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Prove that the torque of any current loop is m X B

  1. May 30, 2015 #1
    1. The problem statement, all variables and given/known data
    Problem 6.2 of Griffith's "Introduction to Electrodynamics": Starting from the Lorentz force law ##\vec F=\int I (d\vec l \times \vec B)##, show that the torque on any steady current distribution (not just a square loop) in a uniform field ##\vec B## is ##\vec m\times \vec B##. (##\vec m## is the magnetic moment.)

    2. Relevant equations
    Let the torque be ##\vec N##.

    ##d\vec N = \vec r\times d\vec F##.

    3. The attempt at a solution
    Useful identity: ##\oint \vec r\times d\vec l = 2\vec a##, where ##\vec a## is the area of the loop and points perpendicularly to its surface.

    My question: the solution says that ##d\vec r = d\vec l##, which I don't understand. They are clearly pointing in different directions. ##d\vec r## points in the direction from the origin to the point ##r##, while ##d\vec l## points in the direction of the wire of the loop, which in general is different from the direction of ##d\vec r##.

    The solution:
    Screen Shot 2015-05-30 at 11.27.44 pm.png
     
  2. jcsd
  3. May 30, 2015 #2
    It is ##\vec{r}## that points from the origin to the point r =(x,y,z) of the loop and ##\vec{r'}=\hat{i}(x+dx)+\hat{j}(y+dy)+\hat{k}(z+dz)## that points from the origin to the point r'=(x+dx,y+dy,z+dz) of the loop. Their difference is the ifinitesimal vector ##d\vec{r}=\hat{i}dx+\hat{j}dy+\hat{k}dz## which is the same as dl.
     
  4. May 30, 2015 #3
    Thanks a lot! I've got it now.
     
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