1. Limited time only! Sign up for a free 30min personal 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!

Please help, Solenoid and calc laps of a proton

  1. Apr 30, 2014 #1
    1. The problem statement, all variables and given/known data
    Hello,

    The question of the problem is: Solenoid, calculate laps of a proton?

    Someone could help me solve this exercise, but I try not to do this achievement . Would appreciate your help . This is the information that I provided to do the exercise:

    Picture of the year : http://prntscr.com/3eiwy0

    A proton with velocity v inside a solenoid of 10,000 laps , Long L and 1 A current (see figure) is triggered. The proton velocity vector has an angle of 40 ° with respect to the magnetic field direction . Calculate the turns while supplementing the proton travels a distance L , the time to complete each lap and total time T to travel L.

    The values ​​of the magnitude of the velocity ( in m / s ) and the length of the coil ( L in m) chosen according to the first letter of your name using the following code :
    A- 10 B- 20 C- 30 D -40 E -50 F- 60 G -70 H-80 ...
    N -140 O- ​​150 P -160 Q -170 R -180 S -190 U - 200 U -210 V -220 W -230 ...

    If your name is Domingo Pérez then v = 40 m / s and L = 160 m . If your name will begin with the same letter then use the number of the next letter to the family name. So if your name is Candido Colón use v = 30 m / s and L = 40 m .


    2. Relevant equations



    3. The attempt at a solution
    I try so many times to solve the problems, but for me is impossible to solve, i really appreciate all the help.
     
  2. jcsd
  3. Apr 30, 2014 #2
    Ok, well the first thing to do will be to find the magnetic field inside the solenoid:
    ##B = \mu_{0}in##, where ##n = \frac{turns}{length}##

    Then since the proton moves across field lines, it'll go into circular motion, so think of these equations:
    ##F_{r} = ma_{r} = m\frac{v^{2}}{r}##

    Remember that only the vertical component of the velocity will cause the circular motion.

    So from there you can find the period of the motion, pitch of the helix, etc.
     
  4. Apr 30, 2014 #3
    Data:

    Turns: 10,000
    Lenght: 180m
    Electic current: 1A
    Velocity vector: 40°
    Magnitude of the velocity: 100 m/s

    This is what it got at the moment:

    n=[itex]\frac{10,000}{180}[/itex]
    n=55.6m

    B=(4πx[itex]10^{-7}[/itex]T*m/A)(1A)(55.6m)
    B=6.99x[itex]10^{-5}[/itex]

    Now i need to Calculate the turns the proton complete while it travels a distance L, the time to complete each turn T and the total time to travel L.

    I dont have any clue, someone can give a help to aproach the problem?

    I aprreciate your help, thank you guys.
     
  5. Apr 30, 2014 #4
    Well, try starting with the period, T. Period is defined as: ##T = \frac{2\pi r}{v}##, which is just the circumference of the circular path divided by the speed, so can you solve for T from here?
     
  6. Apr 30, 2014 #5
    Well, i dont have r. But i saw a video in which they calculate R in the following way:

    R=(lenght/2∏)
    R=(180m/2∏)
    R= 28.65

    T= [itex]\frac{(2∏)(28.65)}{100}[/itex]
    T=1.80

    Turns: 10,000
    Lenght: 180m
    Electic current: 1A
    Velocity vector: 40°
    Magnitude of the velocity: 100 m/s

    Im not sure if the calculates are correct :(
     
  7. Apr 30, 2014 #6
    Well, r is the radius of the circular path (which will be a helix), so you don't want to use the 180m -- that's the length of the solenoid, not the circumference of anything. Look at the equation for circular motion I posted above -- you can solve for the radius from that.

    Also, since the particle is launched at an angle, it won't be the entire magnitude that goes into creating the circular path -- it will only be the component that's perpendicular to the field (since magnetic fields only act on things moving perpendicularly to them). So that's where the angle comes into play. The component that's instead parallel to the field is responsible for moving the particle down the solenoid, and pulling the circular path out into a helix.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Please help, Solenoid and calc laps of a proton
  1. Calc Help (Replies: 1)

  2. Calc two help please (Replies: 4)

  3. Calc Help (Replies: 1)

Loading...