Magnetic Fields from Currents in a Wire and a Cylindrical Shell

diethaltao

1. The problem statement, all variables and given/known data
A solid cylindrical conducting shell of inner radius a = 5.3 cm and outer radius b = 7.9 cm has its axis aligned with the z-axis as shown. It carries a uniformly distributed current I2 = 7.1 A in the positive z-direction. An infinite conducting wire is located along the z-axis and carries a current I1 = 2.7 A in the negative z-direction.


What is [itex]\int[/itex][itex]^{P}_{S}[/itex] [itex]\vec{B}[/itex] . [itex]\vec{dL}[/itex], where the integral is taken on the straight line path from point S to point P as shown?

Link to the picture: http://i89.photobucket.com/albums/k2...cylindersD.png

2. Relevant equations


3. The attempt at a solution
I'm not even sure how to approach this problem. At first I found the difference between the values of the magnetic field at P and at S, but this was wrong.
Then I thought to use
∫[itex]\vec{B}[/itex] . [itex]\vec{dL}[/itex] = μoI
but B is not constant over the interval [S,P] so I can't pull it out of the integral.

I was able to calculate the integral along the dotted path in the picture, which I basically did by realizing R to S was perpendicular to the field so it didn't count, and that P to R was 1/8 of a larger circle drawn around the diagram. But in this case, B was a constant distance from the centre.

Any input is appreciate. Thanks!

tiny-tim

hi diethaltao!

how about doing it along a path 8 times as long, in a square?

diethaltao

Hi tiny-tim!

So, I would find r using the Pythagorean Theorem:
r = [itex]\sqrt{((0.21)(0.6))^2+(0.21-(0.21)(0.6))^2}[/itex] = 0.151.
So I would multiply the field (which I found at point P to be 4.19E-6) by the perimeter of the square?
And the perimeter would be 8R...? Obviously I messed up somewhere.

tiny-tim

hi diethaltao!

(just got up )

why do you want to know the field?

try using the Ampere-Maxwell law

diethaltao

Hi tiny-tim!

Ampere-Maxwell's law says that [itex]\vec{B}[/itex].[itex]\vec{dL}[/itex] around a closed loop is proportional to the current enclosed in that loop, or
[itex]\oint[/itex][itex]\vec{B}[/itex].[itex]\vec{dL}[/itex] = μo*Ienc.
But I can't just plug in the values of μo and Ienc to solve for the integral. And if I choose a square to be my enclosed area, B wouldn't be the same along the side, correct? For example, the field at point S would be stronger than at point P.

tiny-tim

yes, but if eg you extend PS to a point T on the y axis,

then ∫ B.dl along PS will be the same as ∫ B.dl along ST, won't it?

diethaltao

Ok, I understand that.
So now I have something like the picture attached, where R is the hypotenuse of a 0.21 by 0.21 triangle (not sure if I need that value, but I calculated it anyways.)
The integral from P to S is the same as the integral of T to S.
And therefore, the integral from P to T would be the same as it would be for the other three sides of the square.
I tried integrating around the whole square and dividing it by 8, but that was incorrect.

http://i89.photobucket.com/albums/k2...o/Untitled.png

tiny-tim

hi diethaltao!

use the Ampere-Maxwell law …

∫ B.dl = … ?

diethaltao

Oh, wow, I get it now.
Can't believe I overlooked something so simple! Thanks so much!