# Magnetic field: the Banebridge velocity selector

1. Feb 6, 2010

### Karol

1. The problem statement, all variables and given/known data

The electric field intensity between the plates of the Banebridge velocity selector (I am not a native English speaker …) is 120 [Volt/cm] and the magnetic field magnitude of both fields is 0.6 [Weber/m2].
A singly charged Neon ion beam makes a radius of 7.28 [cm]. What is it's mass number.
The result must be 21, 10 times smaller than mine.

2. Relevant equations

Velocity of beam: V=$\frac{E}{B}$
radius of the beam: $$R[m]=\frac{mV}{qB}$$
mass number=mass/proton mass: $$N=\frac{m}{m_{p}}$$

3. The attempt at a solution

120[Volt/cm]=12,000[Volt/m]

$$V=\frac{E}{B}=\frac{12,000}{0.6}=20,000[m/sec]$$

$$R=0.728=\frac{mV}{qB}=\frac{m*20,000}{1.602\ \times\ 10^{-19}*0.6}\Rightarrow m=3.5\ \times\ 10^{-25}$$

$$N=\frac{m}{m_{p}}=\frac{3.5\ \times\ 10^{-25}}{1.672\ \times\ 10^{-27}}=209$$

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2. Feb 6, 2010

3. Feb 6, 2010

### Karol

My result is 10 times bigger, according to the book and the periodic table of elements, it should come out 20.9

4. Feb 6, 2010

### thebigstar25

I cant see whats wrong with your solution , it seems okay ..

5. Feb 6, 2010

### Karol

Thanks for your effort, Ihope I'll get another answer before I decide there was a mistake in the book. By the way, it's taken from a translation to Sears & Zemansky, 1964...