Finding current density given B-field

1. May 24, 2013

Saitama

1. The problem statement, all variables and given/known data
Find the current density as a function of distance $r$ from the axis of a radially symmetrical parallel stream of electrons if the magnetic induction inside the stream varies as $B=br^a$, where $b$ and $a$ are positive constants.

2. Relevant equations

3. The attempt at a solution
Honestly, I don't have any idea for this problem. I don't see any equations in my notes which relates the B-field and current density.

Any help is appreciated. Thanks!

2. May 25, 2013

haruspex

No, but I'm sure you have an equation for the field based on the current in a wire. Won't it just be a matter of integrating the density, viewed as many wire elements?

3. May 25, 2013

Saitama

Ampere's circuital law?

$$\int \textbf{B}\cdot d\textbf{l} = \mu_o\int \textbf{J} \cdot d\textbf{S}$$
What should I replace dl with? $2\pi r$?

4. May 25, 2013

ehild

Because of radial symmetry, the vector B is tangential to the circles around the axis and its magnitude depends only on the distance from the axis. dl=rdθ and yes, the line integral is B(2πr).

ehild

5. May 25, 2013

Saitama

That gives me
$$2\pi b r^{a+1}=\mu_o \int \textbf{J} \cdot d\textbf{S}$$
Is $dS=2\pi rdr$?

And how am I supposed to solve the above equation?

6. May 25, 2013

ehild

Plug in the expression for dS. The integral is valid for all r. You get an equation of form f(r)=0r∫g(r')dr'. How do you get g(r)?

ehild

7. May 25, 2013

Saitama

Thank you ehild!

I get
$$br^{a+1}=\int_0^{r}\textbf{J}rdr$$
Differentiating both the sides,
$$\Rightarrow (a+1)br^{a}=\textbf{J}r$$
$$\Rightarrow \textbf{J}=(a+1)br^{a-1}$$