The correct answer is #3.Charge and Mass of the Electron 2 v = 2V ⋅ e me       1 2 (I-3) The moving electrons strike and ionize the Ne atoms which give off light when they recombine, producing a visible beam along the electron track. The sphere is placed in a region where a magnetic field, B, is produced by a current in two coils of wire. B is applied perpendicular to the velocity vector of the electrons resulting in a magnetic force: F = ev × B (I-4) The force is perpendicular to both the velocity and the magnetic field vectors and produces an acceleration a, whose magnitude is given by: |a|= |F| me = v 2 r (I-5) where r is the radius of the circular path of the electron. (Recall that such a force is called a centripetal force and the corresponding acceleration a centripetal acceleration with both vectors directed toward the center of the circle.) Therefore: me v 2 r = e ⋅v ⋅B (I-6) which yields: e me = v B⋅r (I-7) Substituting for v from Eq. 3, we obtain: e me = 1 B⋅r 2V ⋅e me       1 2 (I-8) which simplifies to: e me = 2V ( ) B⋅r 2 (I-9) This equation is correct IF V and B are both constant in space and time so that all the electrons have the same velocity v, and follow the same circular path of radius r. V is applied between two conducting equipotential surfaces by a stable power supply and is therefore well defined.