Ampere's Law - finding Magnetic Field

[smileandbehappy]

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

 

[Hesch]

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) = μ_r * μ₀ * H(r)

 

[Hesch]

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) = μ_r * μ₀ * H(r)

 

[smileandbehappy]

Sorry I am confused - what is H and what is N?

 

[Hesch]

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 = μ_r * μ₀ ( In air/vacuum μ_r = 1 )

 

[rude man]

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.

 

[smileandbehappy]

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.

Thanks for all those who have given me hints and help.

 

[rude man]

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'? Seriously, you're doing OK, main thing is recognizing the right unit vector. You're not 'completely stupid' by a long shot!

 

[smileandbehappy]

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'? 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

 

[rude man]

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.

 

[Hesch]

A long, straight, cylindrical conductor with relative permeability μ has a radius a and carries a current I.

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 "μ_r".
μ_r is dimensionless and μ_r ≥ 1.
So μ = μ_r * μ₀ ( at least in Europe ).

 

[rude man]

I don't think you are right here.

"Normally" you will call the absolute permeability "μ", and the relative permeability "μ_r".
μ_r is dimensionless and μ_r ≥ 1.
So μ = μ_r * μ₀ ( at least in Europe ).

That in no way contradicts what I said. μ_r ≠ μ₀.

 

[Hesch]

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 "μ_r".

What I meant was:

So you don't want to say μ = μ0μ ever.

in my terminology:

So you don't want to say μ = μ₀ * μ_r 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 ).

 

[rude man]

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

What I meant was:

in my terminology:

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