- #1

egoot247

- 2

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I'm looking for why method 1 does not equal method 2.

**1. Homework Statement : 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)= μ

_{0}*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)= μ

_{0}*I/4∏ ∫dl`x A/(A

^{2}+R

^{2})

^{(3/2)}

Taking the integral of dl around my circle, this is obviously the circumference of the circle,

so

B(z)= μ

_{0}*I/4∏ 2*pi*A^2/(A

^{2}+R

^{2})^(3/2)

And this is correct.

B(z)= μ

_{0}*I/2*A^2/(A

^{2}+R

^{2})^(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)= μ

_{0}*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 *u

_{z}

r`=A *u

_{r}

B(z)= μ

_{0}*I/4∏ ∫dl` x (z*u

_{z}-A*u

_{r})/(z^2+A^2)

^{3}

dl` I think should be A*dθ u

_{θ}. I believe this. I see it.

B(z)= μ

_{0}*I/4∏ ∫ A*dθ u

_{θ}x (z*u

_{z}-A*u

_{r})/(z^2+A^2)

^{3/2}

And the integral is from 0 to 2 pi.

This gives me:

B(z)= μ

_{0}*I/2 [A*u

_{θ}x (z*u

_{z}-A*u

_{r})/(z^2+A^2)

^{3/2}

B(z)= μ

_{0}*I/2 A^2 /(z^2+A^2)

^{3/2}u

_{z}

But if you look closely, my r component didn't go away... :S I actually have a component along u

_{r}when I took the cross-product:

B(z)= μ

_{0}*I/2 A*z /(z^2+A^2)

^{3/2}u

_{R}.

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?