Formulas/physics/Moving Charges/Cyclotron — Maximum Kinetic Energy

Cyclotron — Maximum Kinetic Energy

Maximum kinetic energy of a particle accelerated in a cyclotron of dee radius R. Obtained by substituting v = qBR/m into ½mv².

Class 12

Derivation

Limitation of the Cyclotron

The particle spirals outward as it gains energy. It exits when its orbital radius equals the dee radius $R$ . This sets the maximum speed.

Derivation

From $r = mv/qB$ , at $r = R$ :

v_{\max} = \frac{qBR}{m}

Maximum kinetic energy:

KE_{\max} = \frac{1}{2}mv_{\max}^2 = \frac{1}{2}m\left(\frac{qBR}{m}\right)^2

\boxed{KE_{\max} = \frac{q^2B^2R^2}{2m}}

Dependence

$KE_{\max} \propto B^2$ : stronger field gives much more energy.
$KE_{\max} \propto R^2$ : larger dees give much more energy.
$KE_{\max} \propto 1/m$ : lighter particles reach higher energies in the same cyclotron.

Role of the Accelerating Voltage

The voltage $V$ between the dees determines how quickly the particle spirals out — more voltage means fewer revolutions to reach $R$ . But $KE_{\max}$ is set entirely by $B$ and $R$ , not by $V$ .

Note

For a proton in a cyclotron with $B = 1.5$ T and $R = 0.5$ m: $KE_{\max} \approx 27$ MeV. Relativistic effects become significant well before this, which is why synchrotrons replaced cyclotrons for high-energy physics.

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