# Ampere's Law - Magnetic Field Strength

1. May 14, 2014

### teme92

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

Use Ampere's Law to show that the magnetic fi eld strength at a distance R from a long, straight wire, carrying a current I, is:

H = I/2$\pi$R

2. Relevant equations

F=qVB
B=$\mu$0I/2$\pi$R

3. The attempt at a solution

I'm not sure how to answer this question. I got the integral form of Ampere's Law as:

∫B.dl=$\mu$0I

However I don't understand what the question is asking me. Any help would be much appreciated.

2. May 14, 2014

### Gauss

Ampere's Law is used to described the relationship between enclosed current and magnetic field.

∫B.dl=μ0I

Here, B refers to the magnetic field given off by the wire. Since the length of the wire can be assumed to be infinite, we assume the magnetic field strength is independent of the length of the wire. dL refers to an infinitesimal distance along the path you're integrating over. μ0 is the permeability constant and I refers to the current enclosed within the region of space your integrating over.

What you have to do is integrate the region at a radius R around the wire (assume B is constant). Than substitute this quantity into Gauss Law.

3. May 14, 2014

### teme92

Hey Gauss,

So if B is constant:

B∫dl=$\mu$0I
Bl=$\mu$0I

Gauss' Law: ∫B.dA=Q/$\mu$0

So BA=Q/$\mu$0

How do I equate these?

4. May 14, 2014

### Gauss

Oh I'm sorry I misspoke. I meant to say substitute this into Ampere's Law.

Since you recognized B is constant you got B∫dl which is correct. However, the integral of dl isn't only l.

For this integral you are integrating a cylinder around the wire. The way Ampere's Law works is that you integrate a shape that encompasses the object, in this case, a wire. Think of a cylinder that completely surrounds the wire and expands as long as the wire does. You take the integral of this shape (where dl is a small portion on the cylinder) then set this equal to μ0I(enclosed).

5. May 14, 2014

### teme92

Ok makes a bit more sense now :P Thanks for the help by the way.

So the integral of dl is l+c?

Last edited: May 14, 2014
6. May 14, 2014

### Gauss

Actually the integral of dl is definite, so you don't need to include + C.

Ignore the wire for a second. Imagine all you are doing is integrating a cylinder (while not considering the two end faces, just considering the body). This cylinder has a radius R. l is an infintesimal slice of this cylinder, and you are concerned about the outer edge of this cylinder only.

dl is a few small portion of this edge. It can be defined as dl = Rdθ because R is the radius of the cylinder (and remains constant) while θ changes. Since the cylinder edge is effectively a circle, you will integrate from 0 to 2π.

so use this definition of dl as well as the bounds of integrating (0 to 2π) in Ampere's law.

B∫dl = μ0I

7. May 14, 2014

### teme92

Ah so the integral is:

2$\pi$R - 0

So I can then get B = $\mu$0I/2$\pi$R

But in the question there is no $\mu$0. Can you explain why this is?

8. May 14, 2014

### Gauss

Yes thats the correct integral. So you've derived B = μ0I/2πR. The magnetic field strength "H" is related to B by H = B/μ0 + M. You aren't concerned about the magnetization "M" for this problem so H = B/μ0. All you have to do is divide your derived equation by μ0, and you have proven what was asked.

9. May 14, 2014

### teme92

Brilliant! Thanks a million Gauss, you explained the process very clearly and you were quick responding. This is why I love using these forums.

10. May 14, 2014