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This should be easy

  1. Sep 29, 2004 #1
    Grr I've been trying to figure this out and i'm just getting frustrated because I can't find any help in my book...

    A long straight wire carries a current of 20 A. An electron, traveling at 2X10^7 m/s, is 3 cm from the wire. What force (magnitude and direction) acts on the electron if the electron's velocity is directed toward the wire?

    given: ok, so i know that the wire is long and straight (so i know how its magnetic field will act) and i know the current in it..I also know the charge on the electron..it's velocity and its distance from the wire...

    I'm supposed to find the magnitude and direction of the force acting on the electron... Ok- so let's say the wire is coming out of the page- i know the magnetic field will be counter-clockwise so in that case, the force would be "up" (perpendicular to the current.) I can find the force on the electron due to the magnetic field at it's current point using the equation B= (uoI/2pi r)
    where uo is the constant. I really dont know though how to adjust this for the charge on the electron or the velocity. I know how a charge would act in a uniform magnetic field and i remember that when a charged particle moves it creates an electric field- so i know if a electron moves from infinity to any location close to a proton or electron i can find the force...

    gah i cant figure it out! please help!
     
    Last edited: Sep 29, 2004
  2. jcsd
  3. Sep 29, 2004 #2

    Doc Al

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    Staff: Mentor

    You need to know how to find the magnetic force on a moving charge:
    [tex]\vec{F} = q \vec{v}\times\vec{B}[/tex]
     
  4. Sep 29, 2004 #3
    yeah i tried that- here's what i have done:
    i got the value of B to be 1.33X10^-4 I then used F=qvBsin(90) and got the force to be -4.272X10^-16 which it says is wrong

    I also put in that the force was going to be perpendicular to the current but apparently that was wrong because I missed it (and i only get one shot)- grrrr!!!
    it turns out the force is parallel or the current?? i dont understand that- i look at it again and figured it to be antiparallel (using the right hand rule)


    .oh wait- duh...MAGNITIDE no it's not a negative number...nevermind...

    still does anyone understand why the force would be parallel and not antiparallel to the current in the wire?
     
    Last edited: Sep 29, 2004
  5. Sep 29, 2004 #4

    Doc Al

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    [edit: This is correct.]
    [edit: This is the correct magnitude.]

    It's certainly not perpendicular to the current! How did you get anti-parallel? If the current is out of the page, the B field is counterclockwise, and -v x B is out of the page and thus parallel to the current.

    Edit: I had originally said that GreenDinos's numbers were wrong--my mistake: I dropped an exponent in doing the arithmetic. :uhh:
     
    Last edited: Sep 30, 2004
  6. Sep 29, 2004 #5
    off by a trillion? it was the right answer- it just wanted magnitude which means the absolute value.- i realized this then it said it was correct

    perpendicular- i got this at first because i was thinking about a point in space- if the current in the rod was coming out of the page towards you the magnetic field is counter-clockwise- if there was a point there (let's say north of the wire) it would feel the force perpendicular to the wire- to the west.

    Ok then i realized it's different for a moving particle- I drew a figure where the current ran in a wire from south to north (from left to right on the page) which means that above the wire, the magnetic field it out of the page and under it it is into the page. Here, no matter if i put the electron above or below the wire i get antiparallel- i use the right hand rule where my fingers are in the direction of v then i curl them towards B- where my thumb points is the force. every time it points to the opposite of where the current is running.


    ...oh wait- because it's negative that you do the opposite of the right hand rule....ok
     
  7. Sep 30, 2004 #6

    Doc Al

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    Oops... my mistake. I dropped an exponent. Your answers are numerically correct. Sorry about that!

    That's it!
     
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