Breakthrough in the construction of nuclear fusion reactors - new calculations - a charged ionized Kawecki ring could replace the strongest electromagnets
Calculations for my nuclear fusion reactor ring - the charge accumulated in half the ring and the average angle of interaction is 45° - the average value of the sine function from 0 to 180 degrees is 2 ÷ pi, so we multiply the charge interaction of half the ring by 2 ÷ pi and compare the Coulomb electric force from half the ring to the Lorentz magnetic force of the strongest magnets.
ITER magnet to be 13 T
Wendelstein 7-X magnets 6-15 T
MIT magnet 20 T
Chinese magnets 21.7 - 35.1 T
Tokamak Energy magnets 24.4 T
The velocity of a deuteron with kinetic energy of 1 MeV is approximately 9.79 × 10⁶ m/s
Maximum Lorentz magnetic force F = q × V × B
Deuteron charge 1.602176634×10^−19 C
The velocity of a deuteron with kinetic energy of 1 MeV is approximately 9.79 × 10⁶ m/s
Magnetic flux density also called magnetic B-field strength from 6 to 35.1 T
Checking the Lorentz magnetic force for 22 and 35 T
F 22 T = 1.602176634×10^−19 × 9.79 × 10^6 × 22 = 3.45076803 × 10^-11 newton N
F 35 T = 1.602176634×10^−19 × 9.79 × 10^6 × 35 = 5.48985824 × 10^-11 newton N
F 35.1 T = 1.602176634×10^−19 × 9.79 × 10^6 × 35.1 = 5.50554355 × 10^-11
Coulomb electric force F = q × E
Electric field intensity due to half the charged ionized ring E = 2 ÷ π × k × Q ÷ r^2 = 2 ÷ π × 1 ÷ ( 4 π × ε₀ ) × Q ÷ r^2
Q electric charge of half the charged ionized ring
Electric field intensity required to produce the equivalent magnetic force on a deuteron with a velocity of approximately 9.79 × 10⁶ m/s in a magnetic field of 35.1 tesla T
E = F ÷ q
F = 5.50554355 × 10^-11 newton N
q Charge of deuteron 1.602176634×10^−19 C
E = 5.50554355 × 10^-11 ÷ ( 1.602176634×10^−19 ) = 343,629,000.27164 N / C newton / coulomb
E = 2 ÷ π × 1 ÷ ( 4 π × ε₀ ) × Q ÷ r^2
Q = E × π ÷ 2 × 4 π × ε₀ × r^2 = E × π^2 × 2 × ε₀ × r^2
E = 343,629,000.27164 N / C
ε₀ = 8.854187812813 × 10^−12 F/m
Q = 343,629,000.27164 × π^2 × 2 × 8.854187812813 × 10^−12 × 0.16^2 = 343,629,000.27164 × π^2 × 2 × 8.8541878128 × 10^−12 × 0.16^2 = 0.0015374756 C
It's a bit low, I'll check later, but if there's no mistake, it's shockingly low
For comparison, I previously calculated that a mini piece of copper measuring 1 mm by 1 mm by 1 mm can be charged
Q = 13.5664942388 coulombs C
At an ionization voltage of
U = 33,928,277.445413 V
13.5664942388 C ÷ 0.0015374756 C = 13.5664942388 ÷ 0.0015374756 = 8,823.8761244731 times less charge, so we calculate how much we can reduce the ionization voltage and the dimensions of the copper piece to maintain a charge of 0.0015374756 C
cube root cbrt(8,823.8761244731) = 20.6642573027 times
U = 33,928,277.445413 V ÷ 20.6642573027 = 33,928,277.445413 ÷ 20.6642573027 = 1,641,882.2582595 volts V
So instead of using expensive, energy-intensive electromagnets with a magnetic flux density also called magnetic B-field strength of, for example, 35.1 Tesla T, simply use a charged ionized ring - a voltage of 1,641,882.2582595 volts V is enough.
For comparison, small pocket-sized stun guns available on the market can have a voltage of 50-98 million volts V, and a lightning bolt can have a voltage of 100 million to 1 billion volts V.
5 Liters of Copper × 8.935 kg/L = 44.675 kg
44.675 kg × 80 PLN/kg = 3,574 PLN + cost of ionizer, inverter and other accessories