Your F = q(E + v x B) is spot on. There is only a B field, so F = q (v x B) . Right hand thumb to the right, index finger out of the paper pointing to your nose: middle finger points in the direction of F.
F is perpendicular to v, so no change in magnitude of v, only in direction. And this remains so: , |v| constant, |F| constant and ##\vec F \perp \vec v##. The sure sign of a path that is characterized by a radius (what a giveaway in the original problem tekst!), for which we know (right?) F = mv
2/R. So there is your ## R={mv\over QB}##. As long as you have no clue about m, no value for R.
To collect 3 Coulomb of charge you need at least 6 x 10
18 protons and a lot of energy to squeeze them together. That many protons still is only 10 nanograms, though. But to avoid instant discharge by a lightning flash you need a substantial charge carrier. A 1 meter conducting sphere has a capacitance of 1 ##\mu##F, with 3 C the surface potential is 3 x 10
9 Volt. 4 Tesla of nicely homogenetic magnetic field isn't all that trivial either and barely possible above the, say, 1 m
2 scale. (Others will correct me if I am wrong).
I think this is just a physics thought exercise and somewhat not-good: good students get distracted by the missing of m and average students put m=1 kg or something arbitrary.
I was triggered by this thread because as a 17 year old I had an oral grammar school exam where they put a
Teltron tube in front of me that I had never seen before. The thing measures e/m as a ratio, using your formula. But googling 'e over m experiment' shows that JJ Thomson stole the cake in 1897.
