Worked exercise — kilometre-scale cyclotron

Problem

You want to build a cyclotron of radius $R = 1\;\text{km}$ to circulate fully ionised helium ions (He$^{++}$, charge $q = 2e$, mass $m \approx 6{,}6\cdot 10^{-27}\;\text{kg}$) with kinetic energy $K = 10^{-3}\;\text{J}$. What magnetic field value is needed?

Solution

From kinetic energy to velocity. From $K = \tfrac{1}{2}mv^2$: $v = \sqrt{\frac{2K}{m}} = \sqrt{\frac{2\cdot 10^{-3}}{6{,}6\cdot 10^{-27}}} \approx 5{,}5\cdot 10^{11}\;\text{m/s}$

This velocity is greater than $c$: this means that at this energy the ions are already ultra-relativistic and the classical formula is no longer valid. We use it only as a dimensional check.

From the balance of the Lorentz force as centripetal force: $qvB = \frac{mv^2}{R} \quad\Rightarrow\quad B = \frac{mv}{qR}$

$B = \frac{6{,}6\cdot 10^{-27}\cdot 5{,}5\cdot 10^{11}}{2\cdot 1{,}6\cdot 10^{-19}\cdot 10^3} \approx \ev{11{,}3\;\text{T}}$

The classical field $B = mv/(qR) \approx 11{,}3\;\text{T}$ is already enormous, and it is actually an underestimate: at this energy ($K = 10^{-3}\;\text{J} \gg mc^2$) the ions are ultra-relativistic and the correct momentum is $p \approx K/c \approx 3{,}3\cdot 10^{-12}\;\text{kg·m/s}$, much larger than the classical $mv$, so the truly required field is $B = p/(qR) \approx 1\cdot 10^{4}\;\text{T}$ — many orders of magnitude larger. No real magnet reaches that: this is why a classical $1\;\text{km}$ ring cannot reach PeV energies, and CERN uses $\sim 8\;\text{T}$ superconducting magnets with much larger rings.

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Topics: Magnetismo Concepts: Forza di Lorentz · Moto circolare uniforme · Energia cinetica