Integrals over vector fields and Ampere's Law

1. Feb 26, 2014

Agent 47

1. The problem statement, all variables and given/known data

Experiments show that a steady current I in a long wire produces a magnetic field B that is tangent to any circle in the plane perpendicular to the wire and whose center is the axis of the wire. Ampere's Law relates the electric current to its magnetic effects and states that

$\int B \cdot dr = \mu I$

where $I$ is the net current that passes through any surface bounded by a closed curve $C$ and $\mu$ is a constant called the permeability of free space. By taking $C$ to be a circle with radius $r$, show that the magnitude $B = |B|$ of the magnetic field at a distance $r$ from the center of the wire is

$B= \frac {\mu I} {2 \pi r}$

and

Here: http://sites.fas.harvard.edu/~math21a/handouts/hw28.pdf

but they weren't in depth enough.

2. Feb 26, 2014

Dick

What you've found should be plenty to work with. What about the explanation confuses you? Try to ask a specific question.

3. Feb 26, 2014

Agent 47

I guess I'm just looking for the intermediate step in a more detailed manner. How do you determine $\int B \cdot dr = B 2 \pi r$? What are the limits of integration? Are there any limits of integration? Is B interpreted as a constant? I know the beginning of the problem. I know the end. I just don't know how to connect the dots.

4. Feb 26, 2014

Dick

$\vec B$ isn't constant as you move around the circular curve. It's direction, at least, will change since the problem states it's always parallel to the direction of the circular curve $d \vec r$. You can also argue by the symmetry of the problem that its magnitude doesn't change. Why should it be larger at one particular point along the circumference than another? So shouldn't that mean $\vec B \cdot d \vec r$ must be a constant as you integrate around the circle? Further, what's the magnitude of $|\vec B \cdot d \vec r|$ since the two vectors are parallel? Is that helping?

Last edited: Feb 26, 2014
5. Feb 26, 2014

Agent 47

I think I understand. So is this the integral I was looking for?

$B \int_0^{2 \pi r} dr$

6. Feb 26, 2014

exclamationmarkX10

That equation is a line integral over a vector field; see http://en.wikipedia.org/wiki/Line_integral#Line_integral_of_a_vector_field

Imagine the plane perpendicular to the wire where the current (positive current) is pointing at you. Put a coordinate system on the plane with the origin at the point where the wire intersects the plane (you assume the wire is infinitely thin). Now you can represent each point on the circle by (rcost, rsint) for t in the interval [0, 2π]. The derivative is (-rsin(t), rcos(t)). From experiment, the direction of the magnetic field vector is tangent to the circle (Right Hand Rule). Since the magnetic field vector and the derivative are 'pointing in the same direction', the dot product is just the product of the magnitudes (recall the dot product of two vectors is the product of the magnitudes of the vectors and cosine of the angle between the vectors; so if they are pointing in the same direction, the cosine factor is 1). Hence you get ∫|B||r|dt where the limits are from t = 0 to t = 2π. And that is how you get Br2π = μI.

7. Feb 26, 2014

Dick

Sure, you are integrating $B dr$ over a circle of circumference $2 \pi r$. It's only stated that $\vec B$ is parallel to $d \vec r$ so $\vec B \cdot d \vec r$ might be $-|B|dr$, but that's not going to matter in the absolute value.

Last edited: Feb 26, 2014
8. Feb 26, 2014

Agent 47

So I guess this leads into another question for me.

When do you interpret $dr$ as just $dr$ and when do you interpret it as $r'(t)$?

Does it depend on whether or not I decide to parameterize it?

9. Feb 26, 2014

Dick

If you parameterize is it's $dr = r'(t)dt$. But it doesn't matter whether you parametrize it or not, $dr$ is still arc length.

10. Feb 26, 2014

Agent 47

Okay I appreciate all your help. I get it now.

Out of curiosity would it be possible to solve this integral by parameterizing it?

I just tried and I know I have to get it into the form $\int_C f(r(t)) \cdot r'(t)dt$

So to get a circle of radius r

$r(t) = r<cost,sint>$
$r'(t) = r<-sint,cost>$

But then how would you interpret $f(r(t))$ in relation to the problem?

11. Feb 27, 2014

Dick

Do you mean $B(r(t))$? That's supposed to be parallel to the tangent $r'(t)$ and have magnitude |B|. So it must be $B(r(t))=\pm|B|<-sin(t),cos(t)>$, right?

Last edited: Feb 27, 2014