Homework Help: 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...