# Magnetic Force parabolic wire

1. Nov 24, 2003

### discoverer02

I'm stuck on this problem. I think it's the 3D geometry that's giving me the biggest problem.

The current i exists in the direction shown (from left to right) in a parabolic wire segment whose equation is y = (x^2)/b in the XY plane between x = -a and x = +a. Magnetic flux density, given by B = py^2 and the B field lines are everywhere out of the paper making an angle 60 degrees to the +X axis and 60 degrees to the +Y axis.

a) Show that B can be written as [(py^2}/2)i + [(py^2}/2]j + [(py^2)/(2^(1/2))]

b) Find X, Y, and Z components of the total force on the wire.

Part a) was easy the projections in the x and y directions are just Bcos60 and z^2 = B^2 - x^2 - y^2.

Part b) is really giving me fits.

I know that dF(magnetic) = I(dl)Bsin(theta) and when B is perpendicular to dl life is great because dF = IdlB and the dFx and dFy are dFsintheta and dFcostheta respectively.

But I'm totally thrown by B not being perpendicular and having trouble with the 3-dimensionality of the problem.

Thanks

2. Nov 24, 2003

### Staff: Mentor

Consider it this way: dF = I dl X B (where X is the vector cross product)

The length of wire, dl, has components in the x and y directions.

Also, take advantage of symmetry about the y-axis. The current though the y component of a wire segment (dl) on one side of the y-axis is exactly opposite the current through the y component of the segment on the opposite side of the y-axis. What does that tell you about its contribution to the total force on the wire?

3. Nov 24, 2003

### discoverer02

Thanks for your help, but the real problem I'm having is with the angle between B and dl, since the magnitude of dF is IdlBsin(theta).

Symmetry will make Fx = 0. So now I only have the Y and Z components to figure out.

Thanks again.

4. Nov 24, 2003

### discoverer02

Do I need to use the direction cosines in my equation? The angle between dl and B is changing depending on where you are in the wire right?

5. Nov 24, 2003

### Staff: Mentor

Symmetry will make the contribution to the force due to the y-component of dl equal to zero. So all you care about is the x-component of dl.

6. Nov 24, 2003

### Staff: Mentor

Yes, the angle keeps changing. But one way to attack this problem is as I suggested above. Rather than try to visualize the angle, use components and take the cross product.

7. Nov 24, 2003

### discoverer02

I'm still stumped. I understand what you're saying about dy not contributing anything to the total force on the wire, but there's still a y component in the total force.

If I integrate the x component, I'm going to integrage for -a to +a and Fx = 0,

I have the y component of B from part a) of the problem. The y component of dl is just dy right?

So is dFy = Idly X By or |dFy| = IdlyBysin(angle between the dly and By) = IdlyBySin60.

That can't be right because Fy = [2^(1/2)Ipa^5]/[5b^2] and
Fz = [Ipa^5]/[5b^2]

8. Nov 24, 2003

### Staff: Mentor

Yes!
Well, yes, but: I thought we agreed that the y-component of dl does not contribute to the force? Forget about it!
You are correct... That can't be right!

What you have is this:

I dx X (Py2/2 [j] + Py2/&radic;2 [k]) ...Right? (Don't go beyond this point until you get this!)

Now do the cross product --- like I've been telling you! (i x j = k, etc.)

Then integrate from x= -a to +a. (Don't forget: y = x2/b) And you will see the light!

9. Nov 24, 2003

### discoverer02

Thanks.

I see the light now. It's not a super bright light, but it is bright enough for me to think clearly about what's going on in the problem.

I took the cross product and integrated and, low and behold, I got the correct answer.

I was thinking, incorrectly, that I should isolate both the dl and B into its components and insisting on finding the magnitude of the cross product instead of taking the actual cross product which in this case makes life much easier.

I need a lot more practice, but this is a good start.

Your help is very much appreciated. Thanks.