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What is the wire's linear charge density?

  1. Oct 20, 2012 #1
    1. The problem statement, all variables and given/known data

    A proton orbits a long charged wire, making 1.30*10^6 revolutions per second. The radius of the orbit is 1.20cm.

    What is the wire's linear charge density?

    2. Relevant equations

    - q E = m w^2 r
    - 9*10^9 [2 λ /r] q = m w^2 r

    3. The attempt at a solution

    λ = linear charge dens
    w = 2pi/period = 2pi/T
    w = d theta/dt = 1.4* 2pi /1

    I'm confused on how to solve for λ here though. Can anyone help me out?
     
  2. jcsd
  3. Oct 20, 2012 #2
    You said you know:
    F=m v^2/r=m r ω^2=mr(2π/t)^2
    F=q E= q 2 k λ/r

    You correctly said that you have to set them equal to each other
    2qkλ/r=4π^2mr/t^2
    You were given r and t. q, k, and m are known physical constants. The only unknown variable is λ. Just solve for it and plug in numbers.
     
  4. Oct 21, 2012 #3
    Ah, okay, so I have it set up like this:

    (2*1.6*10^-19*9*10^9*λ)/(.012)=(4π^2*1.67*10^-27*.012)/(1.30*10^6)^2

    I got 1.95056*10^-33 nC/m but that seems to be wrong, anywhere I went wrong?
     
    Last edited: Oct 21, 2012
  5. Oct 21, 2012 #4
    Don't plug in numbers until the final answer. Solve for λ, then plug in numbers. Also, just because it's small doesn't mean it's wrong.
     
  6. Oct 21, 2012 #5


    Alright so I got λ = 2π^2mr^2/kqt^2. I plugged in the numbers and got the same answer as before though, lol.
     
  7. Oct 21, 2012 #6
    Ok, you got the same answer. Why did you think it was wrong?
     
  8. Oct 21, 2012 #7
    I submitted it through Mastering Physics and it said it was wrong, lol. I'm not sure if its the units because the answer is suppose to be in nC/m, but I am pretty sure it cancels out accordingly. Hmm..
     
  9. Oct 21, 2012 #8
    I see the problem. You weren't given the period, you were given the rotation speed. "1.30*10^6 revolutions per second"
     
  10. Oct 21, 2012 #9
    Oh, so I have to convert the rotation speed to its period? So it would be 1/1300000?
     
  11. Oct 21, 2012 #10
    Or you could use the frequency 1.6*10^6 instead of 1/period. Either way, you're multiplying by (1.3*10^6)^2 instead of dividing.
     
  12. Oct 21, 2012 #11
    Hmm I got 5.57*10^-33 and that still seems to be wrong, lol.
     
  13. Oct 21, 2012 #12
    Let's go back to the equation that you found:
    [tex]\lambda = \frac{2\pi^2mr^2}{kqt^2}[/tex]
    Replace t by 1/f to get
    [tex]\lambda = \frac{2\pi^2mr^2f^2}{kq}.[/tex]
    Now, what you have been plugging in as t (1.6*10^6s) isn't t, it's the frequency f (1.6*10^6 Hz). Also, make sure that when you're plugging in units, you're using only base units. Only m should be in kg, r should be in m, f should be in Hz, k should be in N*m^2/C^2. Your answer should be in C/m.
     
  14. Oct 21, 2012 #13
    Alright I got 8.439 nC/m and that is still wrong, lol. Damn, I don't know where I'm going wrong. I'm plugging in everything:

    I got:
    m = 1.67*10^-27 kg
    r = .012m
    f = 1.6*10^6 Hz
    k = 9*10^9 N*m^2/C^2
    q = 1.6*10^-19 C
     
  15. Oct 21, 2012 #14
    That's because I had a typo and I wrote 1.6 MHz instead of 1.3 MHz.
     
  16. Oct 21, 2012 #15
    Ah, I finally got it. Thanks alot for all of your help, I really appreciate it.
     
  17. Oct 21, 2012 #16
    You're welcome.
     
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