# Force on a Current-carrying loop

by ninjarawr
Tags: currentcarrying, force, loop
 P: 12 1. The problem statement, all variables and given/known data 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 2. Relevant equations t = NIABsin(theta) B=unI (n= # of turns, u = 4 * 3.14 * E-7) F=ILBsin(theta) 3. 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?
 HW Helper P: 4,433 you have not used the number of turns in the solution.
P: 12
 Quote by rl.bhat 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.

 HW Helper P: 4,433 Force on a Current-carrying loop It will be 4 times the force on each turn.
P: 12
 Quote by rl.bhat 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?
HW Helper
P: 4,433
 Quote by ninjarawr 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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