# With what speed does the ion enter the magnetic field in a mass spectrometer?

1. Jul 23, 2012

### pinkyaliya

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

A mass spectrometer has a potential difference of 120 V applied to its accelerator plates. It is accelerating a positively charged ion that has a mass of 2.32 x 10^-26 kg. the ion then is directed into a uniform magnetic field of 0.17 T. The magnetic field is perpendicular to the direction that the particle is traveling. The radius of the circular path that the ion takes, while in the magnetic field, is 2.5 cm.
with what speed does the ion enter the magnetic field?

Potential difference = 120 V
Mass = 2.32 x 10^-26 kg
Magnetic field=0.17 T
Theta=90 degrees
Charge=3.08 x 10^-19 C

2. Relevant equations

3. The attempt at a solution

since it's traveling in a circular path a=v^2 / r
I also am confused about the free body diagram (do you even need me to figure this question out?)

edit: Is it that the magnetic force = the force of gravity or something like that?

Last edited: Jul 23, 2012
2. Jul 23, 2012

### pgardn

So for your equations, you dont want to use the strength of the B-field, voltage or mass? Do you need any of them? ohhh charge also?

Last edited: Jul 23, 2012
3. Jul 23, 2012

### pinkyaliya

Some of the information was used to solve for the charge so some of the given is not necessary. I'm pretty sure you need the charge to figure this one out.

4. Jul 23, 2012

### pgardn

Yea that is a bit of a strange charge. How did you get it? Its close to what I would expect, but not quite...

5. Jul 23, 2012

### pinkyaliya

I used the equation

m=qβ^2R^2 / 2ΔV

q=m2ΔV / β^2R^2

q=(2.32 x10^-26kg)(2)(120V) / (0.17T^2)(0.025m^2)
q=3.08...x10^-19 C

Is that right?

6. Jul 23, 2012

### pgardn

Yes but...
Probably rounded a bit short since you have a very wide range of numbers some much smaller than the others. Since the amount of charge in 1 electron or proton is 1.6 x 10^-19C.

That would mean 2 extra protons in your ion should have about 3.2 x 10^-19C of charge.

Anyways...

Since you went through all that and got close, how about using cons. of Energy? What do you know about this situation. Any thoughts on kinetic and potential energy?
.

7. Jul 23, 2012

### pinkyaliya

Is it maybe that

Ek + Ee = Ek' + Ee'

I don't see how potential energy would play role since it's not changing height.

8. Jul 23, 2012

### pgardn

Potential Energy comes in many forms. Sometimes unit analysis can help you. I am going to tell you that before the ion is accelerated it has electric potential energy. Do you have an equation for this, or can you just look a the units on what you have and come out with Joules?

What does that Ee mean for example?

9. Jul 24, 2012

### pinkyaliya

I was working around with it and the only problem I have is finding out the velocity in the end

What I mean is
v=?
ΔV=120 V
β=0.17 T
q=3.08...x10^-19 C
m=2.32 x10^-26 kg

v'=?

ΔE = qΔV
ΔE = (3.08...x10^-19 C)(120 V)
ΔE = 3.696...x10^-17 J

So I know that there was a potential energy of 3.696...x10^-17 J. If it started at rest, I know that this energy would be transformed into kinetic energy and that
Ek'=3.696...x10^-17 J
but I don't know what kinetic energy it ended with.

10. Jul 24, 2012

### ehild

The ion is accelerated by U=120 V potential difference. Its final KE is equal to the initial potential energy qU, the accelerating voltage multiplied by the charge of the ion: mv2/2 =qU

The magnetic field exerts a force F, which is perpendicular to both the magnetic field B and the velocity of the ion. The magnitude of the force when v is perpendicular to B is F=qvB. The ion moves along a circle of radius R, and the centripetal force is equal to the magnetic force, mv2/R=qvB. Substitute 2qU for mv2, you get 2qU/R=qvB. q cancels. Isolate v.

ehild

11. Jul 24, 2012

### pgardn

So as Ehilde stated:

All potential energy starting from rest

to all kinetic as the ion enters perpendicular to the B field at which point it turns a circle.

Sorry but I was attempting to "draw" it out of you.

All electrical pot qV = all kinetic 1/2mv^2