Answer to Question #168409 in Astronomy | Astrophysics for Bethv

Question

A 10 keV electron is moving in a circular orbit of radius 1 m in a plane at right angles to a uniform magnetic field. What is:

(a) the speed of the electron?

(b) the magnitude of the magnetic field?

(c) the frequency of revolution?

(d) the period of motion?

Answer 1

The force exerted by the magnetic field is the Lorentzian force $F = qvB$ .

(a) The kinetic energy of an electron is $E = \dfrac{m_ev^2}{2} \Rightarrow v = \sqrt{\dfrac{2E}{m_e}}= \sqrt{\dfrac{2\cdot10^4\cdot1.6\cdot10^{-19}\,\mathrm{J}}{9.1\cdot10^{-31}\,\mathrm{kg}}} = 5.9\cdot10^7\,\mathrm{m/s}.$

(b) The electron is moving in a circular orbit, so the centripetal acceleration is equal to the acceleration from the Lorentz force

$\dfrac{m_ev^2}{R} = qvB \Rightarrow B = \dfrac{m_ev}{qR} = \dfrac{9.1\cdot10^{-31}\,\mathrm{kg}\cdot5.9\cdot10^7\,\mathrm{m/s}}{1.6\cdot10^{-19}\,\mathrm{C}\cdot 1\,\mathrm{m}} = 3.4\cdot10^{-4}\,\mathrm{T}.$

(c) The frequency is $\nu = \dfrac{1}{T} = \dfrac{v}{2\pi R} = \dfrac{5.9\cdot10^7\,\mathrm{m/s}}{2\pi \cdot 1\,\mathrm{m}} = 9.4\,\mathrm{MHz}.$

(d) $T = \dfrac{1}{\nu} = \dfrac{1}{9.4\cdot10^6\,\mathrm{Hz}} = 1.1\cdot10^{-7}\,\mathrm{s}.$