What is the solution to the magnetic field problem in Purcell's book?

[fluidistic]

Homework Statement

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).

 

[fluidistic]

Sorry to bump this thread, but I'm extremely worried about this issue!
Any kind of help is greatly appreciated.
And feel free to infract me, I deserve it as I know I shouldn't bump.

 

[UgOOgU]

If I had understand the problem, what you are looking for is calculate the integral $$\oint_C \vec{B}\cdot\vec{dl}$$.

But from his argument I don't understand at all since CD and AB are also orthogonal to B!).

CD and AB are not orthogonal to B.

That was what my intuition believed before I solved the exercise.
Precisely, I get that $$\vec B = \frac{\mu _0 I}{4\pi r_1}-\frac{\mu _0 I}{4 \pi r_2} \hat k$$.

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.

Notice that I took $$\hat k$$ pointing into the sheet of paper.

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.

 

[fluidistic]

If I had understand the problem, what you are looking for is calculate the integral $$\oint_C \vec{B}\cdot\vec{dl}$$.

Yes.

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.

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!

 

[UgOOgU]

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?

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?"

 

[fluidistic]

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?"

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.

 

[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.
$$B = \oint_{wire}...$$

 

[fluidistic]

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.
$$B = \oint_{wire}...$$

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.