Ampere's Law - finding Magnetic Field

AI Thread Summary
The discussion focuses on applying Ampere's Law to find the magnetic field inside and outside a long, straight cylindrical conductor carrying a uniform current. The magnetic field inside the cylinder is expressed as B = μ₀I/(2πa), while outside it is B = μ₀I/(2πr), where r is the radial distance from the center. Participants clarify the distinction between the magnetic field strength H and magnetic induction B, emphasizing the correct use of variables and units. There is confusion regarding the use of relative permeability (μ) and absolute permeability (μ₀), with participants discussing their definitions and applications. The thread highlights the importance of accurate notation and understanding of magnetic concepts in solving the problem.
smileandbehappy
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Homework Statement


A long, straight, cylindrical conductor with relative permeability ## \mu ## has a radius a and carries a current I. The current density is uniform. Using Ampere's law in it's integral version, finding the magnetic field B:

a) Inside, and
b) outside the cylinder.

Homework Equations



Amperes law: ## \int \textbf{B}.d\textbf{l} = \mu_0 I ##

The Attempt at a Solution


I am using cylindrical geometry so: ##\textbf{B} = \mu_0 I \textbf{e_r} /(2 \pi r l) ##

Now I think that the magnetic field: ##\textbf{B} = \mu_0 I \textbf{e_r} /(2 \pi a l) ## inside the cylinder.

And just - ##\textbf{B} = \mu_0 I \textbf{e_r} /(2 \pi r l) ## Outside the cylinder.

But that seems too simple. What am I missing here? Thanks
 
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smileandbehappy said:
Amperes law: ∫B.dl=μ0I \int \textbf{B}.d\textbf{l} = \mu_0 I

Well, I have learned:

circulation H ⋅ ds = N * I ( N=1 as for a long straight conductor ).

and using this you will find H(r) = N * I / ( 2πr ).

Outside the conductor you will find: B(r) = μ0 * H(r).

Inside the conductor you will find: B(r) = μr * μ0 * H(r)
 
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Hesch said:
Inside the conductor you will find: B(r) = μr * μ0 * H(r)

Sorry, that was wrong.

Inside the conductor you will find:

H(r) = N * I / ( 2πr ) * r2 / R2, ( where R = radius of the conductor and r is the variable radius. )

You must only integrate around a part of the current, therefore the factor r2 / R2

B(r) = μr * μ0 * H(r)
 
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Sorry I am confused - what is H and what is N?
 
smileandbehappy said:
Sorry I am confused - what is H and what is N?

H is the magnetic strength field. Unit = [ A/m ]

N = number of turns through the integration path ( here N = 1 ).

B is the magnetic induction. Unit = [ Tesla ].

B = μ * H where μ is the absolute permeability = μr * μ0 ( In air/vacuum μr = 1 )
 
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smileandbehappy said:

Homework Statement


A long, straight, cylindrical conductor with relative permeability ## \mu ## has a radius a and carries a current I. The current density is uniform. Using Ampere's law in it's integral version, finding the magnetic field B:

a) Inside, and
b) outside the cylinder.

Homework Equations



Amperes law: ## \int \textbf{B}.d\textbf{l} = \mu_0 I ##

The Attempt at a Solution


I am using cylindrical geometry so: ##\textbf{B} = \mu_0 I \textbf{e_r} /(2 \pi r l) ##

Now I think that the magnetic field: ##\textbf{B} = \mu_0 I \textbf{e_r} /(2 \pi a l) ## inside the cylinder.

And just - ##\textbf{B} = \mu_0 I \textbf{e_r} /(2 \pi r l) ## Outside the cylinder.

But that seems too simple. What am I missing here? Thanks
What is ## I \textbf{e_r} ## inside and outside the cylinder? It's not constant. It's a function of r.
 
Sorry guys - I was being silly. I have posted what I think is the correct answer... But I might be completely stupid here. Also apologies for not typing it. I needed the diagram to understand it... Let me know if I'm being completely stupid and it's wrong.

17441524910_a0eda1756a_b.jpg


Thanks for all those who have given me hints and help.
 
smileandbehappy said:
Sorry guys - I was being silly. I have posted what I think is the correct answer... But I might be completely stupid here. Also apologies for not typing it. I needed the diagram to understand it... Let me know if I'm being completely stupid and it's wrong.
You have the wrong unit vector for B.
You have the wrong μ for inside the conductor..
You should not use I for current density, use j. I is the usual symbol for current.
See why I'm known by 'rude man'? :rolleyes: Seriously, you're doing OK, main thing is recognizing the right unit vector. You're not 'completely stupid' by a long shot!
 
rude man said:
You have the wrong unit vector for B.
You have the wrong μ for inside the conductor..
You should not use I for current density, use j. I is the usual symbol for current.
See why I'm known by 'rude man'? :rolleyes: Seriously, you're doing OK, main thing is recognizing the right unit vector. You're not 'completely stupid' by a long shot!

Actually that's one of the best posts I've got on here. I'd rather people were brutal.

e(phi) should be unit vector.

μ should be inside the conductor - although I'm unsure if it should be just μ or μμ_o
 
  • #10
smileandbehappy said:
Actually that's one of the best posts I've got on here. I'd rather people were brutal.
e(phi) should be unit vector.
right.
μ should be inside the conductor - although I'm unsure if it should be just μ or μμ_o
You have to be careful with your μ's.
B = μ0H in vacuum (or other non-magnetic material). B = μH in magnetic material.
In other words, μ0 is not a dimensionless number. μ0 = μ in a vacuum.
So you don't want to say μ = μ0μ ever.
μ > μ0 inside the conductor here.
 
  • #11
smileandbehappy said:
A long, straight, cylindrical conductor with relative permeability μ has a radius a and carries a current I.
rude man said:
So you don't want to say μ = μ0μ ever.

I don't think you are right here.

"Normally" you will call the absolute permeability "μ", and the relative permeability "μr".
μr is dimensionless and μr ≥ 1.
So μ = μr * μ0 ( at least in Europe ).
 
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  • #12
Hesch said:
I don't think you are right here.

"Normally" you will call the absolute permeability "μ", and the relative permeability "μr".
μr is dimensionless and μr ≥ 1.
So μ = μr * μ0 ( at least in Europe ).
That in no way contradicts what I said. μr ≠ μ0.
 
  • #13
rude man said:
That in no way contradicts what I said. μr ≠ μ0

That's right, but there is some confusion in the thread, because in #1 the relative permeability here is called "μ" ( not "μr".

What I meant was:
rude man said:
So you don't want to say μ = μ0μ ever.
in my terminology:

So you don't want to say μ = μ0 * μr ever.

Maybe the problem is that things are called by unconventional names, like if you are reading some text concerning trigonometry, and the symbol "π" is substituted by "θ", the text will indeed be very difficult to read/understand ( though the author has mentioned this substitution ).
 
  • #14
Hesch said:
That's right, but there is some confusion in the thread, because in #1 the relative permeability here is called "μ" ( not "μr".

What I meant was:

in my terminology:

So you don't want to say μ = μ0 * μr ever.

Maybe the problem is that things are called by unconventional names, like if you are reading some text concerning trigonometry, and the symbol "π" is substituted by "θ", the text will indeed be very difficult to read/understand ( though the author has mentioned this substitution ).
I am confused by your post. Sorry.
 
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