# B inside a cylinder of radius R

1. Mar 13, 2005

### 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 in to be able to get it in cartesian coords regardless of whether or not ampere's law can be used?

Last edited: Mar 13, 2005
2. Mar 13, 2005

### Andrew Mason

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

3. Mar 13, 2005

### Murr14

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

4. Mar 14, 2005

### Andrew Mason

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

5. Mar 14, 2005

### Gamma

IN a right handed coordinate system, Should that be

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