# 14 Tesla magnet strength? Too much?

1. Jan 2, 2009

### Cspeed

I used the formula here:
http://en.wikipedia.org/wiki/Magnet#Force_between_two_bar_magnets

The values came from an experiment in which I found the force between two small Neodymium magnets by balancing with the force of gravity (it was around .15 newtons). Am I using this equation incorrectly? I solved for "magnetic flux density very close to each pole, in T" and got a result of over 14 Teslas. How can I interpret this data? Is it completely bogus? My goal is to find out the magnetic field strength At certain distances from the magnet.

2. Jan 2, 2009

### Topher925

14 sounds pretty high to me. If I remember correctly, a rather strong NdBFe magnet is in the 0.25-0.4 Tesla range. I thought field strengths in the 10+ range could only be produced by bitter solenoids. Although I could be wrong about this.

3. Jan 2, 2009

### Cspeed

I found a math mistake, and now I'm getting a much more reasonable ~1 T. This still seems somewhat high for a small half-inch magnet. Could anyone tell me what this value actually means though? Can I use it to estimate the magnetic field strength 2 cm from the magnet?

4. Jan 2, 2009

### Gokul43201

Staff Emeritus
Last edited: Jan 2, 2009
5. Jan 2, 2009

### Cspeed

Aside from using a Hall Effect sensor, is there a poor man's way to measure B, through experiments or something? And am I correct in saying that using that equation which gave me ~1 T means that B at the surface of the magnet is 1 T?

6. Jan 3, 2009

### clem

That depends on the dimensions of L and R compared to 2 cm.
B near the axis is roughly constant as long as 2cm<<L or R.
The B field outside a uniformly magnetlzed magnet is the same as the electric field outside a uniformly charged disk. On the axis, this is fairly simple. Off axis, either Legendre polyinomials or elliptic integrals are needed.

7. Jan 3, 2009

### Cspeed

The L=R which is less than 1 cm. Can I find the B field on the axis 2cm from the magnet with simple math?

I found an equation for the electric field due to a charged disk, but it is considered a flat disk. The variables are the charge per unit area, radius, and distance from disk. Could this help me at all? And how would I find the charge per unit area if I know the magnetic flux density very close to each pole of my magnet?

Last edited: Jan 3, 2009
8. Jan 3, 2009

### Gokul43201

Staff Emeritus
I guess that depends on what you've got in your basement/garage/home/world-destruction-headquarters to use for experiments. Do you have a small magnetic compass that works reasonably well?

More likely, this is the field inside the magnet, along the axis, between the poles. At the surface, it could already be a few times smaller.

Surely, this isn't right. The field from a disk falls away like 1/r^2 for large r, while the field from a magnet is a dipole field, and falls away like 1/r^3.

Likely, your best efforts will leave you with an error bar of a couple orders of magnitude, using only the data you have so far.

If you do have a tiny compass, you can find the null points of your magnet, look up the value of the terrestrial field at your location (specifically, the horizontal component of it if that is how you are doing the experiment), and scale that up by (r/2)^3 to get a rough estimate of the field at 2cm (here, r=distance to null point, in cm, and is hopefully large enough to give to a reasonable estimate).

9. Jan 3, 2009

### Cspeed

I do have a compass. Funny enough, I think I demagnetized it using the magnets, but now it seems to generally point north. I'm afraid I can't understand exactly how your idea would work. I haven't found a website which would tell me the value of the terrestrial field at my location yet, but I'll keep looking. But once I get that, I'm not sure what to do.

