A deuteron (the nucleus of an isotope of hydrogen) has a mass of $\mathbf{3}\mathbf{.}\mathbf{34}\mathbf{\times }{\mathbf{10}}^{\mathbf{27}}\mathbf{kg}$and a charge of +e. The deuteron travels in a circular path with a radius of in a 6.96mm magnetic field with magnitude 2.50 T. (a) Find the speed of the deuteron. (b) Find the time required for it to make half a revolution. (c) Through what potential difference would the deuteron have to be accelerated to acquire this speed?

The term potential difference may be defined as the work done of unit charge in moving from one point to another.

The force acting on it

${\overrightarrow{\mathrm{F}}}_{\mathrm{B}}=\mathrm{q}\left({\overrightarrow{\mathrm{v}}}_0\times \overrightarrow{\mathrm{B}}\right)$

The magnetic field and velocity perpendicular to each other so

${\mathrm{F}}_{\mathrm{B}}=\left|\mathrm{q}\right|\mathrm{vB}$ ……(1)

According to Second law of motion the magnitude of the force is product of mass and acceleration

${\mathrm{F}}_{\mathrm{B}}=\mathrm{ma}$

And radial acceleration is$\mathrm{a}=\frac{{\mathrm{v}}^2}{\mathrm{r}}$ so ${\mathrm{F}}_{\mathrm{B}}=\frac{{\mathrm{mv}}^2}{\mathrm{r}}$ ……(2)

From (1) and (2)

$$\begin{gathered} \left|\mathrm{q}\right|\mathrm{vB}=\frac{{\mathrm{mv}}^2}{\mathrm{r}} \\ \mathrm{v}=\frac{\left|\mathrm{q}\right|\mathrm{BR}}{\mathrm{m}} \end{gathered}$$

Put all the values

$$\begin{gathered} \mathrm{v}=\frac{\left(1.602\times {10}^{-19}\mathrm{C}\right)\left(2.50\mathrm{T}\right)\left(6.96\times {10}^{-3}\mathrm{m}\right)}{3.34\times {10}^{-27}\mathrm{kg}} \\ \mathrm{v}=8.35\times {10}^5\mathrm{m}/\mathrm{s} \end{gathered}$$

Hence, the speed of the deuteron is$8.35\times {10}^5\mathrm{m}/\mathrm{s}$

The time can be calculated using relation

$$\begin{gathered} \mathrm{t}=\frac{\mathrm{d}}{\mathrm{v}} \\ \mathrm{t}=\frac{\mathrm{\tau \tau R}}{\mathrm{v}} \\ \mathrm{t}=\frac{\mathrm{\tau \tau }\left(6.96\times {10}^{-3}\mathrm{m}\right)}{8.35\times {10}^5\mathrm{m}/\mathrm{s}} \\ \mathrm{t}=2.62\times {10}^{-8}\mathrm{s} \end{gathered}$$

Hence, the time required for it to make half a revolution is$2.62\times {10}^{-8}\mathrm{s}$

The potential difference can be calculated using the relation

$$\begin{gathered} \frac{1}{2}{\mathrm{mv}}^2=\left|\mathrm{q}\right|\mathrm{V} \\ \mathrm{V}=\frac{{\mathrm{mv}}^2}{2\left|\mathrm{q}\right|} \\ \mathrm{V}=\frac{\left(3.34\times {10}^{-27}\mathrm{kg}\right)\left(8.35\times {10}^5\mathrm{m}/\mathrm{s}\right)}{2\left(1.602\times {10}^{-19}\mathrm{C}\right)} \\ \mathrm{V}=7.27\times {10}^3\mathrm{V} \end{gathered}$$

Hence,$7.27\times {10}^3\mathrm{V}$potential difference would the deuteron have to be accelerated to acquire this speed.