Finding current density given B-field

[Saitama]

Homework Statement

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.

Homework Equations

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!

 

[haruspex]

I don't see any equations in my notes which relates the B-field and current density.

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?

 

[Saitama]

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?

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? $2\pi r$?

 

[ehild]

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? $2\pi r$?

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

 

[Saitama]

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

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?

 

[ehild]

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?

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

ehild

 

[Saitama]

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

ehild

Thank you ehild!

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