Force on a Current-carrying loop

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Homework Help Overview

The problem involves a rectangular loop of wire with multiple turns carrying a current in a uniform magnetic field. The loop's dimensions and the magnetic field's direction are specified, and the task is to determine the torque and forces acting on the loop.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the calculation of torque and forces, questioning the application of the number of turns in the force calculation. There is an exploration of how to correctly incorporate the number of turns into the force equations.

Discussion Status

Some participants have provided guidance on the relationship between the number of turns and the forces acting on the loop. There is an ongoing exploration of the reasoning behind multiplying the force by the number of turns, with some participants seeking clarification on this approach.

Contextual Notes

Participants are navigating the constraints of the problem, including the specific dimensions of the loop and the uniform magnetic field's orientation. There is also a focus on ensuring that all relevant factors, such as the number of turns, are properly accounted for in the calculations.

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Homework Statement



A rectanglular loop consists of 4 turns of wire carrying a current of 2.7 A. The loop is in the x-y plane, and the direction of flow of the current is shown in the figure. The loop has dimensions a = 1 cm and b = 6 cm. Consider a uniform magnetic field of strength 2.3 × 10-4 T in x, y, or z directions.

If the uniform field of 2.3 × 10-4 T is along the +x axis, find the magnitude of the torque acting on the loop and the total force on side a and b.

Diagram:
http://uploadpic.org/view-pic.php?img=100601"
http://uploadpic.org/view-pic.php?img=100601

Homework Equations


t = NIABsin(theta)
B=unI (n= # of turns, u = 4 * 3.14 * E-7)
F=ILBsin(theta)


The Attempt at a Solution


-I found torque using the above equation, and it was 1.49E-6
-Fb is simply 0 because it is parallel to the magnetic field
-but I have NO IDEA how to solve for Fa. I tried using F=ILBsin(theta), (=2.7*.1m*2.3E-4) and it rejects my answer. I also tried =2.7*.1m*2.3E-4*2.

Please help me find Fa?
 
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you have not used the number of turns in the solution.
 
rl.bhat said:
you have not used the number of turns in the solution.
I did use it to find torque, but did not use it when finding force. how do I use it in the solution? please tell me.
 
It will be 4 times the force on each turn.
 
rl.bhat said:
It will be 4 times the force on each turn.

thanks! that worked for me! But why do we do 4 times for each turn when we are only looking at one side?
 
ninjarawr said:
thanks! that worked for me! But why do we do 4 times for each turn when we are only looking at one side?

Because each side contains four terns.

What is torque? FXb. In that calculation you have used no. of terns to calculate the force. Same thing is true for Fa.
 

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