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Triangular loop in magnetic field

  1. May 13, 2004 #1
    Sup all. I need help setting this problem up.

    A right-triangular current loop has two 45° angles and a hypotenuse a = 5 cm. The loop lies in the x-y plane, with its hypotenuse parallel to the y-axis. A net current I = 0.42 A circulates in the loop in the direction shown in the figure. A spatially uniform magnetic field B = 0.3 T points in the +x direction. The +z direction is OUT of the screen.

    Calculate:
    1) F_Z (AB)=?
    2) F_Z (BC)=?
    3) F_Z (CA)=?

    Where F_Z is the magnetic force in the Z direction on side AB, BC, or CA.

    Now, I know that Force_B = iLxB
    where L is length of side
    and B is the magnetic field.

    Obviously, I have to do some cross product action here.
    I know so far that the answer for 1) is (.42A)(.05m)(.3T) sin theta=-.0063N.
    Now, I'm not sure why that is the answer. Is it because the angle is 270 degrees, and thats why the answer is negative?

    I'm stuck on the last two. I do know that the answer for 3 is either going to be the same as 2, or equal in magnitude but opposite in sign is my guess.

    Any help greatly appreciated.
     

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    Last edited: May 13, 2004
  2. jcsd
  3. May 13, 2004 #2

    Doc Al

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    Staff: Mentor

    Right.
    What makes you think that is the answer? What's the angle θ? What's the sine of that angle? (Did you check the arithmetic?) Use the right hand rule to find the direction of the force (the direction of the cross-product). Does it point in the positive or negative z direction?
     
  4. May 13, 2004 #3
    Ok, I figured out why its negative. Right hand rule shows me the angle between the magnetic field and the current is -90 degrees. Also, I know thats the RIGHT answer because the computer told me so. Sorry if I did not indicate that earlier. So how do I setup these last two parts? Do I have to create vectors and cross them?
     
  5. May 13, 2004 #4

    Doc Al

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    Staff: Mentor

    I don't think you're quite done with the first problem yet. :smile: (And all three are solved in exactly the same manner anyway.) Some comments for #1:

    (a) Use the right hand rule to find the direction of the force, not the angle between the magnetic field and the current. The right hand rule tells me that the force points into the page, thus along the negative z-axis. That's why it's negative.

    (b) The angle between the magnetic field and the current is simply 90 degrees. (But it doesn't matter whether you call it 90, -90, or 270.) Use F = iLBsinθ to find the magnitude of the force. (The direction is given by the right hand rule.)

    (c) Regardless of what the computer says, do you think that (.42A)(.05m)(.3T) sin 90 = 2.52N ?? :uhh:
     
  6. May 13, 2004 #5
    My bad, I computed the answer wrong. Now I feel like an idiot. Supposed to be -.0063N, not 2.52N. :(


    Okay, now is part 2 and 3 done the same way? Sorry, but I suck at using the right hand rule. I know that I curl my fingers in the direction of the magnetic field, but sometimes I dont get it right. Now when the current is traveling down BC, right hand rule tells me the direction of the force points diagonally to the right? Likewise, right hand rule tells me the direction of the force along side AC is down to the left?

    Am I even remotely close?
     
  7. May 14, 2004 #6

    Doc Al

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    Staff: Mentor

    learn the right hand rule!

    One version of the right-hand rule (the one I use) is to curl your fingers as if rotating the first vector (iL) into the second (B): then your thumb will point in the direction of the force. Remember that any cross-product of two vectors is always perpendicular to both vectors.

    This may help: http://www.physics.brocku.ca/faculty/sternin/120/slides/rh-rule.html
    No! Since iL and B are in the x-y plane, the magnetic force must be perpendicular to that plane: that means along the z-axis. Use the right hand rule to find out if it's pointing towards + z or - z. (Didn't you wonder why you are asked for the z-components of the force?)
     
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