Electric Current and Resistance

[zoyash]

Homework Statement

The current density across a cylindrical conductor of radius R varies according to the

equation J=α(r−r2/R), where r is the distance from the axis. Thus the current density is a

maximum J0 at the axis r=0 and decreases linearly to zero at the surface r=R in which α is a constant.

What is the current through the outer portion of the wire between radial distance R/3

and R. Take R=2mm and α=2*105 A/m2

Relevant Equations

J=α(r−r2/R)

I have attached my solution to this problem but I think I got the wrong answer.

 

[Delta2]

I read your post and the attached attempt at the solution but I have problem understanding which exactly is the formula for the current density. Can you write it in Latex? is it $$J(r)=\alpha r (1-\frac{r}{R})$$

Though in the solution it seems that you use as current density the formula below
$$J(r)=ar(1-\frac{1}{R})$$

I have to say that both formulas seem dimensionally incorrect since the units of $\alpha$ are given as A/m^2

 

[rude man]

Homework Statement:: The current density across a cylindrical conductor of radius R varies according to the

equation $J=α(r−r^2/R)$ , where r is the distance from the axis. Thus the current density is a maximum J0 at the axis r=0 and decreases linearly to zero at the surface r=R in which α is a constant.

Your formula says J=0 at r=0. I think you may have meant $J =J_0 \alpha (R - r^2/R)$. Furthermore, in order for J to decrease linearly with r the formula would have to be something like $J = J_0 \alpha(R-r).$ Try again.

 

[haruspex]

in the solution it seems that you use as current density the formula below
$$J(r)=ar(1-\frac{1}{R})$$

I have to say that both formulas seem dimensionally incorrect since the units of $\alpha$ are given as A/m^2

Well, @zoyash started with $$J(r)=aR(1-\frac{1}{R})$$
But then switched to $$J(r)=ar(1-\frac{1}{R})$$
Neither makes sense dimensionally and neither decreases linearly with r. Probably should be $$J(r)=a(1-\frac{r}{R})$$