# Using ampere's law with current density

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?

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## Answers and Replies

dx
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It's very difficult to understand what you've written. Also, these questions should be posted in the homework section.

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

dx
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Write out the question in its original form and draw a labeled diagram.

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.

dx
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I can't make sense of it the way you're formulating it. Maybe you dont remember it correctly.

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?

wht abt the diagram...has it helped?

dx
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I can't see the diagram, it must be approved first. Maybe you can host it somewhere else and give me a link.

oh god...can I send it to ur e-mail?I dunno anywhere where I can host it....

sent it

dx
Homework Helper
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Nope, I'm sorry I'm unable to understand it. Maybe some one else here will be able to help you.

there is no one else communicating with me.wht exactly cant u get?

look there is juz a current going out of the page and into the page basically and the wire is a staright long one.

dx
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In that case, if you consider a small circular loop of radius r around the axis as the path of integration, you will get $$B = 2\mu_{0} J r$$. 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.

In that case, if you consider a small circular loop of radius r around the axis as the path of integration, you will get $$B = 2\mu_{0} J r$$. 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.

dx
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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.

oh yeah u r right...it shud be zer h wait I think the question was whts the magnetic field on the surface of the sheath.sorry abt tht. dx
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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.

dx
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I suggest you move on to other problems instead of trying to solve a problem that you can't remember .

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.

Can you think of anywhere where the magnetic field will be (1/2)(J)(pi)(R^2)

dx
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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.