Ampere's Law - finding Magnetic Field

• smileandbehappy
In summary: English.I think the confusion is more because you said μ = μ0μ which is not correct. You meant μ = μrμ0. The μr is missing.
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

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)

Last edited:
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)

Last edited:
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 )

Last edited:
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.

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'? 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'? 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

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.

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 ).

rude man
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.

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 ).

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.

What is Ampere's Law and how is it used to find the magnetic field?

Ampere's Law is a mathematical equation that relates the magnetic field around a closed loop to the electric current passing through that loop. It is used to find the magnetic field by integrating the current density over the enclosed area.

What are the assumptions made when using Ampere's Law to find the magnetic field?

The main assumptions are that the magnetic field is constant and that the current is steady and uniform over the enclosed area. Additionally, the loop must be closed and the current must be enclosed by the loop.

How is Ampere's Law different from Gauss's Law?

Ampere's Law is used to find the magnetic field, while Gauss's Law is used to find the electric field. Additionally, Ampere's Law applies to magnetostatics (steady magnetic fields), while Gauss's Law applies to electrostatics (steady electric fields).

Can Ampere's Law be used to find the magnetic field for non-closed loops?

No, Ampere's Law only applies to closed loops, as it requires the current to be fully enclosed by the loop. For non-closed loops, the Biot-Savart Law or the Ampere-Maxwell Law must be used.

What are some practical applications of Ampere's Law?

Ampere's Law is used in various areas, such as in designing electromagnets, transformers, and electric motors. It is also used in medical imaging, such as in magnetic resonance imaging (MRI) machines, to create strong and uniform magnetic fields.

Replies
3
Views
649
Replies
10
Views
531
Replies
6
Views
358
Replies
4
Views
1K
Replies
5
Views
1K
Replies
5
Views
411
Replies
5
Views
387
Replies
15
Views
2K
Replies
16
Views
2K
Replies
7
Views
562