Magnetic Force due to current Parallel Wires

In summary: So the net force on wire #4 is just F4=B1.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... so it doesn't contribute anything. So the net force on wire #4 is just F4=B1.
  • #1
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3

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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  • #2
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
Delphi51 said:
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
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
Delphi51 said:
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
Yes, agree. Tricky business getting the directions of the forces!
 
  • #7
Delphi51 said:
Yes, agree. Tricky business getting the directions of the forces!

Yes. Perhaps easier to first rotate the coordinate system.
 
  • #8
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
Phrak said:
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
jegues said:
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
gneill said:
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
jegues said:
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
jegues said:
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
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.
 

What is magnetic force due to current parallel wires?

Magnetic force due to current parallel wires is the force exerted between two current-carrying wires that are parallel to each other. It is caused by the interaction between the magnetic fields created by the two wires.

How is the direction of magnetic force determined?

The direction of magnetic force due to current parallel wires is determined by the right-hand rule. If the current in both wires is flowing in the same direction, the force will be attractive. If the current is flowing in opposite directions, the force will be repulsive.

What factors affect the strength of magnetic force between parallel wires?

The strength of magnetic force between parallel wires is affected by the distance between the wires, the amount of current flowing through the wires, and the permeability of the medium between the wires.

Can magnetic force between parallel wires be used for practical applications?

Yes, magnetic force between parallel wires is commonly used in electromagnets, electric motors, and generators. It can also be used to measure the strength of electric currents.

What is the formula for calculating the magnetic force between parallel wires?

The formula for calculating the magnetic force between parallel wires is F = (μ0 * I1 * I2 * L) / (2π * d), where μ0 is the permeability of free space, I1 and I2 are the currents in the wires, L is the length of the wires, and d is the distance between the wires.

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