# Finding mass of ions in magnetic field

1. Aug 4, 2012

### s-f

A device is used to measure the charge-to-mass ratio (e/m) of ions by accelerating them using an electric field described by a potential difference , and once they have acquired kinetic energy of known amount, they enter a uniform magnetic field. The particles move in a plane perpendicular to the magnetic field. What is the mass of the ion that enters the detector?

Strength of magnetic field = 0.8 T
d = 6.00 cm
ΔV = 100 V
Ion charge = 1.6 x 10^-19 C

I'm not really sure how to set up an equation here. I tried using qΔV = 1/2 mv2 and F = ma = qvB sin theta to solve it but I couldn't figure out how to solve for m here. Is there another formula I should use?

2. Aug 4, 2012

### voko

It seems some information is missing. What is d? A sketch of the setup would help, too.

3. Aug 4, 2012

### truesearch

From your first equation you can write v^2 = qV/m.
In the magnetic field the path is circular so Bqv = mv^2/r
You can combine these equations to give m = B^2xqxr^2/2V
Can you see how to do that?
If d in your question is the DIAMETER of the path in the magnetic field you should be able to solve it.
Have a go and come back if you get stuck.

4. Aug 4, 2012

### s-f

@truesearch - No I didn't get that equation at all when I tried to do it. Can you explain how you came up with that? I know it's correct because I plugged in the variables and it came out to 4.6 x 10^-25 kg which is the correct answer.

Last edited: Aug 4, 2012
5. Aug 4, 2012

### voko

The picture is inaccessible.

6. Aug 4, 2012

### s-f

Hopefully you can click to make it bigger. Truesearch is right that d is the diameter and that the magnetic field in the path is circular.

7. Aug 4, 2012

### voko

What is the force acting on a charge in magnetic fields? What is the acceleration in uniform circular motion?

8. Aug 4, 2012

### s-f

Here is what I tried to do originally:

F = qvB = ma
Since a = v2r, qvB = m(v2r)

qΔV = 1/2 mv2, so v = sqrt (2ΔV)/m. I tried to plug this in as v for the previous equation but I must have messed up algebraically because I did not get the final equation Truesearch posted earlier.

9. Aug 4, 2012

### voko

This is not the correct formula for acceleration in uniform circular motion!

10. Aug 4, 2012

### s-f

You're right; I typed it wrong. I was using a = v2/r though which I think it correct?

11. Aug 4, 2012

### voko

From kinetic energy, you get $v^2 = \frac {2 q V} {m}$. Plug that into the $F = ma$ formula and you should get what you need.

12. Aug 4, 2012

### s-f

Hmm I still think I'm fudging the algebra (terrible at math). I ended up getting:

m = qBr/sqrt (2qV/m)

13. Aug 4, 2012

### voko

Get rid of the root by squaring.

14. Aug 4, 2012

### s-f

Still not getting m = B^2xqxr^2/2V or 4.6 x 10^-25. I ended up not getting q in my numerator which makes no sense.

15. Aug 4, 2012

### voko

You wrote $$m = \frac {q B r} {\sqrt {\frac {2 q V} {m}}}$$

Squared, that becomes $$m^2 = \frac {q^2 B^2 r^2} {\frac {2 q V} {m}} = m \frac {q B^2 r^2} {2 V}$$

You really should have more algebraic practice, these are very simple transformations, things can get much worse.

16. Aug 4, 2012

### s-f

Ok I see, I was making a dumb mistake in the last step - thanks for explain that part.

And yeah I know I'm awful at math but this is the only physics class I need to take to finish high school and it's over in a week :) Not planning on doing physics in college!

17. Aug 5, 2012

### truesearch

Well done s-f
This is quite tricky algebra