Magnetic Forces on infinite current-carrying wires

In summary, the magnetic force per length acting upon wire 2 is F12=-I3LB[3(√2)/2](i-hat)-[(√2)/2]I3B3(j-hat)
  • #1
cookiemnstr510510
162
14

Homework Statement


A) Three infinite, current-carrying wires are arranged as shown below. In terms of the given quantities, determine the magnitude and direction of the magnetic force per length acting upon wire 2. (Part A.jpg attached)
B) Three infinite, current-carrying wires are arranged as shown below in terms of the given quantities, determine the magnitude and direction of the magnetic force per length acting upon wire 5.(Part B.jpg attached)

Homework Equations


B=(μ0I)/(2πd)
F=ILxB

The Attempt at a Solution


A) Both wires 1 and 3 create a magnetic field at the location of wire 2.
Wire 1's magnetic field points straight upward in the j-hat direction.
Wire 3's magnetic field points at a -45° angle from the positive i-hat direction which is from left to right.

Using right hand rule:

F1 on 2 is in the positive i-hat direction
and
F3 on 2 is -45° from the negative i-hat direction
These forces are shown in (forces.jpg)

F1on2=I2LB1i-hat
B1=(μ0I1)/(2πd)
therefore:
F1on2=(I2I1LB1)/(2πd) i-hat

F3 on 2=I2LxB2= here is where I am stuck- when doing the cross product here is the θ the angle between both wires? I guess I am confusing the cross product we do for magnetic field and the one we do for forces..

Any help would be appreciated!
 

Attachments

  • Part A.JPG
    Part A.JPG
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  • Part B.JPG
    Part B.JPG
    12.9 KB · Views: 268
  • Forces.JPG
    Forces.JPG
    21.7 KB · Views: 273
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  • #2
The magnetic force directions you show are incorrect. The magnetic force due to each wire is not along the line joining the wires but perpendicular to it as prescribed by the right hand rule.
 
  • #3
Hello kuruman!
So I was thinking about the magnetic field as (magneticfield.jpg) (Small error in my picture where it says B2 it should say B3)
From there I did the RHR and obtained my forces.
I guess I am still incorrect?
 

Attachments

  • magneticfield.JPG
    magneticfield.JPG
    12.1 KB · Views: 304
  • #4
The directions for the magnetic fields are correct as shown in post #3. Now you have to do another right hand rule diagram for the forces.
 
  • #5
kuruman said:
The directions for the magnetic fields are correct as shown in post #3. Now you have to do another right hand rule diagram for the forces.
Hmm, okay. So for B1 for image in post 3, I point my four fingers in the direction of current of the wire in question (wire with current I2) and curl my fingers up towards B1 which would result in a force directly to the right. Is this correct? or am i confusing RHR's
I can try to draw a picture of what I am representing if that would make it easier to understand.
 
  • #6
There are many ways to use one's fingers for the right hand rule. Please post a drawing of the two forces acting on the wire and we'll take it from there.
 
  • #7
I guess I am not seeing how my Forces.jpg image is incorrect in post 1. I realize you said that the forces need to be perpendicular to the line adjoining wires, so does that mean into or out of the page since those are both perpendicular to the lines adjoining them?
 
  • #8
cookiemnstr510510 said:
I guess I am not seeing how my Forces.jpg image is incorrect in post 1.
Sorry, your forces in that diagram are correct. I got confuse with the mislabeled B2 that should have been a B3. You now need to write F3 in component form and add the two forces.
 
  • #9
kuruman said:
Sorry, your forces in that diagram are correct. I got confuse with the mislabeled B2 that should have been a B3. You now need to write F3 in component form and add the two forces.
ahh, okay. Yes my images were confusing!

So I am getting the total force acting on wire 2 is:
From here on out F1 on 2= F12
and F3 on 2=F32

Ftotal=F12(i-hat) - F32cos(45°)(i-hat) - F32sin(45°)(j-hat)
 
  • #10
After thinking about a bit more:
F=ILxB
F12=I1LB1sin(90)(i-hat)=I1LB(i-hat)
The question is asking for Force per length so
(F12/L)=I1B1(i-hat)
and
F32=I3LxB3sin(90) (-cos(45)-sin(45)) I am saying sin(90) for the cross product because the angle between current I2 and B32 is 90 degrees... Therefore
F32=-I3LB[3(√2)/2](i-hat) - I3LB3[(√2)/2](j-hat)
(F32/L)=-[(√2)/2]I3B3(i-hat)-[(√2)/2]I3B3(j-hat)

(Ftotal/L)=I1B1(i-hat)-[(√2)/2]I3B3(i-hat)-[(√2)/2]I3B3(j-hat)
 
Last edited:

Related to Magnetic Forces on infinite current-carrying wires

1. What is a magnetic force?

A magnetic force is a type of force that arises due to the interaction between moving charged particles and a magnetic field. It can cause objects to move or experience a change in direction.

2. How does a current-carrying wire create a magnetic force?

When a current (flow of electric charge) passes through a wire, it creates a magnetic field around the wire. This magnetic field interacts with other magnetic fields, such as those created by other wires, and can result in a magnetic force.

3. Can the strength of the magnetic force on a current-carrying wire be changed?

Yes, the strength of the magnetic force on a current-carrying wire can be changed by altering the current, the length of the wire, or the distance between the wire and the source of the magnetic field.

4. What is the direction of the magnetic force on a current-carrying wire?

The direction of the magnetic force on a current-carrying wire can be determined using the right-hand rule. If the thumb of your right hand points in the direction of the current, and your fingers wrap around the wire in the same direction as the current, the direction in which your palm faces is the direction of the magnetic force.

5. Are there any real-world applications of magnetic forces on infinite current-carrying wires?

Yes, there are many real-world applications of magnetic forces on infinite current-carrying wires. Some examples include electric motors, generators, MRI machines, and particle accelerators.

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