Using ampere's law with current density

  • #1
K sorry for the bad drawing using paint...but anyways...so the question is :A long wire is in a conducting sheath and viewed from the top.The wire and the conducting sheath each carry opposite currents I into and out of the page with current density as J.The outer sheath has a radius of 2R and he inner wire has a radius of R from the centre.Calculate the magnetic field at the centre of the wire.

J=I/A->I=JA
Taking the amperian loop as a circle of radius 2R
A= pir^2....I1= pi (R^2)I and I2= 4pi(R^2)I
therefore the net current wud be I2-I1 which is 3(pi)(R^2)
then using ampere's law B(2(pi)r)=(mu)I -> B(2(pi)(2R))=3(pi)(R^2)
And I get (3/4)(pi)R^2)J...but the answer is (1/2)(pi)R^2)J
Can you please tell me wht I am doing wrong?Please help me quickly I have an AP C exam tomorrow and I want to make sure I knw how to use Ampere's law porperly.Thank You

Btw if possible can you explain to me wht exactly is happening...if there is a current going in one direction...how come the sheath carries a current in the opposite direction?
 

Attachments

  • Question.JPG
    Question.JPG
    8.7 KB · Views: 428

Answers and Replies

  • #2
dx
Homework Helper
Gold Member
2,112
41
It's very difficult to understand what you've written. Also, these questions should be posted in the homework section.
 
  • #3
k sorry abt the homework section thing but can u please solve it for me.I have an AP exam tomorrow.Wht dont u understand...I will explain
 
  • #4
dx
Homework Helper
Gold Member
2,112
41
Write out the question in its original form and draw a labeled diagram.
 
  • #5
the problem is I dont have the question...the question was on the AP practise test which I cant bring home.So I am writing the question from memory.K I will make a labelled diagram.But basically the two circles u see is one with a radius of 4R(sheath) and smaller one with radius of R(wire).They both have current density of J.Find magnetic field at centre of wire.
 
  • #6
hope this clarifies.here is a labelled diagram.Note:the diagram is being viewed from top.So your looking at the end of the wire from the top.
 

Attachments

  • Question.JPG
    Question.JPG
    13.4 KB · Views: 418
  • #7
dx
Homework Helper
Gold Member
2,112
41
I can't make sense of it the way you're formulating it. Maybe you dont remember it correctly.
 
  • #8
wht do u mean "formulating it"...yes the question isn't structured exactly the way it was...but the info is all the same...which part is confusing u?
 
  • #9
wht abt the diagram...has it helped?
 
  • #10
dx
Homework Helper
Gold Member
2,112
41
I can't see the diagram, it must be approved first. Maybe you can host it somewhere else and give me a link.
 
  • #11
oh god...can I send it to ur e-mail?I dunno anywhere where I can host it....
 
  • #14
dx
Homework Helper
Gold Member
2,112
41
Nope, I'm sorry I'm unable to understand it. Maybe some one else here will be able to help you.
 
  • #15
there is no one else communicating with me.wht exactly cant u get?
 
  • #16
look there is juz a current going out of the page and into the page basically and the wire is a staright long one.
 
  • #17
dx
Homework Helper
Gold Member
2,112
41
In that case, if you consider a small circular loop of radius r around the axis as the path of integration, you will get [tex] B = 2\mu_{0} J r [/tex]. As r goes to zero, B goes to zero. So, at the centre of the wire B should be zero. But you said the correct answer was something else.
 
  • #18
In that case, if you consider a small circular loop of radius r around the axis as the path of integration, you will get [tex] B = 2\mu_{0} J r [/tex]. As r goes to zero, B goes to zero. So, at the centre of the wire B should be zero. But you said the correct answer was something else.

But u havent considered the current in the sheath.I think the loop has to be 4R not just R.
 
  • #19
dx
Homework Helper
Gold Member
2,112
41
You asked for the magnetic field at the centre of the wire. No matter what you choose the initial radius as, when you take the limit of r going to zero, B will go to zero.
 
