# Ampere's Law question

1. Mar 25, 2012

### guss

I am studying AP Physics, and confused about the application of Ampere's Law. When we calculate the magnetic field due to a wire, we use:
$\mu_{0}I = \oint \vec{B}•\vec{dl}$

Does this only apply for straight wires that go off to infinity in both directions? We only focus on the current going through the membrane that the dls surround, but it seems like how the wire behaves after that should also matter.

Additionally, when we calculate the magnetic field at the center of a solenoid, we can use this equation even though the wire is looping. This brings up the same question, along with another one: during this calculating, we looked only at a cross section of the solenoid (like a slice of the center). Additionally, we only accounted for the current of the wires on one side of the solenoid and completely ignored the second side of the "slice", along with the fact that the solenoid is 3 dimensional and obviously many more slices can be looked at by just rotating where we take the slice from.

I am learning this from a book right now and unfortunately these issues are glossed over. Thanks for the help!

Last edited: Mar 25, 2012
2. Mar 25, 2012

### Redbelly98

Staff Emeritus
Ampere's law applies to any current configuration. But that does not mean it is always useful. It is only useful when there is enough symmetry or knowledge about B to easily calculate the integral.

For a long straight wire, we know that B is independent of distance from the wire. So B·dl is constant for a circular path around the wire, and the integral is simple to evaluate.

For a solenoid, we know that B is constant and parallel to the axis inside the solenoid, and negligible outside of it. So for a rectangular path that is part inside and part outside the solenoid, the integral is again easy to calculate.

3. Mar 25, 2012

### guss

Thanks, but I'm still having trouble understanding.

Can you be more specific?

Yes, but why does this allow us to only consider one cross-section? The part of the cross section we are considering may as well be straight wires, because we would get the same answer for the magnetic field. Why can we just completely ignore the magnetic field caused by the other side of the cross section, and all other cross sections? And why can we say that the current due to a circular wire (like in a solenoid) is the same as the current due to parallel long straight wires?

4. Mar 25, 2012

### Bob S

B(r) is proportional to 1/r surrounding a long straight wire.

5. Mar 26, 2012

### guss

Still looking for help.

6. Mar 26, 2012

### Dr. Philgood

Hello Guss I am actually learning the same material in my class at this time. I'll see if I can make it a little clearer.

The solenoid's magnetic field is a sum of all the fields produced by the individual turns. If you look at a cross section of a solenoid points very close to a turn, magnetically, look like that of straight wire. This is because it is much longer than it is wide and magnetic fields must loop unlike electric fields. This allows B(magnetic field) to be considered parallel to the central axis.

Now outside of an ideal solenoid the magnetic field can be considered zero(if you would like elaboration just ask).

These two things allow us to use ampere's law to integrate a rectangular loop placed partially into an ideal solenoid. This can be done using four simple integrals along the rectangle. (3 of which are conveniently zero). And the last of which being B*h.

in the formula for ideal solenoid "n"(number of turns per unit length) is what takes care of the current throughout. current enclosed(from ampere's law) is equal to i(n*h) this is what allows us to to use ampere's law(2d) and apply it to solenoid(3d).

I'm not sure if I answered your question properly but if you want elaboration on any part or there is something I didn't answer correctly just let me know.

7. Mar 26, 2012

### guss

Thanks, you helped, but I am still confused. Specifically about how we can apply Ampere's law by seemingly ignoring whether the wire curves or not.

8. Mar 26, 2012

### K^2

The way B is distributed does depend on configuration of the wire, but the integral around the contour does not. If you bend the wire, you'll increase B on one side of the contour, and reduce it on another. Also, note that component perpendicular to the contour can be anything, and it won't affect the integral. It's consequence of Stoke's Theorem and the fact that curl of magnetic field is proportional to current density. If that sounds Greek to you right now, don't worry about it.

9. Mar 26, 2012

### guss

Ah, okay, thanks. Just so I vaguely understand, can it be summarized by saying that the B in Ampere's law represents some form of average magnetic field, but the density of that magnetic field can change while still preserving the average, which allows us to make these simplifications and still get the right answer without really knowing what's going on.

10. Mar 26, 2012

### K^2

Almost. It's the average along the contour times the length of the contour.

11. Mar 26, 2012

### Redbelly98

Staff Emeritus
Yes, I mistyped. I meant B only depends on distance from the wire. It also has zero radial component, since it has zero divergence.

12. Mar 26, 2012

### K^2

For sake of clarity, you can only use zero divergence to state that radial component is zero if you have cylindrical symmetry and circular contour concentric with the wire. If you start bending the wire, you can get radial component.