# Homework Help: Magnetic field of a bent wire

1. Mar 28, 2013

### Octavius1287

1. The problem statement, all variables and given/known data
An infinite wire of parabolic shape carries a current I, find the magnetic flux density B, at the focus

2. Relevant equations
B=(μ0*I)\(2∏r) I know this is for a straight wire, i'm not sure what i would use for a U shaped wire

3. The attempt at a solution
Well I figure the current is running for one end to the other and the B field around the wire but I don't know what at the focus means, and I'm not sure what the bent U shape/parabolic effects the B field

2. Mar 28, 2013

### rude man

First, draw a parabola y = x2 and draw the d l and r vectors at some point up the parabola on both sides. The focus in this case is at (0, 0.25).

Express the d l and r vectors in Cartesian terms. Wind up with everything in terms of x, with no y terms. Then integrate the Biot-Savart expression for dB from x =-∞ to +∞.

I must say the math looks awful. Perhaps someone else has a better approach.

3. Mar 29, 2013

### Octavius1287

oh great

4. Mar 29, 2013

### rude man

Awful, not impossible! In fact, looking at it again, may not be bad at all.

First, notice that you only need to address the parabola for x >= 0, then multiply by 2.

You know you need dl and r going from a point on the parabola (x,y). You also have y = x2.

dl = dx i + dy j and points upwards & to the right (assume current flows in the +y direction).
r= -x i + (0.25 - y) j and points from a point (x,y) on the parabola to the focus.

That should get you going.

5. Apr 1, 2013

### Octavius1287

alright cool thank you

6. Apr 1, 2013

### Octavius1287

why is the focus at 0, 0.25? and would i use this? even though it for a closed curve?

7. Apr 1, 2013

### Octavius1287

or this http://pms.iitk.ernet.in/wiki/images/math/5/c/8/5c8f0ff0e0d4749758d0abaf2faa2af2.png [Broken]

Last edited by a moderator: May 6, 2017
8. Apr 1, 2013

### rude man

You have the right formula, aka the Biot-Savart law.

The expression for a parabola with vertex at the origin and open end upward (along the y axis) is
y = x2/4p
where p = distance along y from the vertex to the focus. So if you choose y = x2 then p = 1/4.

You can choose p to be any positive real number you want. The problem did not specify p so presumably it makes no difference to the solution.

BTW the "C' by the integral sign does not belong. This is not a contour integration. You are integrating d B to get B at one point.

Last edited: Apr 1, 2013
9. Apr 1, 2013

### rude man

EDIT:

This is the differential form of Biot-Savart. The " k " indicates that B wound up in the +z direction, just as it will in your case. θ is the angle between d l and r.

However, I earlier suggested vector forms for d l and r and I suggest you ignore this formula and retain the vector format for dB which is

d B = (μ0/4π)d l x r / r3.

Last edited: Apr 1, 2013
10. Apr 3, 2013

### Octavius1287

i guess ,y last question would be, how did you figure the
dl = dx i + dy j
r= -x i + (0.25 - y) j

11. Apr 3, 2013

### rude man

the first is just pythagoras: you have a vector dl with x component dx and y component dy. keep in mind dl is a differential stretch along the parabola.

The second is derived by the usual formula for a vector going from (x1,y1) to (x2,y2). (The direction is important).

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