10. Jan 3, 2009

Could you explain a little about how you would use the Hall Effect Sensor to measure a magnetic field? Seems like a pretty useful thing for only $2.... 11. Jan 3, 2009 ### Vanadium 50 Staff Emeritus Using a Hall probe is quite simple. Usually there are two wires coming out that carry the signal: a voltage that is proportional to the field, and you get a calibration from the factory. There are really only two complications - one is that probes tend to be fragile, and the other is that they measure B along a particular direction, so you need to make sure it's oriented correctly. 12. Jan 3, 2009 ### Gokul43201 Staff Emeritus If you believe the compass still works fairly well, here's the basic idea. With the magnetization of your Nd-magnet pointing along the horizontal plane, towards the Earth's S-magnetic pole, slowly move the compass along the axis of the magnet, starting at its NP and moving away. When the compass is close to the magnet, it responds to a field which is dominated by the magnet. When it is far away from the magnet, the field it sees will be the Earth's field, which (as designed) is pointing in the opposite direction. Somewhere along the way, the compass needle will have to flip around. This is roughly the point where the earth's field is equal and opposite to the magnet's field, and it's called the null point of the magnet. So, at this distance (r_null), you know that the magnet's axial field is roughly equal to the terrestrial field (typically around 0.025mT). If you assume that the field falls away like $B \propto 1/r^3$ over most of the region from 2 cm to r cm, then you can scale accordingly to find the field at 2cm. Hopefully, you find 3cm < r_null < 20cm. Otherwise, the errors might get kinda large. You will also want to keep the ~1T number in mind for the upper bound. Terrestrial fields in the US are tracked by USGS: http://geomag.usgs.gov/observatories/ [Broken] http://geomag.usgs.gov/realtime/ Last edited by a moderator: May 3, 2017 13. Jan 3, 2009 ### Gokul43201 Staff Emeritus Just looked up the spec sheet for the sensor I pointed to...it has 3 leads: 2 for a power supply (it operates at 4.5-6V, so even a simple$3 lantern battery will work for that), and one for the signal out (measured relative to the ground lead). You can read off the signal using a regular hand-held multimeter. You then look up the calibration table that comes with the sensor to convert the voltage into a magnetic field.

I'd be surprised if you can't get a fully assembled probe for a "little" more money.

http://www.laboratorio.elettrofisico.com/eng/products.asp?famiglia=Instrumentation [Broken]

Last edited by a moderator: May 3, 2017
14. Jan 3, 2009

### clem

I am not sure whether you want to measure B or calculate it.
The F~MA^2 Wikiformula is for touching magnets, but only if L>>R, which is not your case.
For L=R, do you even know whether M is along the axis or perpendicular to the axis?
For a kitchen magnet, it is perpendicular.
The long wikkiformula with three terms is for x>>L and R, which is not your case.
The B field on the axis (with M along the axis) is the same as the E field of two disks with (in gaussian units) the charge density equal to the magnetization M. For your case, you would have to consider each end as a uniformly charged disk.

15. Jan 3, 2009

### Cspeed

Gokul43201, thanks, I'll try that out.

I just want to find out what it is, whatever method it takes (minus fancy equipment). So, if I can find it through experiments, then that's fine. But I used this equation, solving for B, hoping that I would get on the right track of finding the max B due to the magnet. [This will then be used as a comparison of the B I calculate in a separate project].

I didn't use that equation at all. Just to clarify, I found the force between two identical magnets, 5 cm apart. The L=R=.6cm. And also what exactly does L>>R mean?

I am pretty sure the axis goes directly through the center of the two circular surfaces of the cylindrical magnet. I hadn't even used M, and I don't know anything about 'magnetization.'

16. Jan 3, 2009

### Cspeed

I got around 28cm for the null point. Somehow I forgot that I already put all of the magnets in my generator (duh) so I can't do a lot of testing at this point. If the other magnets didn't interfere too much, it is probably somewhere around 28 cm. Using .0211 mT, I plotted 422/x^3 (because this gives a y-value of .021 for x=28). When I'm 2 cm away this yields .052 T, but at .5 cm it yields 3.3 T, which is very unreasonable.

17. Jan 4, 2009

### Gokul43201

Staff Emeritus
The inverse cubic scaling is only valid for r>>L. For r<<L, the expression is closer to inverse square. 2cm is not really "much" larger than your magnet thickness, is it? So, I would expect your number to be a bit of an overestimate, but I'd guess that 0.01T at 2cm is probably within a factor of 3 from the correct answer.

Or, you could fit to the function for the E-field from a pair of oppositely charged disks, as suggested above. It's just the vector sum of the expression you have for one disk with a similar one for the second disk (except that it is displaced by L, and has the opposite "charge").

See what that gets you.

18. Jan 4, 2009

### Cspeed

Out of curiousity, as far as balancing out with the magnetic force of the Earth, what would happen using two magnets instead of one? Could the force be expected to double from the magnets?

19. Jan 6, 2009

### heliotrope

1 T is right on for the surface field of a neodymium magnet. If you look up those magnets for sale you will find the surface fields vary from 1 to 1.5 T.