Question

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?

Short answer

1. Two nuclei would have to move at $\mathbf{8}\mathbf{.}\mathbf{3}\mathbf{.}{\mathbf{10}}^{\mathbf{6}}\mathbf{m}\mathbf{/}\mathbf{s}$1
2. 0.14 T magnetic field is needed to make deuterium nuclei with this speed travel in a circle of diameter 2.50 m

Step-by-step solution

Kinetic energy and potential energy

Kinetic energy is given by

$\frac{{\mathbf{mv}}^{\mathbf{2}}}{\mathbf{2}}$

The potential energy is given by

$\mathit{k}\frac{{\mathbf{e}}^{\mathbf{2}}}{\mathit{r}}$

Determine the speed of the nuclei

(a)

Total energy of two nuclei is equal

${\mathit{K}}_{\mathbf{1}}\mathbf{+}{\mathit{U}}_{\mathbf{1}}\mathbf{=}{\mathit{K}}_{\mathbf{2}}\mathbf{+}{\mathit{U}}_{\mathbf{2}}$

Kinetic energy at the beginning is 0

Therefore,

$\begin{array}{r}\mathbf{2}\frac{\mathbf{m}{\mathbf{v}}^{\mathbf{2}}}{\mathbf{2}}\mathbf{=}\mathbf{k}\frac{{\mathbf{e}}^{\mathbf{2}}}{\mathbf{r}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\\ \mathbf{v}\mathbf{=}\mathbf{e}\sqrt{\frac{\mathbf{k}}{\mathbf{m}\mathbf{r}}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\\ \mathbf{=}\mathbf{1}\mathbf{.602}\mathbf{\cdot }{\mathbf{10}}^{\mathbf{-}\mathbf{19}}\sqrt{\frac{\mathbf{8}\mathbf{.99}\mathbf{\cdot }{\mathbf{10}}^{\mathbf{9}}{\mathbf{Nm}}^{\mathbf{2}}\mathbf{/}{\mathbf{C}}^{\mathbf{2}}}{\mathbf{3}\mathbf{.34}\mathbf{\cdot }{\mathbf{10}}^{\mathbf{-}\mathbf{27}}\mathbf{kg}\mathbf{1}\mathbf{\cdot }{\mathbf{10}}^{\mathbf{-}\mathbf{15}}\mathbf{m}}}\\ \mathbf{\approx }\mathbf{8}\mathbf{.3}\mathbf{\cdot }{\mathbf{10}}^{\mathbf{6}}\mathbf{m}\mathbf{/}\mathbf{s}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\end{array}$

Therefore, two nuclei would have to move at$\mathbf{8}\mathbf{.}\mathbf{3}\mathbf{.}{\mathbf{10}}^{\mathbf{6}}\mathbf{m}\mathbf{/}\mathbf{s}$

Determine the magnetic field

(b)

The magnetic field is given by

$\begin{array}{r}\mathbf{R}\mathbf{=}\frac{\mathbf{m}\mathbf{v}}{\mathbf{\left|}\mathbf{q}\mathbf{\right|}\mathbf{B}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\\ \mathbf{B}\mathbf{=}\frac{\mathbf{m}\mathbf{v}}{\mathbf{\left|}\mathbf{q}\mathbf{\right|}\mathbf{R}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\\ \mathbf{=}\frac{\mathbf{3}\mathbf{.34}\mathbf{\cdot }{\mathbf{10}}^{\mathbf{-}\mathbf{27}}\mathbf{kg}\mathbf{8}\mathbf{.3}\mathbf{\cdot }{\mathbf{10}}^{\mathbf{6}}\mathbf{m}\mathbf{/}\mathbf{s}}{\mathbf{1}\mathbf{.602}\mathbf{\cdot }{\mathbf{10}}^{\mathbf{-}\mathbf{19}}\mathbf{C}\mathbf{1}\mathbf{.25}\mathbf{m}}\\ \mathbf{\approx }\mathbf{0}\mathbf{.14}\mathbf{T}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\end{array}$

Therefore, 0.14T magnetic field is needed to make deuterium nuclei with this speed travel in a circle of diameter 2.50 m