physics-hall probe and flux density

sarah12345

hi i need to devise a method for callibrating a hall probe and using it to measure the magnetic flux density mid way between two permenant magnets

im thinking that i would use helmholtz coils to set up a uniform magnetic field of known magnetic flux density to callibrate the probe-by putting the probe in the field and measuring the voltage then using the equation for voltage produced in a hall probe change the probe until i got the right voltage and then repeat for several different mag flux densities
would this work?

also i have to say how i would mount the permenant magnets and i think that all i have to do is mount them on a non ferromagnetic material is this correct?

thanks for any help and pointers its much appriciated!

Are you using

\mathrm B \ = \ \frac{B_0}{V_0}V

?

Also can't you hold the magnets using a clamp?

Reply 5

A clamp would draw magnetic flux line into it. This is because a clamp has a high permerability (if it happens to be made of iron, which I would bet it was).
Your plan seems to be ok. You would probabily want to place the two permenent magnets lined up to each other so it produces a dipole and a near uniform field between them. You are right in that you would want to mount the magnets on non-magnetic surfaces. Ideally you would want to find some matterial that has a relative permability as close to 1 as possible. A good example is aluminium, which is a paramagnetic material. I would suggest that the hall probe be attached to a DC current source (a type of power supply that limits the voltage so that a constant current is flowing through the terminals).
By using helmholtz coils you should be able to produce a known flux density by altering the current. Use this to produce a response graph of voltage out vs flux density of the hall probe. Then you replace the helmholtz coils with the magnets, measure the voltage out of the probe, look up the flux density from the graph, and hey presto you know how strong the magnets are.
Hints: The more current that is flowing through the probe the larger the output voltage, however the current will generate heat, which would change the charactoristics of the probe (the Vout is related to the velocity of the charge carriers, increasing the temperature will cause more charge carriers to be liberated if the probe is a semiconductor, thus reducing the velocity of the charge carriers). Keep the probe at a constant temperature.

Since I am bored out of my skull I decided to do the intergration for you on the Biot Bavart Law for a pair of solenoids arranged as Helmholtz coils
B = 2mju0 I / sqrt(10)pi r
Where mju0 is 4pi x 10^-7Hm^-1
pi is erm ...3.141blah blah blah
r is the radius of the Helmholtz coils and the seperation
I is the current flowing through the coils
n is the number of windings of each coil.

This was from the intergration (All bolds are vectors)

dB = mju0 I (dL v a)/4pi |a|^3
a^2 = r^2 + (r/2)^2 (a being the seperation between the wire dL and the point in space we are measuring the flux at)
a = r sqrt(5)/2
Using symetry we can reduce the intergration down to just dL . univector r .cos45
=dLcos45
leaving the intergration being dB = [cos45mju0 I/2sqrt(5)pi r^2] dL
dL is replaced by r sintheta dtheta
Intergrate the whole thing over 2pi and 0
B = mju0 I / sqrt(10)pi r
There are two coils so
B = 2mju0 I / sqrt(10)pi r
Then we multiply by n for each current loop there is...easy...

Reply 6

Mehh

By using helmholtz coils you should be able to produce a known flux density by altering the current. Use this to produce a response graph of voltage out vs flux density of the hall probe. Then you replace the helmholtz coils with the magnets, measure the voltage out of the probe, look up the flux density from the graph, and hey presto you know how strong the magnets are.
Hints: The more current that is flowing through the probe the larger the output voltage, however the current will generate heat, which would change the charactoristics of the probe (the Vout is related to the velocity of the charge carriers, increasing the temperature will cause more charge carriers to be liberated if the probe is a semiconductor, thus reducing the velocity of the charge carriers). Keep the probe at a constant temperature.

Since I am bored out of my skull I decided to do the intergration for you on the Biot Bavart Law for a pair of solenoids arranged as Helmholtz coils
B = 2mju0 I / sqrt(10)pi r
Where mju0 is 4pi x 10^-7Hm^-1
pi is erm ...3.141blah blah blah
r is the radius of the Helmholtz coils and the seperation
I is the current flowing through the coils
n is the number of windings of each coil.

..

What about using a simple straight wire, with known current to find magnetic flux density, to calibrate the probe, using that equation?

Reply 7

o i see someones doing the same plannign exercise as me. :smile:

i need help with it, the things easy except i dnt get what the calibration of the hall probe is, using helmholtz coils? this produces a uniform magnetic field but i dnt see how this helps

Reply 8

Simple reason for this a simple straght wire does not produce a uniform magnetic field, thus there will be different field strengths at different parts of the probe. Second don't you think that a simple wire is going to produce an insanely small magnetic field?! Remember there is a background magnetic field to worry about (we usually call it Earth's magnetic field, read about it), your magnetic field should be a few orders of magnitude higher than this field (placing the equipment perpendicular to it might help things too, try a verticle experiment :biggrin:

).
The magnetic field drops off as an inverse square law, this means that at large distances your field is really ickle, at small distances, small deviations cause large fluxtuations in the flux density, in short, it'll be a biach to get a good reading on the distance...

Reply 9

Mehh

Simple reason for this a simple straght wire does not produce a uniform magnetic field, thus there will be different field strengths at different parts of the probe. Second don't you think that a simple wire is going to produce an insanely small magnetic field?! Remember there is a background magnetic field to worry about (we usually call it Earth's magnetic field, read about it), your magnetic field should be a few orders of magnitude higher than this field (placing the equipment perpendicular to it might help things too, try a verticle experiment :biggrin:

).
The magnetic field drops off as an inverse square law, this means that at large distances your field is really ickle, at small distances, small deviations cause large fluxtuations in the flux density, in short, it'll be a biach to get a good reading on the distance...

