Ampere's Law - Magnetic Field Strength

AI Thread Summary
Ampere's Law is used to derive the magnetic field strength around a long, straight wire carrying current I, leading to the equation H = I/2πR. The discussion emphasizes integrating a cylindrical path around the wire, where the magnetic field B is considered constant. The integral of the path length dl is evaluated from 0 to 2πR, resulting in B = μ0I/2πR. The relationship between B and H is clarified, showing that H can be derived by dividing B by the permeability constant μ0. The conversation highlights the importance of understanding the integration process and the application of Ampere's Law in this context.
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Homework Statement



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\piR

Homework Equations



F=qVB
B=\mu0I/2\piR

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=\mu0I

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

So if B is constant:

B∫dl=\mu0I
Bl=\mu0I

Gauss' Law: ∫B.dA=Q/\mu0

So BA=Q/\mu0

How do I equate these?
 
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).
 
Ok makes a bit more sense now :P Thanks for the help by the way.

So the integral of dl is l+c?
 
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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
 
Ah so the integral is:

2\piR - 0

So I can then get B = \mu0I/2\piR

But in the question there is no \mu0. Can you explain why this is?
 
Yes that's 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.
 
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Brilliant! Thanks a million Gauss, you explained the process very clearly and you were quick responding. This is why I love using these forums.
 
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No problem, glad to help!
 
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