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Magnetic Force due to current Parallel Wires

  • Thread starter jegues
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  • #1
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Homework Statement



See figure attached

Homework Equations





The Attempt at a Solution



I've attempted to draw the magnetic fields resulting at the location of wire 4 due to wires 1, 2, 3.

I denoted them as B4,1, as in the "The magnetic field at 4 as a result of 1".

Are these correct?

I've been having a really tough time finding the resulting magnetic force for this problem, I want to make sure I've got the first few steps right.

I know that the net magnetic field at 4 will be perpendicular to the current in wire 4 regardless.

So to find the force all I need to do is apply,

[tex]F_{B} = iB[/tex] (We are calculating per meter of wire length)

Is this correct?

How do I get the force in its components if I want?

Thanks again!
 

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Answers and Replies

  • #2
Delphi51
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It all looks good. You'll have to resolve B4,2 into its x and y components and add them to the purely x B4,1 and purely y B4,3 to get the total force in each direction.
 
  • #3
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It all looks good. You'll have to resolve B4,2 into its x and y components and add them to the purely x B4,1 and purely y B4,3 to get the total force in each direction.
Okay so,

EDIT: Whoops calculator was in radians,

[tex]B_{4,1} = 1.11*10^{-5}T[/tex]

[tex]B_{4,2} = 7.86*10^{-6}T[/tex]

[tex]B_{4,3} = 1.11*10^{-5}T[/tex]

Breaking B4,2 into its components,

[tex]B_{4,2x} = 5.55*10^{-6}T[/tex]

[tex]B_{4,2y} = 5.55*10^{-6}T[/tex]

So the net B's of each component,

[tex]B_{xnet} = 5.54*10^{-6} \hat{i} T[/tex]

[tex]B_{ynet} = 1.67*10^{-5} \hat{j} T[/tex]

Do I now use,

[tex]F_{B} = iB[/tex]

to calculate the forces?
 
Last edited:
  • #4
Delphi51
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I agree with the 7.86 x 10^-6.
But after multiplying by cos(45) I get 5.55, not 6.98.
Add the x components, add the y components, then use F = i*B.
 
  • #5
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I agree with the 7.86 x 10^-6.
But after multiplying by cos(45) I get 5.55, not 6.98.
Add the x components, add the y components, then use F = i*B.
I'm getting this as the final answer,

[tex](-125 \mu N/m)\hat{i} + (41.7 \mu N/m)\hat{i} [/tex]
 
  • #6
Delphi51
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Yes, agree. Tricky business getting the directions of the forces!
 
  • #7
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Yes, agree. Tricky business getting the directions of the forces!
Yes. Perhaps easier to first rotate the coordinate system.
 
  • #8
gneill
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If you're comfortable with vector operations, then let the math do the work and calculate the force as F = I x B. In this case the current is the vector 7.5A<0, 0, 1>, since it's "coming out of the page", so it's flowing in the +z direction.
 
  • #9
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Yes. Perhaps easier to first rotate the coordinate system.
Can you explain?

I'd love to know easier methods to solve this problem, I get confused with the directions myself.
 
  • #10
gneill
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I'd love to know easier methods to solve this problem, I get confused with the directions myself.
For a foolproof method, work strictly with vectors and trust that the vector operations will take care of the direction machinations. In the case of this problem, the currents can be given vectors in the z-direction and the magnetic fields computed using cross products with radius vectors (with the usual constants included). Similarly, the net force is given by a cross product of the wire 4 current and the net magnetic field produced by summing the fields from the other wires.

Attached is an example showing how this problem can be worked in this manner. Note that vectors are shown in "column format" (the components are stacked vertically for ease of alignment and viewing).
 

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  • #11
1,097
2
For a foolproof method, work strictly with vectors and trust that the vector operations will take care of the direction machinations. In the case of this problem, the currents can be given vectors in the z-direction and the magnetic fields computed using cross products with radius vectors (with the usual constants included). Similarly, the net force is given by a cross product of the wire 4 current and the net magnetic field produced by summing the fields from the other wires.

Attached is an example showing how this problem can be worked in this manner. Note that vectors are shown in "column format" (the components are stacked vertically for ease of alignment and viewing).
So can the formula for the magnetic field be written as,

[tex]B = \frac{\mu_{0}i \times \hat{r}}{2 \pi}[/tex]

?
 
  • #12
gneill
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So can the formula for the magnetic field be written as,

[tex]B = \frac{\mu_{0}i \times \hat{r}}{2 \pi}[/tex]

?
Try:

[tex] \vec{B}(\vec{r}) = \frac{\mu_0}{2 \pi} \left( \frac{\vec{i} \times \vec{r}}{\vec{r} \cdot \vec{r}} \right)[/tex]

and r is the vector from the current i to some location.

If you want the radius vector to extend from the destination to the source of the field, change the sign.
 
  • #13
4,239
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Can you explain?

I'd love to know easier methods to solve this problem, I get confused with the directions myself.
Rotate the drawing 45 degrees clockwise, putting wire #1 at the top of the page. First group the wires #1 and #3 together and solve for their forces on wire #4 together. Notice that the contributions in the vertical direction cancel. That is B1y + B3y=0. The contribution from wire #2 on wire #4 has only a vertical component and bobs your uncle.
 
  • #14
SammyS
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The force per unit length for two parallel conducting wires is a result commonly covered in introductory physics texts.

"Principles of Physics: a calculus-based text", Volume 1
By Raymond A. Serway, John W. Jewett

gives the result as: [tex]\frac{F}{\ell} = \frac {\mu_0 I_1 I_2} {2 \pi a}\,,[/tex] where the separation is a and the currents are I1 & I2.
 

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