Finding Magnetic Field Solutions for a Current-Carrying Cylinder?

[adrian116]

the question is that:

A long, straight, solid cylinder, oriented with its axis in the z-direction, carries a current whose current density is \vec J. The current density, although symmetrical about the cylinder axis, is not constant but varies according to the relation

(the relation is in the attachment)

where a is the radius of the cylinder, r is the radial distance from the cylider axis, and I_0 is a constant haveing units of amperes.
a) show that I_0 is the total current passing through the entire cross section of the wire.
b). Using Ampere's law, derive an expression for the magnitude of the magnetic field \vec B in the region r>=a .
c). Obtain an expression for the current I contained in a circular cross section of radius r<=a and centered at the cylinder axis.
d). Using Ampere's law, derive an expression for the magnitude of the magnetic field \vec B in the region r<=a.


For a, Since for the entire cross section of the wire, i subt. r=a into the relation. But it will give zero. I shown nothing. If I subt. J=I/A,
then I=2 I_0 [1- (\frac{r}{a})^2]. Anything wrong,
and how to proof that?

 

[siddharth]

Since the current density is not constant, you need to integrate over the cross section.

Also, have you taken a look at the https://www.physicsforums.com/showthread.php?t=8997"l? If you post the question that way, you won't need to wait till the attachment is approved.

 

[adrian116]

i am sorry since i do not familiar that tutorial yet...

Should i integrate \frac{2 I_0}{\pi a^2} [1- (\frac{dr}{a})^2]
from 0 to a? if yes, how to integerate (dr)^2

 

[siddharth]

i am sorry since i do not familiar that tutorial yet...

Should i integrate \frac{2 I_0}{\pi a^2} [1- (\frac{dr}{a})^2]
from 0 to a? if yes, how to integerate (dr)^2

No, that's completely wrong.

If you take a small elemental area da, then the current which flows through that bit is \vec{J}.\vec{da}

To find the net current through the whole wire, in a sense you add up the current through all the small elemental areas.
So your net current will be

I=\int \vec{J}.\vec{da}

Now,
(i) Can you tell me what elemental area you will take?
(ii) What will the limits of integration be?

 

[adrian116]

elemental area is the small cross section area dA=2 \pi r da,
and the limits of integration is from 0 to a?

 

[adrian116]

I have got the ans.
and the following problems are also be solved,
thank you so much