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Direction of Force on Circuit

  1. Dec 9, 2013 #1
    1. The problem statement, all variables and given/known data

    A wire of negligible resistance is bent into a rectangle as shown in the figure, and a battery and resistor are connected as shown. The right hand side of the circuit extends into a region containing a uniform magnetic field of 38 mT pointing into the page. Find the magnitude and direction of the net force on the circuit.

    3. The attempt at a solution
    I already got the magnitude to be 15.2 mN.

    I know that the horizontal portions of the wire will cancel out, and that the magnetic field points into the page. But I don't know how to figure out the rest...

    The right hand rule says to point the fingers toward the direction of the magnetic field, and the thumb in the direction of the velocity (in this case the direction of the current?) and that the palm will indicate the direction of the force. But when I try that, the direction of force is to the left, and the answer says it should be to the right. Can someone explain this to me?
     

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    Last edited: Dec 9, 2013
  2. jcsd
  3. Dec 9, 2013 #2

    rude man

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    F = i L x B. L = vector representation of the section of the coil in the B field. The direction depends on direction of current as well as orientation of the section in the B field.

    Write L and B as vectors using a cartesian coordinate system. Assign unit vectors to L and B. Then use the rule for the vector cross-product given the unit vectors of each. That is easier and safer than using the rt-hand rule IMO.
     
  4. Dec 9, 2013 #3
    Can you show me a brief example please?
     
  5. Dec 9, 2013 #4

    rude man

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    Say a wire of length L is lined up in the y direction and the current moves in the +y direction. Say also that the B field points in the positive x direction.

    Then,
    i L = i L j
    B = B i
    Then F = iL x B is in the -k direction (the -z direction).
    Vectors are in caps.

    You need to learn the sequence of ijk unit vectors for a cross-product:

    i x j = k
    i x -j = -k
    k x i = j
    -k x j = i
    etc.

    This sequence is easy to remember if you put i at 12 o'clock on a circle, j at 4:30 and k at 7:30. Then if you go clockwise the cross-product sign is + and if you go counterclockwise the cross-product sign is -.

    A minus sign in front of either unit vector changes the sign of the cross-product from the above rule. A minus sign on both unit vectors is the same as if no minus signs were there.

    This system gets more useful the more cross-products there are in one term.

    So now apply the foregoing to your situation.
     
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