How Do You Calculate the Constant C in a Cylindrical Wire's Current Density?

[Hank_Rearden]

Homework Statement

A long, solid, conducting cylindrical wire of radius a has a current density J(r) = Cr and
carries a total current I

find a value for the constant C in terms of I and a

Homework Equations

J = I/A

The Attempt at a Solution

I believe that I need to integrate along the radius, r, from zero to a, but doing so always yields C = 4*I/(pi)*a^4, which is incorrect. But I'm also unsure as to why an integral is even necessary

 

[jhae2.718]

Could you show your steps?
You'll want to use I = \int\limits_A \vec{\jmath} \cdot d\vec{A} for this. Think carefully about what dA is...

 

[Stephen Tashi]

yields C = 4*I/(pi)*a^4

Do you mean 4*I/(pi * a^4) ?

 

[jhae2.718]

Do you mean 4*I/(pi * a^4) ?

That's still not the answer.

 

[Hank_Rearden]

Sorry, I'm not ignoring you, I'm just still getting the hang of using PF's symbols

 

[jhae2.718]

Sorry, I'm not ignoring you, I'm just still getting the hang of using PF's symbols

https://www.physicsforums.com/showthread.php?t=386951"

 

[Hank_Rearden]

I've been using \pi*r^{2} dr as my dA

 

[jhae2.718]

I've been using \pi*r^{2} dr as my dA

The problem there is that \pi r^2 is the total area as a function of r. (I.e., A(r) = \pi r^2.)

What you want is the infinitesimal area dA, which you can get by taking \frac{dA}{dr} and then multiplying through by dr.

 

[Hank_Rearden]

but where is the expression dA/dR coming from?

 

[Feldoh]

You're trying to take the current density through some area. For any given radius your current density will change, meaning you don't have a constant dot product (meaning you can't use pi*r^2 as the area).

You need to figure out the area element through which the current passes. HINT: Cylindrical wire leads to cylindrical symmetry, also current is moving axially.

 

[klondike]

Should be \frac{2\pi c a^3}{3}

 

[jhae2.718]

but where is the expression dA/dR coming from?

dA/dr is the derivative w.r.t. r of A=\pi r^2.

We can get the infinitesimal area this way:
<br /> \begin{align*}<br /> A &amp;= \pi r^2 \\<br /> \frac{dA}{dr} &amp;= 2\pi r \\<br /> dr \frac{dA}{dr} &amp;= 2\pi r dr \\<br /> dA &amp;= 2\pi r dr<br /> \end{align*}<br />

Now, what do you think you should do next?

 

[Hank_Rearden]

Integrate from 0 to a, (C*r)*(2*pi*r) dr?

 

[jhae2.718]

Right. Here's the integral (you can click on it to see the LaTeX) I = \int_0^a Cr \cdot 2\pi r dr = 2\pi C\int_0^a r^2dr

Now, one thing I want to point out is that the current is the integral of the dot product of the current density vector and the infinitesimal area.

In this problem, they are parallel, which means that \mathbf{j} \cdot d\mathbf{A} = j\,dA. This won't always be the case.

Are you familiar with the area vector?

 

[Hank_Rearden]

I've seen it used before, and I think I understand it as a concept, but I'm not particularly adept at using it myself

 

[jhae2.718]

The area vector is important in cases of electric flux, Gauss's law, current, the continuity equation, etc.

Basically, if you have an area, the area vector is the unit vector that is normal to the surface of the area, pointing out.

Now, I've mentioned the dot product in \int\limits_A \mathbf{j} \cdot d\mathbf{A} several times. This makes the integral equivalent to the integral of the current density times the cross sectional area in the direction of the current. (This is analogous to the way electric flux is defined, which you should at least be familiar with if you are discussing current.)

 

[Hank_Rearden]

That makes sense, thanks

 

[klondike]

\int _{s} J\cdot \hat{n}dS = \int _{s} JdS = C\int^{2\pi}_{0}\int^{a}_{0}r^2drd\theta=\frac{2\pi C a^3}{3}