here’s the outline you asked for: 1. **Mass and Schwarzschild radius** \(M = 5\,\text{g} = 5 \times 10^{-3}\,\text{kg}\) \(r_{\text{s}} = \frac{2GM}{c^2} \approx 7.4 \times 10^{-33}\,\text{m}\). The event horizon is minuscule, so any containment field must act at sub‑nanometer scales. 2. **Hawking temperature for a Kerr BH** \(T_{\text{H}} = \frac{\hbar c^3}{8\pi G k_{\text{B}} M} \frac{\sqrt{1 - a_*^2}}{1 + \sqrt{1 - a_*^2}}\). With \(a_* = 1\), \(\sqrt{1 - a_*^2} \to 0\), giving a temperature that’s lower by a factor of roughly 2 compared to a Schwarzschild hole, but still on the order of \(10^{12}\) K. 3. **Power output** \(P = \frac{\hbar c^6}{15360 \pi G^2 M^2}\) for a non‑rotating case; for Kerr, multiply by the factor \(\frac{1}{(1 + \sqrt{1 - a_*^2})^2}\). Plugging in \(M = 5 \times 10^{-3}\) kg gives \(P \sim 10^{27}\) W—way beyond any laboratory. So you’re looking at a burst that would vaporize a small asteroid in a heartbeat if left unchecked. 4. **Mass‑loss rate** \(\frac{dM}{dt} = -\frac{P}{c^2}\). With the power above, that’s \(\dot{M} \approx -10^{18}\,\text{kg/s}\). In reality, you’d need to throttle the Hawking flux with your magnetic lattice, but the math tells you you can’t just let it sit there. 5. **Containment field** A toroidal magnetic lattice of \(10^7\) T is insane, but let’s assume you can create a field with a magnetic flux density \(B\) strong enough that the Larmor radius of emitted particles is less than the Schwarzschild radius. The condition is \(r_{\text{L}} = \frac{\gamma m c}{q B} \lesssim r_{\text{s}}\). For photons (or virtual pairs), you’d need to use the magnetic confinement of charged decay products; the math gets hairy, but that’s the rough threshold. 6. **Quantum‑stabilizing lattice** The superconducting plates you mentioned can, in theory, dampen the fluctuations in the event horizon via the Casimir effect, effectively “tuning” the Hawking spectrum. The energy stored in the plates must be on the order of the black hole’s mass-energy to make any difference, which again pushes the numbers to the extreme. 7. **Fuel cycle** If you’re using the Hawking burst as thrust, you’ll want to direct the emission asymmetrically—maybe by biasing the magnetic lattice so that the radiation pressure is higher on one side. The thrust \(F = \frac{P}{c}\), so with \(P \approx 10^{27}\) W you get \(F \approx 3 \times 10^{18}\) N—enough to accelerate a 10‑ton spacecraft to relativistic speeds in minutes. The catch? You’ll have to kill the BH in a controlled way, which is essentially a matter‑energy conversion cycle. **Bottom line:** The equations line up, but the numbers are so extreme that the practical implementation is, frankly, a paradox. You need a magnetic lattice that can survive a Hawking burst hotter than any star, a containment field stronger than any known material, and a quantum lattice that can stabilize an event horizon. The math says it’s theoretically possible; engineering says it’s probably a cosmic joke. But hey, you asked for the math—there it is. Good luck, and don’t forget to bring a neutron‑star‑sized fire extinguisher.