Currents in a Wire and a Cylindrical Shell

[CoolBeanz99]

Homework Statement

Hello PF, first time poster here. I don't normally ask for help on the internet, especially for homework, but I've visited this website several times and I've seen nothing but good as I've looked at everyone else being helped, so I decided it'd be worth a try. :)

So here's the question I've been stuck on for a while now; any help with it would be greatly appreciated:

A solid cylindrical conducting shell of inner radius a = 4.2 cm and outer radius b = 6.5 cm has its axis aligned with the z-axis as shown. It carries a uniformly distributed current I₂ = 7.8 A in the positive z-direction. An inifinte conducting wire is located along the z-axis and carries a current I₁ = 2.8 A in the negative z-direction. Also given: d = 27 cm.

What is the integral of B dot dl from S to P, where the integral is taken on the straight line path from point S to point P as shown?

[PLAIN]http://smartphysics.com/images/content/EM/15/h15_cylindersD.png

Homework Equations

Ampere's Law: \oint\vec{B}\bullet\vec{dl} = \mu_{0}I_{encl}

The Attempt at a Solution

Ok. So in a previous problem regarding the same situation, they had me calculate B dot dl from S to P following the dotted path. That was a nice problem because since the B field was perpendicular to the path from S to R, B dot dl for that portion was simply 0. Finding B dot dl for the portion from R to P was relatively simple since the B field is tangential to a circular Amperian loop with radius d that follows the dotted path exactly. Basically every B and dl vector were parallel and could be multiplied directly without any angles involved (also B was of constant magnitude since it was always equidistant from the center). The only thing left to do was to realize that instead of using the entire circumference of the loop, only the portion between R and P is needed, so instead of the integral of dl simplifying to 2\pid, it simplified to \frac{\pi}{4}d, the segment being an eighth of the full loop.

This problem seems a lot more tricky in that we are taking the things that were nice before and making them... not as nice. Now, as we move along the path from S straight to P, two things are happening that did not happen before: both the magnitude of B, and the angle it makes with dl is changing. This means that the integral will not simplify as it did last time such that B can be pulled out as a constant and the dot product can be simplified to scalar multiplication; a dependency on angle is now required.

Past this, I'm not really sure where to go. My first instinct is to try something like use a circular Amperian loop that has some kind of changing radius and integrate B dot dl using the cosine of the angle between the B and dl vectors, which also changes with radius. Polar coordinates are what I'm thinking might be needed for this problem.

Whew, sorry I write a lot. Any suggestions as to how to approach this problem?

Thanks,
CoolBeanz99

 

[CoolBeanz99]

Anyone? Is it really that hard? lol

 

[JD1992]

Hi, I think it's important to look at the problem conceptually. Particularly the idea of independence of path.

 

[CoolBeanz99]

Well, I've been out of this class since last summer, but I guess it would still be interesting to know how to solve this problem. I recall going to class the next day and asking my professor. He essentially said that the problem was beyond the scope of our class and he wasn't sure how it got in our homework or how we would go about solving it, except that it would end up being messy. I told him my instinctive approach and he said it sounded conceptually correct, just setting up and solving the integral would have been a real pain.

 

[Piratejacob]

I found that the integral is the same, it is path independent since you are summing up the contributions that are tangential in both cases.

 

[JMcDe]

Use the Definition of Amperes Law. The Shape formed does not have a current enclosed. Use answer from the previous question. It should be the negative of what that answer was to make it zero, since no current is enclosed by the shape.