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Magnetic Fields, Current, and Tension

  1. Feb 13, 2007 #1
    Hi everyone. I'm stuck on a physics problem and it's gotten me pretty confused.. and I came here wondering if anyone can guide me without giving me the answer and allowing me to learn the material instead of just receiving the answer without a clue as how to do the problem. I'd greatly appreciate any help ^^;; ... and thanks for reading this !!

    1. The problem statement, all variables and given/known data

    A horizontal wire 0.20 m long and 80 grams in mass is hung in a uniform B-field by two massless strings. The magnitude of the magnetic field is 0.070 T. When a 42 amp current passes through the wire, the wire swings upward forming an angle *.

    a. What direction is the current going through the wire?
    b. What does * equal?
    c. What is the tension in each of the two strings?

    Diagram: http://img.photobucket.com/albums/v696/talimtails/pp22A.jpg
    * = theta
    2. Relevant equations
    F = BILsin*
    B = kI/r (possibly)
    k (constant)= 2 x 10^-7 T-m/A (T = tesla, A = amps, m = meters)
    +Application of the right hand rule, with thumb being current, other fingers being magnetic field, and force coming out of the palm.

    3. The attempt at a solution
    For a, I said the current was going out of the page. I figured the magnetic field would be going down (N to S on the diagram) because magnetic fields usually are directed from + to -. So using my right hand, if the magnetic field (my four fingers excluding thumb) is down, the current (my thumb) seemed to be going out of the page towards me.

    For b, I thought there must be some kind of acceleration that is making the wire move. If so, F = ma, but F also equals BILsin*. Therefore, I set the two equal to each other and solved for *.
    sin* = ma/BIL = (.08 kg)(9.8 m/s^2)/(.07 T)(42 A)(.2m) ... * = 4/3.
    However, I quickly realized that 4/3 is not a possible solution for * as it can never go over 1 for sin. This is where I get stuck. I'm not sure, to be truthful, that acceleration is a factor since it seems the magnetic field would be forcing the wire to the N pole rather than acceleration... but I could not find any other applicable equation. Is there some obvious factor I'm missing? Or.. if you'd rather not give that, how should I go about finding the angle (without giving me a direct answer). I'd be helpful if I could understand why and then do the calculations myself rather than being giving an answer.

    For c, recalling early chapters, I would draw a free body diagram of the components I have and solve for T using that. In that case, would I consider the normal force (F_n) and the weight (mg)? I've never done this with magnetic fields before, so it leads me to believe that I should also consider the magnetic field.. as that seems to be the dominating force that determines the angle and therefore tension.
     
  2. jcsd
  3. Feb 13, 2007 #2
    Um... ^^; I guess I'll sleep on this.. but any help would still be appreciated :)
     
  4. Feb 14, 2007 #3

    JK423

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    Some notes (trying to solve this myself):

    a) correct. Simple application of the Right-Hand rule + Laplace force

    b)I understand from the problem that when the wire form an angle θ, it will balance at that position. So draw the forces acting on it:
    1)Weight
    2)The (2) tensions from the two strings
    3)The Laplace force

    Note that the laplace force is always horizontal to the "walls" N and S.
    Also the following is not correct: F=IBLsinθ. because the angle between the current`s direction and magnetic field`s direction is π/2. So, F=IBL.
    Thats why u got sin > 1.

    After drawing the forces, take the balance equations.

    Hope i helped a little
     
  5. Feb 14, 2007 #4
    I didn't know what the laplace force was.. so I looked it up and got the equation F = q(E + v x B) .. but seeing as how I don't have a velocity, electric field, or charge.. I dont know why I'm supposed to use it.

    Weight would be acting down, and tension would be acting on the strings.. but since tension is unknown, I don't know how to find the angle with only one variable.

    How do you know the angle between the magnetic field and current's direction is pi/2? And even if you only use IBL, how are you supposed to find an angle without an angle being used in the equation? Unless.... you mean to say that I should divide IBL by mg to get (42 A)(.07 T)(.2 m) / (.08 kg)(9.8 m/s^2) ... which would give me .75 and make * (theta) 48.5*....
    (I did that, because I drew a free body diagram with the force and weight, and solved for theta... would that work?)
     
