The Lorentz force on a particle of charge q produced by a magnetic field is
[tex]F=qvB[/tex]
where v is the velocity of the particle and B the intensity of the magnetic field.
Due to this force, the particle will acquire a circular motion, so this force will be the centripetal force of the motion:
[tex]m \frac{v^2}{r} = qvB[/tex]
where m is the mass of the particle, and [tex]r[/tex] the radius of the trajectory.
Re-arranging we have
[tex]r= \frac{mv}{qB} [/tex]
So, we have to find r for both the proton and the electron. We know that the two particles have same kinetic energy:
[tex]K_p = K_e[/tex]
and so
[tex] \frac{1}{2} m_p v_p^2 = \frac{1}{2}m_e v_e^2 [/tex]
The mass of the proton is 1833 times the mass of the electron ([tex]m_p = 1.67 \cdot 10^{-27} kg[/tex], [tex]m_e = 9.11 \cdot 10^{-31}kg[/tex]), i.e.
[tex]m_p = 1833 m_e[/tex]
and so from the relationship between the kinetic energies we find:
[tex]v_p = v_e \sqrt{ \frac{m_e}{m_p} }= v_e \sqrt{ \frac{m_e}{1833 m_e} } = \frac{v_e}{42.8} [/tex]
Now we can calculate the ratio between the radius of the proton and electron trajectories. Keeping in mind that q is the same for proton and electron, and the field B is the same, we have
[tex] \frac{r_p}{r_e} = \frac{m_p v_p}{m_e v_e}= \frac{(1833 m_e)( \frac{v_e}{42.8}) }{m_e v_e}=42.8 [/tex]
So, the radius of the proton trajectory is 42.8 times the radius of the electron trajectory.
