B inside a cylinder of radius R

[Murr14]

hey all, this is confusing me a lot:

consider an infinitely long cylinder of cross-section radius R. we choose symmetry axis of the cylinder as the z-axis. The cylinder carries a uniform current density J in the +z direction throughout it's cross section. what is B at r inside of the cylinder? Express you answer in the component form B = Bx i + By j + Bz k


...what I'm confused about is whether or not I can use ampere's law with an amperian loop inside the cylinder or if i have to use Biot-Savart...

i did it using ampere's law and i got |B| = uJs/2 ...and the vector B is in the phi direction...wrapping around the z-axis...did i do that right? how do i get it into cartesian coords? Do i have to use Biot-Savart into be able to get it in cartesian coords regardless of whether or not ampere's law can be used?

 

[Andrew Mason]

i did it using ampere's law and i got |B| = uJs/2 ...and the vector B is in the phi direction...wrapping around the z-axis...did i do that right? how do i get it into cartesian coords? Do i have to use Biot-Savart into be able to get it in cartesian coords regardless of whether or not ampere's law can be used?

It is a fairly simple Ampere's law problem. The line integral of the magnetic field around a circle at radius r is just:

\oint \vec B\cdot ds = \mu_0I_{encl}= \mu_0AJ = \mu_0\pi r^2J

due to symmetry, |B| is constant and always in the direction of ds so:

\oint \vec B\cdot ds = B2\pi r

B2\pi r = \mu_0\pi r^2J

B = \mu_0rJ/2

At a given point \vec r = x\hat i + y\hat j, the magnetic field vector is perpendicular to the radial vector. So By/Bx = x/y. Divide by r to get the unit vectors.

\vec B = B\frac{y}{r}\hat i + B\frac{x}{r}\hat j

where B = \mu_0rJ/2

AM

 

[Murr14]

ok thanks man...that makes sense...ok so thaty's B for inside the cylinder...now whatabout outside?

 

[Andrew Mason]

ok thanks man...that makes sense...ok so thaty's B for inside the cylinder...now whatabout outside?

Outside the enclosed current is the entire current in the cylinder. So:

B = \mu_0J\pi R^2/2\pi D where D is the distance from the centre.

AM

 

[Gamma]

IN a right handed coordinate system, Should that be

\vec B = -B\frac{y}{r}\hat i + B\frac{x}{r}\hat j