  6. Feb 14, 2007 #5
    If anyone wants me to put the diagram I drew on paint I will.. :)
     
  7. Feb 14, 2007 #6

    JK423

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    Well you should study first before trying to solve problems..

    Laplace force is (if E=0): F=q*vxB and if you multiply and divide by "t" (time) you get: F=(q/t)*(vt)xB => F=I*LxB or dF=I*dLxB.

    Using the right-hand rule you get the force`s direction.
    Suppose the magnetic field goes from S to N or N to S. The current`s direction is perpendicular to B. Its easy to see from your diagram..
    So: F=I*L*B*sin(π/2)=IBL.
    Thats how you get the formula..
     
  8. Feb 14, 2007 #7
    Well I didn't know about Laplace force because we never went over that in class.. and it wasn't discussed in the chapter. So I had no idea to even begin using it. I do take my studies very seriously.. so please don't say or think I don't study.
    sin of pi/2 is 1, and then you can get IBL.. I see that.
    I was just wondering, since I know force is going to the right due to the right hand rule, would it be correct to divide the F I get from BIL by mg.

    See.. I did a FBD (free body diagram).
    Force is going to the right because of the right hand rule. We know that mg is going down. so sin* = oppsite / adjacent = (42 A)(.07 T)(.2 m) / (.08 kg)(9.8 m/s^2)
    And then I got * = 48.5*
     
  9. Feb 14, 2007 #8

    JK423

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    Sorry, i didn`t want to insult you or anything.. I just noticed that you trying to solve a problem without knowing its theory. Thats all.

    I don`t think that it`s correct to say: sinθ=FBL/mg. Show me your FBD if you can please.
     
  10. Feb 14, 2007 #9

    JK423

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    From the balance equations you get:

    T*cosθ=mg [1]
    T*sinθ=BIL [2]

    Thats all.
     
  11. Feb 14, 2007 #10
  12. Feb 14, 2007 #11
    The Laplace force (aka Lorentz force) is just the name for the force youre describing with the right hand rule and F=ILBsintheta. In this case, the current and field are at a right angle. JK423's equations are what you need, but remember that the tension is split between 2 strings. Your FBD is okay, but if you want the forces to sum to zero, your tension forces need to be in the opposite direction
     
  13. Feb 14, 2007 #12
    So instead of the way I drew tension in the diagram, it should be in the opposite direction.. right?
    And since tension is in two strings, I divide the answer I get for tension by 2.
    Now.. if my FBD is okay, then does that mean I got the angle correct?
     
  14. Feb 14, 2007 #13

    JK423

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    I dont think you can do that. The direction of the string is not necesserily(oups :P) the same with the ΣF of mg and BIL.
    Even if you could do that(which you cannot) it would be tanθ=f/mg.
    Draw all the forces (+ Tension), and get ΣF=0 on the axis which is best. You`ll get the equation [1] and [2] as i wrote above
     
    Last edited: Feb 14, 2007
  15. Feb 14, 2007 #14
    T*cosθ=mg [1]
    T*sinθ=BIL [2]

    I don't understand how you got those equations.. sin is associated with y axis, and cos is associated with the x axis... so wouldn't the cos and sin in your equations be switched?
     
  16. Feb 14, 2007 #15

    JK423

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    Last edited: Feb 14, 2007
  17. Feb 14, 2007 #16
    Oh.. you don't have to do that oo;! I don't want to trouble you so much.. =(
     
  18. Feb 14, 2007 #17
    That is the wrong way of thinking. Rules like that only work for angles in standard position. Youve got a triangle, go back to SOH CAH TOA and youll see how he got those equations
     
  19. Feb 14, 2007 #18
    Ohh... I understand now.:)
     
  20. Feb 14, 2007 #19
    So I can solve for the angle using this:
    mg/cosθ = BIL/sinθ
    mgtanθ = BIL
    tanθ = BIL/mg
    and.. solve for θ that way..
     
  21. Feb 14, 2007 #20

    JK423

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