# Magnetic field of Diametrically magnatized rod

1. Feb 7, 2017

### Chris Fuccillo

Hello again everyone sorry it’s been a while since last post or question, been working hard and had my nose to the grind stone, face stuck in many books along with heaps upon heaps of magnetic theory and EE lectures muhahahahaha.

Ok to the question, I have been having some trouble finding a formula for calculating the magnetic field of a diametrically magnetized rod, I believe I should use

Formula for the B field on the symmetry axis of an axially magnetised sphere magnet:

B =B r 23 R 3 (R+z) 3 B=Br23R3(R+z)3

Br: Remanence field, independent of the magnet's geometry (see physical magnet data)

z: Distance from the sphere edge on the symmetry axis

R: Semi-diameter (radius) of the sphere

The unit of length can be selected arbitrarily, as long as it is the same for all lengths.

This should work for most any point along the rod ?

2. Feb 7, 2017

Your question isn't completely clear to me, but I think I can guess what you may be trying to solve: You have a rod of very long length $L$ and let's say this rod extends along the z-axis and the cross section in any plane parallel to the x-y plane is circular with radius $a$. The rod is magnetized with magnetization $M$ along the x direction. If this is the case, the "pole" theory will work quite well for this problem. Also, there is a Legendre type solution inside the rod where $B=\frac{1}{2}M$. This comes from $H_{inside}=-\frac{1}{2}(M/\mu_o ) \hat{i}$ and $B=\mu_o H +M$. (The $H_{inside}=-\frac{1}{2}(M/\mu_o) \hat{i}$ holds for every point inside the rod.) For this problem, I'm not familiar with what the Legendre solution looks like in cylindrical coordinates for the magnetic field outside the rod, but I believe a google of this or the analogous electrostatic problem would yield a solution. Meanwhile, using a magnetized sphere to try to solve this problem is incorrect. (The uniformly magnetized sphere will have $H_{inside}=-\frac{1}{3}(M/\mu_o) \hat{i}$.)