# Mistake with Biot Savart, B of a circular loop integral

1. Mar 24, 2012

### egoot247

I hope you don't mind if I'm a little liberal in your cute format.
I'm looking for why method 1 does not equal method 2.

1. The problem statement, all variables and given/known data: Method 1--the "cute math" method.

Alright, so...

I have a circular loop of radius A, let's place it in the xy plane, with a steady current I going counterclockwise as one looks down on it from the z axis (so I can have my B-field going upwards on the z axis). It is centered on the z axis.

I want to find the B-field of the loop anywhere on the z-axis.
I know how to solve this by simple fact that the integral of dl will be 2*∏*A:

The Biot-Savart Law is:

B(z)= μ0*I/4∏ ∫dl x (R)/|R|3

Now, I know this is true-- basically in class we took the unit vector and "un-unitized it." So we have the full vector in the numerator and divided it by the magnitude of the vector, so although you see a cubed in the denominator, it's still an inverse square relationship.

The magnitude of R is √(A^2+z^2), and the non-z components of the magnetic field will cancel out, leaving us with a vector in the z-direction only, so

B(z)= μ0*I/4∏ ∫dlx A/(A2+R2)(3/2)

Taking the integral of dl around my circle, this is obviously the circumference of the circle,
so
B(z)= μ0*I/4∏ 2*pi*A^2/(A2+R2)^(3/2)

And this is correct.

B(z)= μ0*I/2*A^2/(A2+R2)^(3/2)

3. The attempt at a solution without using cute math.

My teacher once commented in reference to one of our homework problems that his only complaint about Griffifth's book is that it uses too many cute tricks, and expects us to adapt.
The homework problem in question is one that was easier to just "use the biot savart law, and let it all fall into place."
We all tried to use a cute trick from the example problem in the book.

So, I've been trying to do this problem from a different approach, and I get a wrong answer. Why?

The Biot-Savart Law is:

B(z)= μ0*I/4∏ ∫dl x (r-r)/|r-r|3

r will be the distance from the origin to the place on the z axis where I'm measuring the B-field. r will be the distance to the ring at the dl. dl will be an infinitessimal part of the ring.
I think I want to work in cylindrical coordinates.

So

r= z *uz
r=A *ur

B(z)= μ0*I/4∏ ∫dl x (z*uz-A*ur)/(z^2+A^2)3

dl` I think should be A*dθ uθ. I believe this. I see it.

B(z)= μ0*I/4∏ ∫ A*dθ uθ x (z*uz-A*ur)/(z^2+A^2)3/2

And the integral is from 0 to 2 pi.

This gives me:

B(z)= μ0*I/2 [A*uθ x (z*uz-A*ur)/(z^2+A^2)3/2

B(z)= μ0*I/2 A^2 /(z^2+A^2)3/2 uz

But if you look closely, my r component didn't go away... :S I actually have a component along ur when I took the cross-product:

B(z)= μ0*I/2 A*z /(z^2+A^2)3/2 uR.

What do I do to get rid of that? What did I do wrong in my math? was there supposed to be a cosine or sine there somehow? But I can't mathematically justify sticking one in there...

What's wrong with this picture?

2. Mar 25, 2012

### rude man

What is "cute" about using Biot-Savart? especially since it's the only way to do it, far as I know.

And why go to cylindrical coordinates? Cylindrical coordinates are potentially more troublesome since two of the three unit vectors vary with the second coordinate: ur = i cosθ + j sinθ, uθ = -i sinθ + j cosθ.

Anyway, we're reduced to handling the same problem but in cylindrical coordinates, more a math than a physics thing. I have to confess I'm too lazy to work this out with you. Hope someoe else is more amenable!

3. Mar 26, 2012

### egoot247

Both methods are Biot-Savart.
The first one just has cuter math, and I wanted to "plug and chug," I guess. Yes, my question is more mathematically based.

I just worked it out in cartesian coordinates, and the x and y components went away because I was integrating sines and cosines from 0 to 2 pi. THat's nice.
Now why doesn't that happen in cylindrical? It should, right? This should work in any coordinate system. What am I doing wrong? X(

4. Mar 26, 2012

### rude man

You're right, of course. Has to give the same result either way. Well, maybe I'll have a shot at it but don't hold your breath! Not sure I trust myself working in cylindrical coordinates anymore. Been a loooong time since I did!

5. Mar 26, 2012

### rude man

OK, here's what I came up with.
Please note all vectors are in bold. er, eθ and k are the unit vectors for the cylindrical coordinate system.

Let c = μ0i/4π to get that mess out of the way for the moment.

Then dB = c(dl x r)/r3

In cylindrial coordinates:
r = k z - er A and r = (A2 + z2)1/2

So dl x r = er zAdθ - k A2dθ (set up the usual determinant to get this)

We recognize by symmetry that B will have only a z component, so
dl x r reduces to -k A2
and dB = -k cA2dθ/(A2 + z2)3/2
Finally B = -k cA2∫dθ/(A2 + z2)3/2 from 0 to 2π
& substituting c = μ0i/4π gets us the correct answer.

My problem in general with cylindrical coordinates , and it has no bearing on the above BTW, is this: a cylindrical coordinate system defines a position vector r as follows:

r = er r + k z
when it seems to me it should be

r = er r + eθ θ + k z. Like, how can you define a point with only two coordinates? Yet Thomas (calculus text) and my old mechanics text say the former.

I would love to hear from anyone who can shed some light on this dilemma for me.