# Magnetic flux density

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.

∫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

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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.

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

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?

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?