Calculating Magnetic Flux Density in a Hollow Cylindrical Conductor

[likephysics]

π²³ ∞ ° → ~ µ ρ σ τ ω ∑ … √ ∫ ≤ ≥ ± ∃ · θ φ ψ Ω α β γ δ ∂ ∆ ∇ ε λ Λ Γ ô

Homework Statement

An arbitrarily long hallow cylindrical electric conductor shown carries a static electric current density J=\hat{z}J₀. Determine the magnetic flux density B in the hallow region of radius a. Your result should show that B is constant in this region.

Homework Equations

∫B.dl = µ₀I

The Attempt at a Solution

current I = \hat{z}J₀pi(b²-a²)

B. 2pia = µ₀ \hat{z}J₀pi(b²-a²)

B = µ₀ \hat{z}J₀(b²-a²) /2a

I am not sure about B.dl being B.2pi a

 

[chrisk]

Utilize the relation

\vec{B}=\frac{\mu_0\vec{J}\times\vec{r}}{2}

and approach the problem with a uniform current distribution throughout the circle of radius b, plus a current in the opposite direction in the hole.

 

[likephysics]

Utilize the relation

\vec{B}=\frac{\mu_0\vec{J}\times\vec{r}}{2}

and approach the problem with a uniform current distribution throughout the circle of radius b, plus a current in the opposite direction in the hole.

I am trying to do exactly that: +J current density in b and -J in region of radius a.
Add both to get result. But in the equation you mentioned, how to calculate the cross product.
seems a bit complicated. Can't I use just ∫B.dl = µ0I

 

[chrisk]

Superposition of B within the hollow region:

\vec{B}=\vec{B_b}+\vec{B_a}

\vec{B}=\frac{\mu_0\vec{J}\times(\vec{r_b}-\vec{r_a})}{2}

So, what is

\vec{r_b}-\vec{r_a}

equivalent to?

 

[likephysics]

ddddddddddddddddddddddddddddddd.
Can't believe I missed it.
I shall go shoot myself in the foot.

Before I do that, can you tell me how to solve Jxr for a single sphere problem.
Just do the cross product or is there some trick?