# I need help in this question: Four Wires

1. Jun 24, 2009

### OsDaJu

1. The problem statement, all variables and given/known data

http://s27.photobucket.com/albums/c191/oscar888/?action=view&current=LONCAPA.gif

The four wires that lie at the corners of a square of side a= 3.50 cm are carrying equal currents i= 2.10 A into (+) or out of (-) the page, as shown in the picture.

Part A: Calculate the y component of the magnetic field at the center of the square. ( I already got the answer for this one) 4.80×10-5 T

Part B: Calculate the x-component of the force on a 1.0-cm long piece of the lower right-hand wire, due to the other three wires.

2. Relevant equations

B=(mu*I)/(2*pi*r)
mu=4*pi*a0^-7 T*m/A

3. The attempt at a solution

I need help in the second part. I don't know where to start.

2. Jun 24, 2009

### LowlyPion

3. Jun 28, 2009

### OsDaJu

Okay, so I know that B is always perpendicular to Force so B2 is zero in the y direction, F2 is zero in the x direction.
then the superposition is:

B total = B1y + B3y
B total = μI/(2π0.0350)*1/√2 + μI/(2π0.0350)
B total = 2.0485*10^-5 T

Using the force formula,

F = BIL
=2.0485*10^-5 *2.1*0.01= 4.3019*10^-7 N

I did all of this and I got the wrong answer, what am I missing??

P.S: sorry for the delay I had a busy week.

4. Jun 28, 2009

### LowlyPion

What you want is the x-directed force component.

The B-Field is a vector field, and the force component from the diagonal has a cos45 factor still needed to normalize it to x.

Btotal = 1.2 * 10-5 + (.707)(.8 * 10-5)

5. Jun 28, 2009

Thank you!