Calculating Magnetic Flux Density in a Hollow Cylindrical Conductor

In summary, the conversation discusses finding the magnetic flux density in a hollow cylindrical electric conductor with a static electric current density, using the equation ∫B.dl = µ0I. The solution involves using the relation \vec{B}=\frac{\mu_0\vec{J}\times\vec{r}}{2} and approaching the problem with a uniform current distribution throughout the circle of radius b, plus a current in the opposite direction in the hole. The discussion also touches on utilizing superposition to find the total B within the hollow region and the calculation of Jxr for a single sphere problem.
  • #1
likephysics
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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=[tex]\hat{z}[/tex]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 = [tex]\hat{z}[/tex]J0pi(b2-a2)

B. 2pia = µ0 [tex]\hat{z}[/tex]J0pi(b2-a2)

B = µ0 [tex]\hat{z}[/tex]J0(b2-a2) /2a

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

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  • #2
Utilize the relation

[tex]\vec{B}=\frac{\mu_0\vec{J}\times\vec{r}}{2}[/tex]

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.
 
  • #3
chrisk said:
Utilize the relation

[tex]\vec{B}=\frac{\mu_0\vec{J}\times\vec{r}}{2}[/tex]

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
 
  • #4
Superposition of B within the hollow region:

[tex]\vec{B}=\vec{B_b}+\vec{B_a}[/tex]

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

So, what is

[tex]\vec{r_b}-\vec{r_a}[/tex]

equivalent to?
 
  • #5
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?
 

What is magnetic flux density?

Magnetic flux density, also known as magnetic field strength or magnetic induction, is a measure of the strength of a magnetic field in a given area. It is measured in units of tesla (T) or gauss (G).

How is magnetic flux density calculated?

Magnetic flux density is calculated by dividing the magnetic flux, measured in webers (Wb), by the cross-sectional area of the magnetic field. It can also be calculated by multiplying the permeability of the medium by the magnetic field strength.

What factors affect magnetic flux density?

The factors that affect magnetic flux density include the strength of the magnetic field, the distance from the source of the field, and the material properties of the medium through which the field is passing. In addition, the angle at which the field is measured can also affect the magnetic flux density.

Why is magnetic flux density important?

Magnetic flux density is important because it helps us understand and measure the strength and behavior of magnetic fields. It is used in a variety of applications, such as in the design of electromagnets, motors, and generators, as well as in medical imaging techniques like MRI.

How is magnetic flux density related to magnetic force?

According to the Lorentz force law, the force on a charged particle moving through a magnetic field is proportional to the magnetic flux density. This means that a higher magnetic flux density will result in a stronger force acting on the particle. Therefore, an increase in magnetic flux density can lead to a more powerful magnetic force.

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