# Magnetic Fields from Currents in a Wire and a Cylindrical Shell

1. Mar 10, 2012

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

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
∫$\vec{B}$ . $\vec{dL}$ = 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!

2. Mar 10, 2012

### tiny-tim

hi diethaltao!

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

3. Mar 10, 2012

### diethaltao

Hi tiny-tim!

So, I would find r using the Pythagorean Theorem:
r = $\sqrt{((0.21)(0.6))^2+(0.21-(0.21)(0.6))^2}$ = 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.

4. Mar 11, 2012

### tiny-tim

hi diethaltao!

(just got up :zzz:)

why do you want to know the field?

try using the Ampere-Maxwell law

5. Mar 11, 2012

### diethaltao

Hi tiny-tim!

Ampere-Maxwell's law says that $\vec{B}$.$\vec{dL}$ around a closed loop is proportional to the current enclosed in that loop, or
$\oint$$\vec{B}$.$\vec{dL}$ = 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.

6. Mar 11, 2012

### 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?

7. Mar 11, 2012

### 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/k211/diethaltao/Untitled.png

Last edited: Mar 11, 2012
8. Mar 12, 2012

### tiny-tim

hi diethaltao!

use the Ampere-Maxwell law

∫ B.dl = … ?

9. Mar 12, 2012

### diethaltao

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