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Force of the magnetic field

  1. Feb 17, 2014 #1
    1. The problem statement, all variables and given/known data

    A long wire carrying 6.50A of current makes two bends, as shown in the figure (Figure 1) . The bent part of the wire passes through a uniform 0.280T magnetic field directed as shown in the figure and confined to a limited region of space.

    http://session.masteringphysics.com/myct/itemView?assignmentProblemID=31713281&offset=next

    Part A: Find the magnitude of the force that the magnetic field exerts on the wire.
    Part B: Find the direction of the force that the magnetic field exerts on the wire.
    2. Relevant equations

    F=IlB

    3. The attempt at a solution

    for part A I calculated the force by the given information.
    I= 6.50A, B= 0.280T, and l=0.8m(I did the Pythagorean theorem to find the lenght)

    F=(6.50A)(0.8m)(0.280T)

    F= 1.456N.

    I found the force, but I got the answer wrong. Can someone explain where did I go wrong? I figured that I have the length incorrect, but I don't know how to find the length.
     
  2. jcsd
  3. Feb 18, 2014 #2

    Larry Gopnik

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    Hi,

    Is it possible to see a screenshot of the question with the diagram? The link doesn't work.
     
  4. Feb 18, 2014 #3

    Simon Bridge

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    Yep - link does not help: "not signed in" message.
     
  5. Feb 18, 2014 #4
  6. Feb 20, 2014 #5

    Simon Bridge

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    OK - so you want ##\vec F = q\vec v \times \vec B = L\vec I \times \vec B##
    Notice the cross product there?
    See: http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/forwir2.html

    But from what I can see of your working - it looks like you left off part of the wire.

    Note: the diagonal section makes a 1-2-√3 triangle,
    so sin(30)=cos(60)=1/2, cos(30)=sin(60)=(√3)/2

    The opposite side is O=40cm by trigonometry: O = Hsin(30) = H/2 means H = 80cm.

    It's a good idea to look out for these special triangles.
     
    Last edited: Feb 20, 2014
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