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Homework Help: Cross product in spherical coordinates.

  1. Apr 30, 2012 #1
    1. The problem statement, all variables and given/known data

    i am trying to solve for the magnetic torque a circular loop of radius R exerts on a square loop of side length b a distance r away. The circular loop has a normal vector towards the positive z axis, the square loop has a normal towards the +y axis. The current is I in both loops.(Griffiths question 6.1)

    2. Relevant equations
    I used a dipole approximation and found the magnetic vector potential of the circular loop. I then took the curl of this in spherical coordinates and found an expression for B1, the magnetic field due to the loop. I know the torque exerted on the square loop is m2 X B1, where m2 is the magnetic dipole moment of the square (m2 = Ib^2 y)

    3. The attempt at a solution
    B1 = curl(A) =( u_0(m1) /(4 *pi* r^3) ) {2cosθ r + sinθ vartheta }
    m1 = I(pi)R^2 z.

    In order to take the curl, I switched B1 to cartesian coordinates. This is where I believed i made my mistake... I just used the following relationships:
    x = Rcos[itex]\phi[/itex]sin[itex]\theta[/itex]
    (for r I used the r component of B1, that is, R = u_0(m1) /(4 *pi* r^3) ) *2cosθ )
    [itex]\phi [/itex] is 0 here because there is no [itex]\phi[/itex] component of B1.

    y = 0 since sin([itex]\phi[/itex] is 0.

    z = R cos [itex]\vartheta[/itex]

    Now i just cross m2 = <0,Ib^2,0> with B1 using its components found above. However, my answer is incorrect. (The book uses the coordinate free form of B_dip in its solution, but I am trying to understand how to convert my expression for B1 from spherical to Cartesian coordinates.
  2. jcsd
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