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. ;)