If two deuterium nuclei (charge + e, mass 3.34 \\times 10^{ - 27} {\ m{ kg}}) ge

Question

If two deuterium nuclei (charge + e, mass 3.34 imes 10^{ - 27} { m{ kg}}) get close enough together, the attraction of the strong nuclear force will fuse them to make an isotope of helium, releasing vast amounts of energy. The range of this force is about 10^{ - 15} { m{ m}}. 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.


V is 8.3*10^6

Part B
What strength magnetic field is needed to make deuterium nuclei with this speed travel in a circle of diameter 2.20m ?

Explanation / Answer

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

B = (3.34*10-27*8.3*106)/((2.20/2)*1.60*10-19)

=>B = 0.1575 T

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