1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Finding mass of ions in magnetic field

  1. Aug 4, 2012 #1

    s-f

    User Avatar

    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. jcsd
  3. Aug 4, 2012 #2
    It seems some information is missing. What is d? A sketch of the setup would help, too.
     
  4. Aug 4, 2012 #3
    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.
     
  5. Aug 4, 2012 #4

    s-f

    User Avatar

    Sorry this is the image https://mail-attachment.googleusercontent.com/attachment/u/0/?ui=2&ik=d3e9366f37&view=att&th=138f2cfebf63fad8&attid=0.1&disp=inline&realattid=1409394640159768576-1&safe=1&zw&saduie=AG9B_P-_1Tbsc3zXwuOtwk13kB5g&sadet=1344103689147&sads=0rBGq4MW_H8NzAdDPKWjvF6cP-g

    @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
  6. Aug 4, 2012 #5
    The picture is inaccessible.
     
  7. Aug 4, 2012 #6

    s-f

    User Avatar

    2012-08-04_14-03-38_594.jpg

    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.
     
  8. Aug 4, 2012 #7
    What is the force acting on a charge in magnetic fields? What is the acceleration in uniform circular motion?
     
  9. Aug 4, 2012 #8

    s-f

    User Avatar

    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.
     
  10. Aug 4, 2012 #9
    This is not the correct formula for acceleration in uniform circular motion!
     
  11. Aug 4, 2012 #10

    s-f

    User Avatar

    You're right; I typed it wrong. I was using a = v2/r though which I think it correct?
     
  12. Aug 4, 2012 #11
    From kinetic energy, you get [itex]v^2 = \frac {2 q V} {m}[/itex]. Plug that into the [itex]F = ma[/itex] formula and you should get what you need.
     
  13. Aug 4, 2012 #12

    s-f

    User Avatar

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

    m = qBr/sqrt (2qV/m) :confused:
     
  14. Aug 4, 2012 #13
    Get rid of the root by squaring.
     
  15. Aug 4, 2012 #14

    s-f

    User Avatar

    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.
     
  16. Aug 4, 2012 #15
    You wrote [tex]m = \frac {q B r} {\sqrt {\frac {2 q V} {m}}} [/tex]

    Squared, that becomes [tex]m^2 = \frac {q^2 B^2 r^2} {\frac {2 q V} {m}} = m \frac {q B^2 r^2} {2 V} [/tex]

    You really should have more algebraic practice, these are very simple transformations, things can get much worse.
     
  17. Aug 4, 2012 #16

    s-f

    User Avatar

    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!
     
  18. Aug 5, 2012 #17
    Well done s-f
    This is quite tricky algebra
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook