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Find magnetic force on semicircle using line integral

  1. Nov 5, 2011 #1
    Here is the question:

    A very thin wire which follows a semicircular curve C of radius R,lies in the upper half of the x-y plane with its center atthe origin. There is a constant current I flowing counter clockwise, starting upward from the end of the wire on the positive x axis and ending downward at the end on thenegative x axis. The wire is in a uniform magneticfield, which has magnitude B0 and direction parallel to the z-axis in the positive z direction. Determine a symbolic answer in unit-vector notation for the total force on the wire due to the magneticfield. Ignore the forces on the leads that carry the current into the wire at the right end and out of the wire at the leftend.

    (Solution check: The numerical value with I =2.00 A, B0 = 3.00 T, and R =4.00 m is 48.0 N in j direction.)

    I am not getting the right answer.

    First, we know that dF[itex]^{\rightarrow}[/itex] = i dL[itex]^{\rightarrow}[/itex] x B[itex]^{\rightarrow}[/itex].

    So [itex]\int[/itex]dF[itex]^{\rightarrow}[/itex] = i [itex]\int[/itex]dL[itex]^{\rightarrow}[/itex] x B[itex]^{\rightarrow}[/itex].

    Then, since B is always perpendicular to L, we have a sin of 90 degrees. This implies that (along with B being constant)
    F = iB[itex]\int dL[/itex].

    The arclength of a circle is 2[itex]\pi[/itex]R. A semicircle's arclength is [itex]\pi[/itex]R.
    Therefore F = iB[itex]\pi[/itex]R.

    This doesn't get you to the right answer. What am I doing wrong? Why isn't my answer right? And is there some physical quantity that my answer does correspond to?
     
    Last edited: Nov 6, 2011
  2. jcsd
  3. Nov 5, 2011 #2

    rl.bhat

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    Homework Helper

    If you consider a small segment dL of the wire, the force acting on it will be
    dF = i*B*dL. The force is radial.
    Due to symmetry component of this force, dF*cos(theta), gets cancelled out.
    dF*sin(theta) adds up to give net force on the wire.
    Write dL as R*d(theta)
    Now find the integration from zero to pi to get the net force.
     
  4. Nov 6, 2011 #3
    It would be most appreciated if you answer the questions that are asked. You answered how to solve it. I didn't ask that. I asked why mine was wrong. There are subtle differences. For example, I'm still confused by your answer because I'm not sure how (or if one can) integrate the following, given some angle that B makes with some tiny L.

    [itex]\int[/itex]dF[itex]^{\rightarrow}[/itex] = i [itex]\int[/itex]dL[itex]^{\rightarrow}[/itex] x B[itex]^{\rightarrow}[/itex].

    So there are reasons I asked what I did.
     
    Last edited: Nov 6, 2011
  5. Nov 6, 2011 #4

    rl.bhat

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    Your method is correct if the the conductor is straight. In the semicircular conductor, the force on the each element of the conductor experiences a force in the radial direction. To find the net force, resolve dF into two components. Net dF*cos(theta) is zero. Net dF*sin(theta) is -i*B*R*cos(theta).Substitute the limits from 0 to pi to find the net force.
     
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