1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Magnetic Forces (LOOP)

  1. Oct 8, 2009 #1
    A square current loop (ABCD) is oriented in the x-y plane. There is a current in the loop in a clockwise direction (as viewed from z > 0). A constant magnetic field is in the +y-direction. The length of each side of the loop is a. The +z-axis points out of the page.

    http://i662.photobucket.com/albums/uu347/TwinGemini14/LOOP.gif

    -------------
    1) The net force on side CD is in the _______ direction and the net force on side AB is in the _______ direction.

    A) +z, -z
    B) +x, +y
    C) -z, -x

    :: I said the answer is A. Simply apply the right hand rule. Fingers in direction of current and curl them in the direction of the magnetic field. Thus the thumb points to the force.
    -------------
    2) Compare the magnitude of the net force on the two sides AB and BC to the magnitude of the net force on the two sides CD and DA.

    A) |FAB+BC| > |FCD+DA|
    B) |FAB+BC| = |FCD+DA|
    C) |FAB+BC| < |FCD+DA|

    :: The net force of a closed loop must be 0. Therefore the answer must be B.
    -------------
    3) The potential energy of the loop is a maximum when it is oriented as shown in the diagram.

    A) True
    B) False

    :: I believe it to be A, true. The loop here is parallel to the magnetic field and the potential energy will be greatest when the loop is 180 degrees from the magnetic field. 0 when perpendicular, and negative when 0<theta<90
    -------------
    4) The net torque on the loop (due to the magnetic field) depends on the magnetic field strength.

    A) True
    B) False

    :: This is A, true. Because t = u x B; where u is the magnetic dipole moment and B is magnetic field. Therefore, t = uBsin(theta)
    --------------
    5) The net torque on the loop (due to the magnetic field) depends on the area of the loop.

    A) True
    B) False

    :: The answer is A, true. Similar logic to question 4, except realizing that u=NIA.
    (#loops)*(current)*(Area)
    --------------
    6) The magnitude of the torque on the loop will be a maximum when the area vector of the loop is perpendicular to the _______ plane and the magnitude of the torque on the loop will be a minimum when the area vector of the loop is perpendicular to the _______ plane.

    A) x-y, x-z
    B) y-z, x-y
    C) x-y, y-z

    :: I stated that the answer should be B. That is, when the area vector is perpendicular to the magnetic field, torque is 0 (minimum). This represents the x-y plane in our picture. If the area vector is parallel to the magnetic field, toque is IAB (maximum). This represents the y-z plane in our picture.
    ---------------
    7) Given that B = .02 T, I = .18 A, and a = 1.5 m, what is the magnitude of the torque on the loop when the area vector of the loop makes a 30° angle with the magnetic field?

    A) .0041 N*m
    B) .0092 N*m
    C) .0135 N*m
    D) .0183 N*m
    E) .0261 N*m

    :: Well, I said that:
    t = u x B
    t = uBsin(theta)
    t = (IA)Bsin(theta)
    t = (0.18)(1.5^2)(0.02)sin(30)
    t = 0.00405

    the answer is A.
    ======================================
    CAN SOMEBODY PLEASE REVIEW MY SOLUTIONS TO THESE PROBLEMS? I FEEL THE LEAST CONFIDENT ON QUESTIONS 6 AND 7. THANK YOU SO MUCH. I REALLY APPRECIATE THE HELP.
     
  2. jcsd
  3. Oct 8, 2009 #2

    Delphi51

    User Avatar
    Homework Helper

    1 to 5 look good to me.
    I stumped on 6. It depends where the turning axis is. If it is a horizontal line through the middle of the diagram, then maximum torque is as it is shown. After rotating 90 degrees, the two forces will try to pull the loop apart instead of rotating it, so zero torque. I chose answer A.

    Agree with your solution to 7.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook