# Homework Help: Magnetic field problem

1. Feb 1, 2010

### fluidistic

1. The problem statement, all variables and given/known data
I was reading Purcell's book (see page 209, figure 6.6a : http://books.google.com.ar/books?id...resnum=1&ved=0CAoQ6AEwAA#v=onepage&q=&f=false) and I fell over an exercise I had done previously, but he says that the curvilinear integral is null because BC and DA are perpendicular to B and they contribute to nothing. (I understand that they don't contribute to the B field thanks to Biot-Savart law, because $$d\vec l$$ and $$\vec r$$ are parallel, so the cross product is null. But from his argument I don't understand at all since CD and AB are also orthogonal to B!!!!).
However he also says that AB cancels out CD, hence the total integral is null. That was what my intuition believed before I solved the exercise.
Precisely, I get that $$\vec B = \frac{\mu _0 I}{4\pi r_1}-\frac{\mu _0 I}{4 \pi r_2} \hat k$$. And because $$r_2>r_1$$, the magnetic field is not null.
Notice that I took $$\hat k$$ pointing into the sheet of paper.

I'd appreciate very much if someone could point me if I did wrong the exercise and explain better the situation. (I don't see how Purcell's argument holds, but he's a Nobel prize and I'm just a second year student).

2. Feb 4, 2010

### fluidistic

Any kind of help is greatly appreciated.
And feel free to infract me, I deserve it as I know I shouldn't bump.

3. Feb 4, 2010

### UgOOgU

If I had understand the problem, what you are looking for is calculate the integral $$\oint_C \vec{B}\cdot\vec{dl}$$.
CD and AB are not orthogonal to B.
Your intuition is wrong. $$B = \frac{\mu_0 I}{2\pi r_1}$$ and $$B = \frac{\mu_0 I}{2\pi r_2}$$ in the parts AB and CD respectively.
B is not pointing into the sheet, is pointing around the wire. In figure, the wire is the origin of the vector r1 and r2.

4. Feb 5, 2010

### fluidistic

Yes.

Oh... I misunderstood totally the situation. I thought the current was carried by ABCD.
In the case it's carried by ABCD, is the magnetic field at point P worth $$\vec B = \frac{\mu _0 I}{4\pi r_1}-\frac{\mu _0 I}{4 \pi r_2} \hat k$$ as my first post suggests?

Thanks a lot for the clarification!

5. Feb 5, 2010

### UgOOgU

Unfortunatelly no. B is equal to $$\frac{\mu _0 I}{2\pi r_1}$$ solely in the case of an infinitely long wire. In the present case, you have to realize the Biot-Savart integration. Try, it is not difficult.

P.S.: "Usted hablas español, my amigo?"

6. Feb 5, 2010

### fluidistic

Ok suppose that the angle APB is worth $$\theta$$.
$$d\vec B = \frac{\mu _0}{4 \pi} I d\vec l \times \frac{\vec r}{r^3}$$. For the AB part, $$\vec B = \frac{\mu _0 }{4\pi} \oint I \cdot d\vec l \times \frac{\vec r}{r^3} =\frac{\mu _0 I \theta r_1}{4\pi r_1^2} \hat k=\frac{\mu _0 I \theta}{4 \pi r_1} \hat k$$.

$$\vec B$$ due to CD: $$-\frac{\mu _0 I \theta}{4 \pi r_2} \hat k$$.

So $$\vec B$$ total is $$\left ( \frac{\mu _0 I \theta}{4 \pi r_1}-\frac{\mu _0 I \theta}{4 \pi r_2} \right ) \hat k$$.
I hope it's right now.
And yes, I do speak Spanish.

7. Feb 5, 2010

### UgOOgU

What you have to do is calculate B in the points of the segments AB and CD. dl is not upper these segments, is upper the wire that generate B.
$$B = \oint_{wire}...$$

8. Feb 5, 2010

### fluidistic

Oh yes I know but I was solving a different problem. The problem where the B field is generated by ABCD and P lies where the wire of page 209 is.