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Lorentz force — radius of trajectory

A proton (m = 1.673×10⁻²⁷ kg, q = 1.602×10⁻¹⁹ C) enters perpendicularly into a magnetic field B = 0.5 T with velocity v = 2×10⁶ m/s.\nCalculate: (a) the radius of the circular trajectory, (b) the revolution period, (c) the cyclotron frequency.
Given data
m = 1.673×10⁻²⁷ kgq = 1.602×10⁻¹⁹ CB = 0.5 Tv = 2×10⁶ m/s
Review the theory: Magnetismo
Steps
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  1. Calculate the radius of the circular trajectory (in cm).
  2. Calculate the revolution period T (in ns = 10⁻⁹ s).
  3. Calculate the cyclotron frequency f_c (in MHz).
Full worked solution
  1. Calculate the radius of the circular trajectory (in cm).
    r=mvqB=1.673×10−27⋅2×1061.602×10−19⋅0.5r = \dfrac{mv}{qB} = \dfrac{1.673\times10^{-27} \cdot 2\times10^6}{1.602\times10^{-19} \cdot 0.5}
    r=mvqB=1.673×10−27×2×1061.602×10−19×0.5=3.346×10−218.01×10−20≈0.04178 m=4.18 cmr = \frac{mv}{qB} = \frac{1.673\times10^{-27} \times 2\times10^{6}}{1.602\times10^{-19} \times 0.5} = \frac{3.346\times10^{-21}}{8.01\times10^{-20}} \approx \mathbf{0.04178\,m} = \mathbf{4.18\,cm}. The Lorentz force qvBqvB provides the centripetal force mv2/rmv^2/r for circular motion.
  2. Calculate the revolution period T (in ns = 10⁻⁹ s).
    T=2πrv=2πmqBT = \dfrac{2\pi r}{v} = \dfrac{2\pi m}{qB}
    T=2πmqB=2π×1.673×10−271.602×10−19×0.5=1.051×10−268.01×10−20≈1.313×10−7 s=131.3 nsT = \frac{2\pi m}{qB} = \frac{2\pi \times 1.673\times10^{-27}}{1.602\times10^{-19} \times 0.5} = \frac{1.051\times10^{-26}}{8.01\times10^{-20}} \approx \mathbf{1.313\times10^{-7}\,s} = \mathbf{131.3\,ns}. The period is independent of the velocity.
  3. Calculate the cyclotron frequency f_c (in MHz).
    fc=1T=qB2πmf_c = \dfrac{1}{T} = \dfrac{qB}{2\pi m}
    f=1T=11.313×10−7≈7.62×106 Hz=7.62 MHzf = \frac{1}{T} = \frac{1}{1.313\times10^{-7}} \approx \mathbf{7.62\times10^{6}\,Hz} = \mathbf{7.62\,MHz}. This is the cyclotron frequency, which is the basis of particle accelerators.
Result:r = 4.18 cm — T = 131.3 ns — f_c = 7.62 MHz.
TheoryMagnetismoFormula sheetPhysics 2

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