All Problems

Motion of Charged Particles in Electric and Magnetic Fields

Problem 3.397

Protons are accelerated in a cyclotron so that the maximum curvature radius of their trajectory is equal to r=50 cmr=50 \mathrm{~cm}. Find: (a) the kinetic energy of the protons when the acceleration is completed if the magnetic induction in the cyclotron is B=1.0 TB=1.0 \mathrm{~T} (b) the minimum frequency of the cyclotron's oscillator at which the kinetic energy of the protons amounts to T=20MeVT=20 \mathrm{MeV} by the end of acceleration.

Reveal Answer
(a) T=(erB)22m=12MeVT=\frac{(e r B)^{2}}{2 m}=12 \mathrm{MeV} (b) vmin⁡=1πrT2m=20MHzv_{\min }=\frac{1}{\pi r} \sqrt{\frac{T}{2 m}}=20 \mathrm{MHz}
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