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Physics / Sears and Zemansky's University Physics with Modern Physics 13 / Chapter 27 / Problem 57P

Sears and Zemansky's University Physics with Modern Physics | 13th Edition | ISBN: 9780321696861 | Authors: Hugh D. Young; Roger A. Freedman; A. Lewis Ford

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CP Fusion Reactor. If two deuterium nuclei (charge +e,

Chapter 27, Problem 57P

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Problem 57P

CP Fusion Reactor. 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 them to make an isotope of helium, releasing vast amounts of energy. The range of this force is about 10-15 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. (a) How fast would two nuclei have to move so that in a head-on collision they would get close enough to fuse? (Assume their speeds are equal. Treat the nuclei as point charges, and assume that a separation of 1.0 × 10-15 is required for fusion.) (b) What strength magnetic field is needed to make deuterium nuclei with this speed travel in a circle of diameter 2.50 m?

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QUESTION:

Problem 57P

CP Fusion Reactor. 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 them to make an isotope of helium, releasing vast amounts of energy. The range of this force is about 10-15 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. (a) How fast would two nuclei have to move so that in a head-on collision they would get close enough to fuse? (Assume their speeds are equal. Treat the nuclei as point charges, and assume that a separation of 1.0 × 10-15 is required for fusion.) (b) What strength magnetic field is needed to make deuterium nuclei with this speed travel in a circle of diameter 2.50 m?

ANSWER:

Solution 57P

Step 1:

$First\ we\ have\ to\ apply\ the\ conservation\ of\ energy\ to\ the\ motion\ of\ both\ nuclei.$

$And\ \ \ according\ to\ newton's\ second\ law\ \Sigma{}\overline{F}=m\overline{a}\ \ \ which\ implies\ \ \ \$ $\left|{q}\right|vB=\frac{m{v}^{2}}{r}$

$where\ m\ is\ the\ mass\ of\ the\ particle\ and\ r\ is\ \ the\ radius\ of\ the\ circular\ path,$

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