Ah that makes sense, thanks.

Reply 10

Mehh

Simple reason for this a simple straght wire does not produce a uniform magnetic field, thus there will be different field strengths at different parts of the probe. Second don't you think that a simple wire is going to produce an insanely small magnetic field?! Remember there is a background magnetic field to worry about (we usually call it Earth's magnetic field, read about it), your magnetic field should be a few orders of magnitude higher than this field (placing the equipment perpendicular to it might help things too, try a verticle experiment :biggrin:

).
The magnetic field drops off as an inverse square law, this means that at large distances your field is really ickle, at small distances, small deviations cause large fluxtuations in the flux density, in short, it'll be a biach to get a good reading on the distance...

sry but that really doesn't make any sense at all, am i being dumb lol?

Reply 11

Arr I found a different formula:
http://en.wikipedia.org/wiki/Helmholtz_coils

Reply 12

endeavour

Arr I found a different formula:
http://en.wikipedia.org/wiki/Helmholtz_coils

Well done...Your right

dB = mju0 I (dL v a)/4pi |a|^3
a^2 = r^2 + (r/2)^2 (a being the seperation between the wire dL and the point in space we are measuring the flux at)
a = r sqrt(5)/2
Using symetry we can reduce the intergration down to just dL . univector r .cos45
=dLcos45
leaving the intergration being dB = [cos45mju0 I/2sqrt(5)pi r^2] dL
dL is replaced by r (sin)theta dtheta [The sine part was the problem]
Intergrate the whole thing over 2pi and 0
B = mju0 I / sqrt(10) r
There are two coils so
B = 2mju0 I / sqrt(10) r

Reply 13

westerman16

sry but that really doesn't make any sense at all, am i being dumb lol?

Erm...nooo
It sounds like you haven't don't ALevel maths.
Okay...So Draw yourself a graph of the function y=x^-2. Then take a look at it. You really want to have a large magnetic field right? This means that the size of the voltage generated by the hall probe is large, thus easy to measure. Also because you need a field that is about the same strength as the magnet. Now look at where the field is function is large (y is large)...when x is close to zero right? Now imagine measuring that distance x. Now imagine there are little error bars for your measurement, say a 1mm each way. Now mark those error bars beside your chosen x. You will find that the closer your value of x is to zero the larger the difference is between the two error bars. Which gets you back to the first problem, you have large errors.
Of course large distances give a very small field strength which means the voltage is very small hence large errors.
Hence using a single piece of wire is a really crappy way to produce a magnetic field. Plus to generate a measurable magnetic field you would probabily kill whatever power supply you are using.

Reply 14

Mehh

Erm...nooo
It sounds like you haven't don't ALevel maths.
Okay...So Draw yourself a graph of the function y=x^-2. Then take a look at it. You really want to have a large magnetic field right? This means that the size of the voltage generated by the hall probe is large, thus easy to measure. Also because you need a field that is about the same strength as the magnet. Now look at where the field is function is large (y is large)...when x is close to zero right? Now imagine measuring that distance x. Now imagine there are little error bars for your measurement, say a 1mm each way. Now mark those error bars beside your chosen x. You will find that the closer your value of x is to zero the larger the difference is between the two error bars. Which gets you back to the first problem, you have large errors.
Of course large distances give a very small field strength which means the voltage is very small hence large errors.
Hence using a single piece of wire is a really crappy way to produce a magnetic field. Plus to generate a measurable magnetic field you would probabily kill whatever power supply you are using.

i sorta get it but still dont understand how thats going to calibrate a hall probe for the specific purpose we need it to do.

surely you can put the hall probe into a bar magnets field to calibrate it?

PLUS i do a level maths :smile:

and on an A atm

Reply 15

also got any links to websites that could help me on probe calibration? or magnetic flux density?

Reply 16

westerman16

i sorta get it but still dont understand how thats going to calibrate a hall probe for the specific purpose we need it to do.

surely you can put the hall probe into a bar magnets field to calibrate it?

PLUS i do a level maths :smile:

and on an A atm

But you need to know what the magnetic field strength is and compare this to the Hall pd you get out.

Reply 17

Mehh you posted this on another thread:

Mehh

I am assuming that you wish to callibrate a magnetic probe so you can work out the magnetic flux density in a space using the hall effect. If that is so then it seems to be right that you want to work out the current flowing through the helmholtz coils, which the ammeter would do, also the variable dc supply would allow you to vary the current. Ideally you would want a current source dc supply.
Also have you found out the flux density to current ratio of the helmholtz coils? Else you would be calibrating to an arbitary unit.

I assume, having calculated B values using:

Unparseable latex formula:

\Large B \ = \ (\frac{4}{5})^{{\fs{1}3/2}}.\frac{\mu_0nI}{R}

you will have say 5 values for B with their corresponding hall pd (

V_h

) values which you can then plot the graph and find the gradient.
Then this will be the ratio

\frac{B_0}{V_0}

, then you can use

B \ = \ \frac{B_0}{V_0}V_h

?

Reply 18

mullally87

I am using a the uniform magnetic field in the core of a toroid (circular solenoid) for calibration. I think you need to keep the hall probe in the same orientation throughout the invesigation - parallel to the earth's magnetic field so that it has no effect. I also think you need to keep temperature constant, but am struggling -i would appreciate it if someone would clarify this.

Reply 19