Question

a proton with velocity v, enters a magnetic induction field of width of 1cm and perpendicular to the proton's velocity. Because of the field, the proton gains a small velocity component that is perpendicular to the initial velocity and the magnetic field lines, wich as the value of 3,3x10^5 m/s. What is the magnitude of the magnetic field?

Answer 1

Lorentz Force with no electric field

\[\vec{F} = q\vec{v}\times\vec{B}\]

Since the field is parallel to motion
\[F=qvB\]
change in momentum given change in velocity \(\Delta v\) and interval in length \(\Delta l\) that the particle passes through the magnetic field:
\[\Delta p = m_e\Delta v= F\Delta t=evB\frac{\Delta l}{v}=eB\Delta l\]
solve for \(B\)