Question

A particle with specific charge α:=q/m=10e8 C/kg enters a bubble chamber where its path can be observed. A magnetic field of induction B=10e−2 T is perpendicular to the particle's velocity. The particle experiences a viscous drag force which is approximately proportional to its velocity, that is
F drag=−kv .
As a consequence of this force, the particle spirals inward. It is observed that after two full rotations the radius of curvature decreases by 2%. Then the magnetic field is switched off and the particle travels L=30cm before coming to a stop. What was the particle's velocity in m/s right before entering the chamber?

Answer 1

i would appreciate your help very much. i set up a force equation getting

m dv/dt = -kv+Bvq and separating the variables i get differential equation

dv/v = (Bq-k)/m dt, and after integrating i get v(t) = v0 exp( (Bq-k)/m *t)

now, i am not sure if i can calculate v0 from this equation

mv0^2/R = Bv0*q which gives v0=BqR/m

when i do that, and when i put that into the equation above, i still have one unknown and that's "k".