How Does Orientation Affect Magnetic Forces in a Current Loop?

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SUMMARY

The discussion focuses on the effects of orientation on magnetic forces in a square current loop situated in a constant magnetic field. The loop, with a current flowing clockwise, experiences net forces and torques that depend on its orientation relative to the magnetic field. Key conclusions include that the net force on a closed loop is zero, the potential energy is maximized when the loop is parallel to the magnetic field, and the torque is influenced by both the magnetic field strength and the area of the loop. The calculations for torque yield a maximum when the area vector is perpendicular to the magnetic field.

PREREQUISITES
  • Understanding of magnetic fields and forces
  • Familiarity with the right-hand rule for determining force direction
  • Knowledge of torque calculations in physics
  • Basic principles of electromagnetism, including magnetic dipole moments
NEXT STEPS
  • Study the right-hand rule in detail for magnetic force applications
  • Learn about magnetic dipole moments and their significance in torque calculations
  • Explore the relationship between current, magnetic field strength, and torque in current loops
  • Investigate the effects of loop orientation on magnetic forces in various configurations
USEFUL FOR

Physics students, educators, and professionals in electromagnetism, particularly those focusing on magnetic forces and torque in current-carrying loops.

TwinGemini14
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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

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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.
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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.
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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
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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)
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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)
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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.
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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.
 
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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.
 

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