Magnetic field and current of solid wire

  • Thread starter gills
  • Start date
  • #1
116
0

Homework Statement


26_77.jpg


This is also a problem on my Masteringphysics:

A solid conducting wire of radius runs parallel to the axis and carries a current density given by J=J(0) * (1 - r/R)[tex]\hat{k}[/tex] , where J(0) is a constant and r is the radial distance from the wire axis.

The parts are the same as in the textbook.


Homework Equations


biot savart law

J=I/A


The Attempt at a Solution



I haven't answered part B, or C yet.

ok, when i'm entering my answer for part A on masteringphysics, it keeps telling me "variables are case senstive, make sure that you have the right case on your variables." I've switched them around, and it keeps saying the same thing.

Anyway, my simplified answer is J(0)pi*R[tex]^{2}[/tex](1-[tex]\frac{r}{R}[/tex])


Since J is the current density of the wire, the current is just J*Area, right? That's basically what i'm doing, but something is off. Any help would be great.

Thanks
 

Answers and Replies

  • #2
Doc Al
Mentor
44,959
1,223
Why does your answer for total current have an r in it?

You have to integrate to find the total current, since the current density is not constant.
 
  • #3
143
0
Since the density is not constant in the wire (depends on the distance from the center), I guess you will need to integrate over the area instead of just mupliplying. Btw, what is ^k?
 
  • #4
116
0
Since the density is not constant in the wire (depends on the distance from the center), I guess you will need to integrate over the area instead of just mupliplying. Btw, what is ^k?
it's [tex]\hat{k}[/tex], the vector direction.
 
  • #5
116
0
Why does your answer for total current have an r in it?

You have to integrate to find the total current, since the current density is not constant.
ok, so i integrate from 0 to R in the given formula and i get

J = J(0)*[tex]\frac{R}{2}[/tex] = [tex]\frac{I}{A}[/tex] -->

I = J(0)*(piR^3/2)


??
 
  • #6
Doc Al
Mentor
44,959
1,223
That's not correct. Show how you did the integration:
[tex]I = \int J dA[/tex]
 
  • #7
116
0
That's not correct. Show how you did the integration:
[tex]I = \int J dA[/tex]
ok, i actually integrated only J from 0 to R then multiplied by the area after the integration. So i'll integrate what you said.

so DA = (pi*r)dr ??
 
  • #8
Doc Al
Mentor
44,959
1,223
so DA = (pi*r)dr ??
Almost. What's the circumference of a circle?
 
  • #9
116
0
Almost. What's the circumference of a circle?
circumference is 2pi*r

Would dA be 2pi*r dr?

why the need for the circumference?
 
  • #10
Doc Al
Mentor
44,959
1,223
circumference is 2pi*r

Would dA be 2pi*r dr?
Yes.

why the need for the circumference?
Because you dividing the disk into circular rings so you can integrate.
 
  • #11
116
0
Yes.


Because you dividing the disk into circular rings so you can integrate.
ahhh indeed!

i've got I = J(0)*((pi*R^2)/3))
 
  • #12
Doc Al
Mentor
44,959
1,223
Looks good!
 
  • #13
116
0
Looks good!
it's correct, thank you..

For part B, using ampere's Law i came up with ([tex]\mu[/tex](0)*J(0)*R^2)/6r which came out to be correct.

Now for part C, do i need to integrate again since the current surrounded will not be from the whole wire? or i'm just replacing R with r for the total current now because the amperian line is inside the wire?
 
Last edited:
  • #14
Doc Al
Mentor
44,959
1,223
Now for part C, do i need to integrate again since the current surrounded will not be from the whole wire?
Yes.
or i'm just replacing R with r for the total current now...
Not sure what you mean... but don't do it! :wink:
 
  • #15
116
0
Yes.

Not sure what you mean... but don't do it! :wink:
gotcha, haha. Thanks for the help, i'll let you know if i get it.
 
  • #16
116
0
SOLVED!

thanks (again) Doc
 

Related Threads on Magnetic field and current of solid wire

  • Last Post
Replies
3
Views
7K
Replies
6
Views
761
Replies
1
Views
7K
Replies
2
Views
2K
Replies
5
Views
12K
  • Last Post
Replies
3
Views
851
Replies
3
Views
5K
  • Last Post
Replies
2
Views
4K
Top