  • #20
oh yeah u r right...it shud be zero_Oh wait I think the question was whts the magnetic field on the surface of the sheath.sorry abt tht.:blushing:
 
  • #21
dx
Homework Helper
Gold Member
2,112
41
I doubt the question gave the current density of the sheat, because if so, to calculate the current in the sheath, you would also need the thickness of the sheath. If all they said was the current passing through the wire and the sheath are equal and opposite, then the total current through a loop enclosing them both will be zero. So B is zero.
 
  • #22
dx
Homework Helper
Gold Member
2,112
41
I suggest you move on to other problems instead of trying to solve a problem that you can't remember :smile:.
 
  • #23
they din say the current is the same.they said they have the same current density J.does tht make a difference?Yeah I want to but in the actual AP questions come similar to the practise test so I am kinda scared if they bring a similar question.
 
  • #24
Can you think of anywhere where the magnetic field will be (1/2)(J)(pi)(R^2)
 
  • #25
dx
Homework Helper
Gold Member
2,112
41
Yes, it does make a difference. If you're given the current density, you also need the area of the cross-section to calculate the current passing through it.
 
  • #26
the cross section of the wire are the circles.Thts why I was multiplying it by (pi)R^2
 
  • #27
and since the sheath and wire dont have equal cross sections(sheath has bigger cross section area)...the net current I assumed was 3pi(R^2)
 
  • #28
555
0
Does the sheet (the thing with radius 4R) have a thickness? Or is it just an infinitely thin sheet?

If it has a thickness, the current density J is a current per unit area A.
If it has no thickness, the current density J is a current per unit length L.

Where do you need to calculate the magnetic field now? Exactly on the sheet of 4R? In other words, with a radius of 4R?

In both your diagrams, you drew one current density (of the sheet, the one into the page) between the two circles. Is 'between the two circles' you drew the sheet, or is this a vaccuum?

Finally, in your very first post you mention that both the wire and the sheet carry a current I and both have a current density J. The way you say it, both have the same current AND the same current density. This is only possible if they have the same cross sectional area, which I cannot see to be the case from the picture...


I will now assume the sheet is infinitely thin and you need to calculate the magnetic field on this sheet (at 4R). Effectively, the sheet can be left out of the calculation.

If you take a circular path around the inner wire with radius 4R, just beneath the outer sheet, the enclosed current is only the total current in the inner wire:
[tex]I_{\text{encl}} = JA = J \pi R^2[/tex]

Ampere's law tells us now:
[tex]\oint \vec{B} \cdot d \vec{l} = \mu_0 I_{\text{encl}}[/tex]
[tex]B 2 \pi (4R) = \mu_0 J \pi R^2[/tex]
[tex]B = \frac{ \mu_0 J R}{ 8 }[/tex]


If you have to take the loop at 4R including the sheet, I can't figure it out since I don't know the current through the sheet.
If this current is also I (which you say in the first post) then the enclosed current is 0 and the magnetic field is 0.


EDIT
You're saying the answer should be [tex]B = \frac{1}{2} \pi J R^2[/tex] and you got 3/4 instead of 1/2 ?
You're first post is completely wrong however. I can't make anything out of it... I'm not sure but the way I read it, your [itex]\pi[/itex] and one R should cancel out...
 
Last edited:
  • #29
OHHHH...."If it has a thickness, the current density J is a current per unit area A.
If it has no thickness, the current density J is a current per unit length L."...I din knw tht!
I think it doesn't have a thickness cuz they din mention it...but in our AP course we took J juz as J=I/A
Ya I dunno whts tht area between the wire and its sheath...the question is weird.
How come the sheet can be left out of the calculation assuming the magnetic field is on the sheath?
Wht if we had to find the magnetic field on the sheath?Then wudn't the net current be
4(pi)R^2-(pi)R^2=3(pi)R^2...so therefore using ampere's law:B(2)(pi)(2R)=3(pi)R^2
which wud give B=(3/4)(pi)(R^2)
I hope this clarifies
 
  • #30
555
0
I don't understand how you get the [itex]3 \pi J R^2[/itex] for the current. (I assume you forgot the J in the above post?)

