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Vector cross products involving magnetic forces

  1. Oct 23, 2004 #1
    magnetic forces and torque

    The plane of a rectangular loop of wire with a width of 5.00cm and a height of 8.00cm is parallel to a magnetic field of magnitude 0.200T. The loop carries a current of 6.60A.

    What is the maximum torque that can be obtained with the same total length of wire carrying the same current in this magnetic field?

    I found the torque that acts around the loop to be 5.28*10^-3 N*m. I figured that since the torque = IBA sin(theta) and that the length, area, and magnetic field must the same, the maximum torque must be when sin(theta)=1. Wouldnt' this mean that the torque I found is the max amount?

    Thanks!!
     
    Last edited: Oct 24, 2004
  2. jcsd
  3. Oct 24, 2004 #2
    can anyone help me out?
     
  4. Oct 24, 2004 #3

    Tide

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    The question is somewhat ambiguous. In any case, if the wire is planar the maximum torque on it will be when the normal to the cross section is perpendicular to the magnetic field. I think the question is what combination of length and width must the wire have in order to maximize the torque - subject to the constraint that the total length of the wire is fixed.

    I say it's ambiguous because it's possible that shapes other than rectangular might lead to yet greater torque - such as a wire formed into a circular loop or something with many loops! My guess would be they are asking for the rectangular case.
     
  5. Oct 24, 2004 #4
    I dont think i understand what they are asking for. I tried changing the area of the rectangle by making one side .07 and the other 006 to find the maximum area possible but i dont know if its rihgt. Anyone?
     
    Last edited: Oct 24, 2004
  6. Oct 24, 2004 #5
    help please!?
     
  7. Oct 24, 2004 #6

    Tide

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    The torque depends on the product of the IL (force = current times length) and W (moment). If L + W is fixed then how can you maximize the product of ILW? (I.e. how do you maximize the AREA of a rectangle if its perimeter is held constant?)
     
  8. Oct 24, 2004 #7
    does it involve some method of calculus? maybe taking the derivative and setting it to zero? Sorry, i am totally stuck on this problem.
     
  9. Oct 24, 2004 #8

    Tide

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    You can do it without calculus. You want to maximize the area A = LW with fixed perimeter. Let P be the perimeter P = 2(L+W). Then A = L(P/2 - L) which you will recognize as the equation for a parabola. You can find the vertex of the parabola by completing the square so that

    [tex]A = \left( \frac {P}{2} \right)^2 - \left( L - \frac {P}{4} \right)^2[/tex]

    so that the maximum occurs when L = P/4 (and therefore W = P/4) which makes the shape a square!
     
  10. Oct 24, 2004 #9
    ok, i found the perimeter to be .26m so each side of the square would be .26/4= .065m. i then found the area of the square with the length and multiplied it by the magnitude of the magnetic field and the current and got .00557 and that is still wrong i am told.
     
    Last edited: Oct 24, 2004
  11. Oct 24, 2004 #10

    Tide

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    You should always write out the units when referring to a physical quantity.

    Who told you it's "wrong" and what is the answer "supposed" to be?
     
  12. Oct 24, 2004 #11
    its a problem on my physics webassign. it only gives my a couple chances before i get the problem wrong.
     
  13. Oct 25, 2004 #12

    Tide

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    Hmm. That's strange. Have you checked your units? I don't know what else to tell you.
     
  14. Oct 25, 2004 #13
    Are you sure that you must keep the shape to be a rectangle....becuase a single circle would give you more torque than a square...I would guess that the ambiguous question is asking for the shape that is planar and makes a single loop. in tis case, a circle will provide you with the maximum torque.
     
  15. Oct 25, 2004 #14
    if thats the case, the length of the wire would just be the perimeter correct?
     
  16. Oct 25, 2004 #15
    yea...the length of the wire would be the circumference....which is 2(pi)r

    So the perimeter is equal to that....divide by 2(PI) and then you know r....then use area of a triangle to get its area
     
  17. Oct 25, 2004 #16
    thanks alot! i didnt use the triangle equation. i just found r and then used the area of a circle equation.
     
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