Magnetic Field Between Coaxial Cylinders

[Mathmos6]

Homework Statement

Two long thin concentric perfectly conducting cylindrical shells of radii a and b (a<b) are connected together at one end by a resistor of resistance R, and at the other by a battery that establishes a potential difference V. Thus, a current I=V/R flows in opposite directions along each of the cylinders.

Using Ampere's law, find the magnetic field B in between the cylinders.

Homework Equations

Ampere's Law: $\nabla \times B = \mu (J +\epsilon \frac{\partial{E}}{\partial{t}})$

The Attempt at a Solution

Assuming I have got what I think is Ampere's law correct, I'm really not sure where to go on this one - I know we can infer a few assumptions about the fact the shells are 'perfectly conducting' but I'm not sure what exactly, and so I don't know how exactly to proceed - is J uniform, for example?

In addition, once I have an equation in Ampere's law, do I have to solve things component-wise to get B out of Curl(B) or is there a smarter way to do it?

I'm revising for an exam on Tuesday and I'm really stuck on this one so any help would be appreciated as urgently as you can manage!

Many thanks :-)

 

[Mathmos6]

Any thoughts, anyone? I hope I put this in the right forum section!

 

[graphene]

yes, assume uniform J.

you'd better use the integral form of ampere's law.

the field between the two cylinders would be the field due to the inner cyl only.

(the field inside a cylinder is zero.)

 

[RoyalCat]

yes, assume uniform J.

you'd better use the integral form of ampere's law.

the field between the two cylinders would be the field due to the inner cyl only.

(the field inside a cylinder is zero.)

The conclusion in parentheses is a product of the integral form of Ampere's Law:

$$\oint \vec B \cdot \vec{d\ell}} = \mu_0 I_{pen}$$

where $$I_{pen}$$ is the current that penetrates whatever surface you attach to your loop (With the sign of the current determined by the right hand rule and the direction you choose to march in).

In this problem you can assume radially symmetry for $$\vec B$$ which makes it especially simple since you don't need to use the concept of current density at all.

What makes coaxial cables interesting isn't the field inside, but rather the field outside, use Ampere's Law to find that, and you may be surprised. ;)

 

[rdl]

I would first clarify the assumptions and boundary conditions of the problem. It is stated that the cylinders are perfectly conducting, which means that there is no resistance to the flow of current on their surfaces. This also means that the current will be uniformly distributed along the cylinders.

Next, I would use Ampere's law to find the magnetic field between the cylinders. Since the current is flowing in opposite directions along each cylinder, we can use the superposition principle and consider the magnetic field due to each cylinder separately.

For the inner cylinder with radius a, the current enclosed by a circular path of radius r will be I, and for the outer cylinder with radius b, the current enclosed by a circular path of radius r will be -I (due to the opposite direction of current flow). Therefore, using Ampere's law in integral form, we can write:

∮B·dl = μ0(I+(-I)) = 0

Where μ0 is the permeability of free space. This means that the magnetic field inside the inner cylinder and outside the outer cylinder is zero.

For the region between the cylinders, we can use the fact that the current density J is uniform and directed along the length of the cylinders. Therefore, we can use Ampere's law in differential form to find the magnetic field at any point between the cylinders:

∇×B = μ0J

Since J is directed along the length of the cylinders, we can write J = I/πr^2, where r is the distance from the axis of the cylinders. Therefore, we can write:

∇×B = μ0(I/πr^2)

Solving this equation using vector calculus, we can find the magnetic field between the cylinders to be:

B = μ0I/2πr

Where r is the distance from the axis of the cylinders.

In summary, the magnetic field between the concentric cylinders is given by B = μ0I/2πr, where I is the current flowing through the cylinders and r is the distance from the axis of the cylinders. This result is independent of the radii a and b, as long as a < b.

I hope this helps with your understanding of the problem. Good luck with your exam!