UCSB Science Line
Sponge Spicules Nerve Cells Galaxy Abalone Shell Nickel Succinate X-ray Lens Lupine
UCSB Science Line
How it Works
Ask a Question
Search Topics
Our Scientists
Science Links
Contact Information
Thanks for your help so far, but when calculating the diamagnetic force created by bismuth, it would be much easier (at least for me) to use a formula, or a couple of basic formulas to calculate the diamagnetic force. So if you guys have a formula that I could plug in all the variables into TO CALCULATE THE DIAMAGNETIC FORCE CREATED BY BISMUTH that would be a big help. Thanks for your help so far.
Question Date: 2008-06-18
Answer 1:

It's already clear that you know more about this than I do. You probably know this, but from what I was able to dig up online, the magnetization of a substance is equal to the magnetic susceptibility of a substance times the strength of the magnetic field applied to it. The magnetic susceptibility is a constant (although it varies with temperature). For bismuth, the number I found is -1.66 x 10-4, the negative number signifying that it is diamagnetic. I do not know if there are units on that constant; because susceptibility is 1 - permeability, I doubt that there are any units, but if there are, then that number is useless until you can find units somehow (I couldn't).

The magnetization of the bismuth will cause it to have a magnetic dipole that is opposite to that of the inducing magnet, which is why the force between them will be repulsive. How strong this force is will depend on their position relative to one-another, and that depends on the shapes of the magnetic fields and their distance. This is a much more complicated area of physics than I understand, and I don't think it's solvable with simple algebra (you may need some calculus to do it). I'm afraid that's all I can help you with.

Answer 2:

The trouble is, the specific question you're asking can't be answered by a single formula.Here's why.

First, any magnet experiences zero force if it's in a uniform external field. It will only experience a torque: it will try to twist to align itself with the external magnetic field. This is why a compass needle turns but doesn't go flying off to the north pole.There's no net force on the compass, only a torque which *turns* the needle. This is true no matter how large or small the "little"magnet is. So if the bismuth is in a *uniform* magnetic field, like a small piece of Bi at the center of a large magnet, the Bi will not experience any force at all.

So let's assume the bismuth is not in a uniform field, but is on the edge of a magnetic field, where the field changes with position.Then we make an approximation: we assume that the problem can be calculated as if the magnet was only defined by its north and south poles, and that they can be represented as small points. (This is analogous to positive and negative charges in an electric field.)Here's the approximate formula for force between two magnetic poles:

where mu_0 is magnetic permeability of free space (4*pi*10-7)Newton/Amp2), r is the distance between the two poles (like the ends of two bar magnets), and qm1 & qm2 are the pole strengths of each magnetic pole. Since every magnet has 2 poles, this needs to be repeated 4 times:
magnet 1 north to magnet 2 north
magnet 1 north to magnet 2 south
magnet 1 south to magnet 2 north
magnet 1 south to magnet 2 south
r and F will be different for all 4 cases.

These forces are also indifferent directions, so you need to do vector addition (in other words, add up the terms in the "x" direction separately from those in the "y" direction).

Obviously, the distance between the north pole & south pole on your bismuth sample might be quite small, but it's important. (If it were zero, then the magnetic field would seem to be uniform at that scale,so the total force would go to zero.) To say the same thing a different way, if a single pole with pole strength qm1 is in a uniform field B, the force on that pole is F=(qm1)*B. Having two opposite poles next to each other (north & south on a small magnet)means F = (qm1)*B+(-qm1)*B = (qm1-qm1)*B = 0, so zero total force.

Now how big is the pole strength of a magnet? Here's the relation between magnetic field B created by a single pole (either north or south) with pole strength qm1:
B (created by pole qm1) = (mu_0) * (qm1) / (4*pi*r2)Obviously, the field changes with distance r from the pole. This gives a roughly approximation of B near a magnet: B(north pole) minusB(south pole) at the location where the bismuth will be. The distances r_north & r_south may be different, depending on where the bismuth is relative to the two poles. And again, the two B's need to be added in vector addition, since they pull in different directions.For a real magnet, you would use this backwards: measure B vs.distance near the pole, calculate qm1, then use that to calculate B everywhere else including both poles.

It sounds like you already know that diamagnetic material is not magnetized when it's by itself but becomes *temporarily* magnetized in the presence of an external magnetic field. The magnetization of diamagnetic materials is extremely weak. Bismuth happens to be the strongest. In the presence of a magnetic field H (in Tesla), them agnetization of bismuth is:
M = -.000166*H
and by definition,
B = mu_0 * (H + M)
= mu_0 * (H - .000166*H)
= mu_0 * 0.999834 * H
So you need to know the actual magnetic field (H) in the location where the bismuth will be. You can get a rough idea from the calculations in the previous paragraph, or from an experiment like this one:

But it's probably more accurate to find a Hall Effect meter and measure the field directly.

So after all that, I'm still not sure if you've got the answer to your question.

Click Here to return to the search form.

University of California, Santa Barbara Materials Research Laboratory National Science Foundation
This program is co-sponsored by the National Science Foundation and UCSB School-University Partnerships
Copyright © 2017 The Regents of the University of California,
All Rights Reserved.
UCSB Terms of Use