Charge-to-mass ratio of electron/proton

[jdc22]

Homework Statement

The Helmholtz coil current = 2.45 A, the magnetic field = 1.91 x 10^(-3) T, the accelerating voltage = 295 V, and the beam radius = .03m.

a) Calculate the e/m ratio of an electron. Show that e/m = 2V / r²B²

b) During the initial use of this experiment, physicists also knew the charge-to-mass ratio of a hydrogen ion, approximately 9.58 x 10^7 Coulombs/kg. If you had little knowledge of atomic structure, what two interpretations could you place on the differences between the two ratios?

Homework Equations

PE = Ve
KE = 1/2mv²
Fmag = evB
Fcent = mv²/r

The Attempt at a Solution

a) Ve = 1/2mv²
v = √(2Ve / m) ; v² = 2Ve/m
Fmag = Fcent
evB = mv²/r
r = mv/Be

I can't seem to get past this part of the derivation. I've tried subbing in √(2Ve / m) into r = mv/Be but it turns out too messy.

b) I'm not too sure about the second part at all.. in what kind of context should I be thinking about the ratios?

 

[kuruman]

In the orbiting stage you have $$evB=\frac{mv^2}{r}\implies v=\frac{eBr}{m}.$$In the accelerating stage you have $$eV=\frac{1}{2}mv^2=\frac{1}{2}m\left( \frac{eBr}{m} \right)^2.$$Can you simplify and solve for $e/m$?