Constants Part A If two deuterium nuclei (charge +e, mass 3.34 × 10-27 kg) get c

Question

Constants Part A If two deuterium nuclei (charge +e, mass 3.34 × 10-27 kg) get close enough together, the attraction of the strong nuclear force will fuse thenm to make an isotope of helium, releasing vast amounts of energy. The range of this force is about 10-15m. This is the principle behind the fusion reactor. The deuterium nuclei are moving much too fast to be contained by physical walls, so they are confined magnetically How fast would two nuclei have to move so that in a head-on collision they would get close enough to fuse? (Treat the nuclei as point charges, and assume that a separation of 1.0 x 10-15m is required for fusion.) Part B What strength magnetic field is needed to make deuterium nuclei with this speed travel in a circle of diameter 3.00 m?

Explanation / Answer

part A

Electrical potential (V) of a point charge at a distance d from the point is ..

V = kq/r

Q = 1.60 x10-19 C

r = 1.0 x 10-15 m ( (the required separation)

V = (9.0 x 109)(1.60 x10-19)/ (1.0 x10-15) = 1.44 x106 J/C

so another point charge (e) brought to this position would give the system an elec.PE, E = eV (J)

E = 1.44 x106 x 1.60 x10-19 C = 2.30 x10-13 J

this is transferred from the KE's of the two moving nuclei as they collide.

KE possessed by each nucleus = E/2 ..

0.5 (2.30 x10-13) = 0.5 mv2

v = sqrt(2.30 x10-13 / 3.34 x10-27)

v = 8.30 x106 m/s

partB

In the B- field centripetal force (mv²/R) is provided by magnetic force (BQv)

mv2/r = Bqv

B = mv/rQ

B = (3.34 x10-27)(8.30 x106) / (3.0/2)(1.60 x10-19)

B = 0.1155 T

Please rate my answer, good luck...

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