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Homework Help: Cylinder On Inclined Plane

  1. Oct 23, 2013 #1
    1. The problem statement, all variables and given/known data

    There is a wire (current I = 1.05 A) that splits into two channels; of resistance R2 = 6.60 and R1 = 1.25 , and re-joins, forming a current in the shape of an isosceles triangle with base distance d = 6.90 cm and height L = 14.0 cm. The loop is entered into the space between the two poles of a magnet with a uniform magnetic field, B = 4.55eāˆ’2 T, that runs from one pole to the other. The loop is placed such that the field lies in the plane of the loop. What is the magnitude of the torque on the circuit about the wire's axis?

    (Edit: Image inserted by moderator)

    2. Relevant equations
    F=I*l x B

    Torque=r x F
    3. The attempt at a solution

    You can picture the triangle with R2 on the left side and R1 on the right side and the wire comes, current from top to bottom right in the middle and the sides meet back up halfway at the base of the triangle.

    The first thing I did was to find the current that is in each side of the wire. Since voltage drop is the same across both sides and both the currents add up to the current going in I derived a system of equation. I also tried to find the voltage using the total current and equivalent resistance and then find the individual currents using the total voltage which is the same for both of them.

    I got .1672 A for the left side and .8828 A for the right side.

    Now for the Force on each wire, since the current is going the same direction in each wire but they are on opposite sides of the axis of the wire the torques will be opposing directions but different strengths because of the difference currents, so using F=I*l*B*sin(theta)

    I found L using pythagoreas theorem side=0.1442 m and then theta using law of cosines theta=arccos(d/2/0.1442)=76.2 deg

    So l*sin(theta) would give you an arm perpendicular to the B field

    F1=I1*l*sin(theta)*B and F2=I1*l*sin(theta)*B

    both resulting in the direction out of the page

    Now for the torque, using the center of the triangle right on the axis of the wire as reference r points straight out a distance d/3. Since the force is directed straight out of the page for both wires but the radius is in the opposite directions this gives the opposing torques, and they are perpendicular so the cross product can go away with sin(pi/2)=1


    I ended up numerically with 1.049e-4 N*m not quite the right answer. Is this the best way of going about this problem?
  2. jcsd
  3. Oct 23, 2013 #2

    Simon Bridge

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    I thin we need the diagram.
    Is the current going around a triangular loop that has a couple of resistors in it or is this two parallel resistors with a current going the same direction through them wrt the magnetic field?

    (I suspect the latter... so where does the current leave the triangle?)

    You do want to divide the current into components parallel and perpendicular to the field.
  4. Oct 23, 2013 #3
    Okay here is a picture of the diagram http://i.imgur.com/LzjNPkV.png

    the current does not go in a loop and the whole wire has the resistances given
  5. Oct 23, 2013 #4

    Simon Bridge

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    OK - so the R1 and R2 are the resistances of the wires on leg 1 and leg 2.
    The point of that is to say that there is a different current going down each one
    ... so you can work that our right away. (Keep the variables.)

    Then realize that only the current that is perpendicular to the magnetic field counts to find the force.
  6. Oct 23, 2013 #5
    Yes I already did that, for I2 (higher resistance) I got It(total current)*R1/(R1+R2) And other other current is I1=It-I2

    And I calculated the angle between the wire and b field, and the length of the triangular side of the wire and found
    F1=I1*l*sin(theta)*B and F2=I1*l*sin(theta)*B
  7. Oct 23, 2013 #6

    Simon Bridge

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    OK great - for "l" would you use the total length of the wire or the perpendicular component of the length?

    You know the relationship between the sine and the lengths given?

    You know the relationship between force and torque?

    So all you have to do is divide the sloping bits of wire into short lengths at different distances from the center of rotation to find the torque on each bit - then add up all the torques.

    Unless you know a shortcut to get the effective moment arm for the total force?
  8. Mar 20, 2017 #7
    Several years late to the party, but I'm curious, was that last comment a hint that there IS a shortcut...?
  9. Mar 22, 2017 #8

    Simon Bridge

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    A lot of these kinds of integrals have been done already - so it is possible to look them up or work out equivalents.
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