Question

In a cyclotron, the orbital radius of protons with energy $300\mathrm{keV}$is $16.0\mathrm{cm}$. You are redesigning the cyclotron to be used instead for alpha particles with energy $300\mathrm{keV}$. An alpha particle has charge$q=+2e$ and mass $m=6.64\times {10}^{-27}\mathrm{kg}$. If the magnetic filed isn't changed, what will be the orbital radius of the alpha particles?

Short answer

$r_{\alpha }=15.95\mathrm{cm}$

Step-by-step solution

Solving for the radius

First we apply our formula for the force on a particle in a magnetic field.

$$\begin{gathered} \stackrel{\rightharpoonup }{F}=q\stackrel{\rightharpoonup }{v}\times \stackrel{\rightharpoonup }{B} \\ F=qvB\sin \left(\varphi \right) \end{gathered}$$

Where $F$is the magnetic force, $q$is the change, $v$is the velocity, $B$is the magnetic field, and $\varphi$is the angle between the velocity vector and the magnetic field. In our case the velocity vector is perpendicular to the magnetic field, so the sin term drops out and we are left with

$F=qvB$

Now we are looking for the radius. Our particles are undergoing uniform circular motion, so we can use that to incorporate the radius. We have that

$F=m{a}_c$

Where $m$is the mass of the particle, and $a_c$is the centripetal acceleration. Thus

$m{a}_c=qvB$

Recall that the centripetal acceleration is given in terms of the tangential velocity by

$a_c=\frac{v^2}{r}$

Putting it together we have an equation for the radius.

$$\begin{gathered} m\frac{v^2}{r}=qvB \\ \frac{mv}{r}=qB \\ r=\frac{mv}{qB} \end{gathered}$$

We are still missing the values for the velocity and the magnetic field Though.

Solving for the velocity.

Recall that we are given the energy. We can use this to find the velocity.

$$\begin{gathered} E=\frac{1}{2}m{v}^2 \\ v=\sqrt{\frac{2E}{m}} \end{gathered}$$

Thus our radius can be written as

localid="1650642773823" $$\begin{gathered} r=\frac{m}{qB}\sqrt{\frac{2E}{m}} \\ r=\frac{\sqrt{2Em}}{qB} \end{gathered}$$

The last step is just to solve for the magnetic field.

Solving for the magnetic field.

We can rewrite our above equation in terms of the magnetic field.

$B=\frac{\sqrt{2Em}}{qr}$

Now we know all of this information for the protons, and because the magnetic field isn't changing, it will be the same field for the alpha particles. Let these values for the proton be subscripted with the letter $p$. Then

$B=\frac{\sqrt{2E_p{m}_p}}{q_p{r}_p}$

Now plug this into our radius equation.

$$\begin{gathered} r=\frac{\sqrt{2Em}}{qB} \\ r=\frac{\sqrt{2Em}}{q\left({\displaystyle \frac{\sqrt{2E_p{m}_p}}{q_p{r}_p}}\right)} \\ r=\frac{\sqrt{2Em}}{q}\frac{q_p{r}_p}{\sqrt{2E_p{m}_p}} \\ r=\frac{q_p{r}_p}{q}\sqrt{\frac{Em}{E_p{m}_p}} \end{gathered}$$

Recall that the variables subscripted with $p$ correspond to the protons, and the variables without subscripts correspond to the alpha particles.

Solving for the radius

Finally we plug in our numbers and solve.

$$\begin{gathered} r=\frac{q_p{r}_p}{q}\sqrt{\frac{Em}{E_p{m}_p}} \\ r=\frac{(1.60\times {10}^{-19}\mathrm{C})·(16.0\mathrm{cm})}{2·1.6\times {10}^{-19}\mathrm{C}}\sqrt{\frac{300\mathrm{keV}·6.64\times {10}^{-27}\mathrm{kg}}{300\mathrm{keV}·1.67\times {10}^{-27}\mathrm{kg}}} \\ \boxed{\mathrm{r}=15.95\mathrm{cm}} \end{gathered}$$