Force per unit length on the wire in the bottom left hand corner.

[missyc8]

1. Homework Statement [/b]
Suppose in the figure that the four identical currents i = 13 A, into or out of the page as shown, form a square of side 64 cm. What is the force per unit length (magnitude and direction) on the wire in the bottom left hand corner? (N/m) Take the positive y direction as up and the positive x direction as to the right.
What is the magnitude of the force per unit length?

(diagram is in following post)

What is the x component of the force per unit length?

What is the y component of the force per unit length?

Homework Equations

F= muo*i1*i2/2*pi*d

The Attempt at a Solution

F= 1.26E-6 * 7A * 7A/ 2*pi* sqrt(2*.64m) = 1.095E-4

i do not know if i am doing this right at all...but this answer is wrong for the the first part..

 

[missyc8]

here is the pic...oops

 

[tomcue]

I would like to clarify a few things about the given problem. First, the diagram is missing and it is important to have a visual representation in order to accurately solve the problem. Second, the given equation is for the force between two parallel wires, but in this problem, we are dealing with four wires forming a square. Therefore, a different equation may be needed to solve for the force per unit length.

In order to provide an accurate response, I would need to see the diagram and understand the orientation of the wires and the direction of the currents. From the given information, it seems like the four wires are forming a square with each side being 64 cm. However, it is not clear which wire is in the bottom left hand corner and what is the distance between this wire and the other wires.

Assuming that the wires are placed at the corners of the square and the wire in the bottom left hand corner is at a distance of 64 cm from the other wires, the magnitude of the force per unit length would be:

F/L = muo * i^2 / 2 * pi * d = 1.26E-6 * (13 A)^2 / 2 * pi * (0.64 m) = 0.0026 N/m

The x component of this force would be zero, as the wires are parallel and the force acts along the y direction. The y component of the force would depend on the orientation of the wire in the bottom left hand corner. If the current is flowing into the page, the y component of the force would be positive, and if the current is flowing out of the page, the y component of the force would be negative.

Again, without a visual representation and more information about the orientation and distances between the wires, it is difficult to provide an accurate response. As a scientist, it is important to have all the necessary information and equations in order to solve a problem accurately.