If you want to know the total current (eg, take a loop with r > 4R) this is:
[tex]I_{\text{encl}} = I_{\text{sheet}} + I_{\text{wire}} = JA_{\text{sheet}} - JA{\text{wire}} = ??? - \pi R^2[/tex]

What is the ??? part? I don't know since I don't know the area of the sheet. You seem to be taking the whole area enclosed by the sheet of 4R (wrongly though, it should be [itex]\pi (4R)^2[/itex]), but (as long as I understood the question well) the current is only running through the (infinitely thin) sheet at 4R, not anywhere < 4R... Right?

How come the sheet can be left out of the calculation assuming...
If the sheet is infinitely thin at radius 4R and you take a loop also at radius 4R, have you enclosed the sheet or not? You can argue this for ever, there is no way to tell since the sheet has no thickness. I don't know wether I should take the sheet into account or not, so I did both, and I showed that both ways yields the (apparently) wrong answer.

EDIT
Also, where does the (2R) suddenly come from?

If you assume the current is running through the entire area enclosed by a loop of radius 4R then the total current would be:
[tex]J ( \pi (4R)^2 - \pi R^2) = 15 J \pi R^2[/tex]

This would yield a magnetic field (on 4R) of:
[tex]B 2 \pi 4R = \mu_0 15 J \pi R^2[/tex]
[tex]B = \frac{15 \mu_0 J R}{8}[/tex] which is also apparently wrong...
 
Last edited:
  • #31
"What is the ??? part? I don't know since I don't know the area of the sheet. You seem to be taking the whole area enclosed by the sheet of 4R (wrongly though, it should be ), but (as long as I understood the question well) the current is only running through the (infinitely thin) sheet at 4R, not anywhere < 4R... Right?"
Isn't the area of the sheath (pi)(2R)^2....so I multiply tht by J which will give the current of the sheath which is (4)(pi)(R^2)
Then the current of the wire is (pi)R^2...agree?
so the net current is the difference of both (3)(pi)(r^2)
 
  • #32
555
0
Why would the area of the sheet be [itex]\pi (2R)^2[/itex]? I don't understand, why?

The area of a wire with a radius of 2R is [itex]\pi (2R)^2[/itex].
The area of a sheet with inner radius 2R and outer radius 3R (thus, thickness R) is [itex]\pi[ (3R)^2 - (2R)^2][/itex] (for example)
The area of a sheet with inner radius = outer radius = 4R (and thus thickness 0) is 0.

If I understand the question correctly, the last example is what you are dealing with, an sheet without any thickness. If this is not correct please tell me.

Also, you keep forgetting to add the J in "the current of the wire is (pi)R^2...". This is an area, not a current. Multiply it with a current density J to make it a current.

Finally about what I said a few posts earlier, that the current density J may be a current per unit length. This is often called a surface current density. I'm not sure if this is what we're dealing with because you haven't specified the question clearly.


EDIT
You may not understand the link between current density, area and current correctly.
In the equation [itex]I = JA[/itex], the area A is the area that the current runs through. In this case, the current does not run through and area of [itex]\pi (4R)^2[/itex], the current only runs through the infinitely thin sheet (right???) which has no area and thus no current.

If the current density J is a current per unit length, then the current through the sheet is simply JL where L is the length of the wire. Since this length is not given and does not cancel out anywhere I still think the question as you have asked it is wrong.


What I want to know is, is the 'sheet' you are talking about infinitely thin or does it have a thickness? Maybe an infinitely small thickness dr or something?
 
Last edited:
  • #33
Oh I got an idea...maybe the area between the smaller and bigger circle is the thickness.Acctually foorget abt thiss qquestion..its getting very messy and I have alrredy done my AP(not very well though!!!)...thank you very much...and appreaciate it
 
  • #34
555
0
If that is the case I get a magnetic field of [tex]B = \frac{14}{8} \mu_0 J R[/tex]

Your original answer specified a magnetic field containing pi however. I can't see any way how the magnetic field can contain pi if you are given the current per area J. The pi will always cancel out...
 

Related Threads on Using ampere's law with current density

  • Last Post
Replies
2
Views
5K
  • Last Post
Replies
4
Views
9K
Replies
9
Views
3K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
1
Views
3K
Replies
3
Views
3K
Replies
2
Views
2K
Replies
3
Views
2K
Replies
5
Views
5K
Top