# Magnetic Susceptibility experiment using Faraday method

1. Good day,
Using 4-inch pole caps and a Mettler M5 balance, I got all the values required in order to calculate the magnetic susceptibility of several inorganic compounds such as Manganese (IV) oxide (MnO2). I have the change in mass of the tube with the sample of the compound when the magnetic field is on and when it is off. I got the mass of the sample as well as the region where the gradient of the magnetic field (dH^2/dx) (x being the vertical direction) is constant.
I know my values to be right but I am having a lot of trouble converting to cgs units.

2. Fx= m/2 * μ0 * χm * dH^2/dx
where χm is the mass susceptibility of the sample.

3. Lets take MnO2 as an example,
Fx = Δm*g = 0.0517 grams * 9.81
dH^2/dx = 6.29169 (This is the value of the slope of B^2 in Tesla squared and x in meteres
m= 1.1765 grams

This gives mass/specific susceptibility which is incredibly larger than the literature value (on the order of 10^6)
The literature value of the MOLAR susceptibility is 2280x 10^-6 in cgs units

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tiny-tim
Homework Helper
hi maqdah! welcome to pf! (try using the X2 and X2 buttons just above the Reply box )
Fx = Δm*g = 0.0517 grams * 9.81
dH^2/dx = 6.29169 (This is the value of the slope of B^2 in Tesla squared and x in meteres
m= 1.1765 grams
i don't understand why you're converting to cgs, or how exactly you got there if you must have a cgs answer, just find the magnetic susceptibility in SI, then divide by 4π, see the pf library cgs (emu) values:

Some books which give values of susceptibility use cgs (emu) units for electromagnetism.

Although susceptibility has no units, there is still a dimensionless difference between cgs and SI values, a constant, $4\pi$. To convert cgs values to SI, divide by $4\pi$ for electric susceptibility, and multiply by $4\pi$ for magnetic susceptibility.​

hi maqdah! welcome to pf! (try using the X2 and X2 buttons just above the Reply box )

i don't understand why you're converting to cgs, or how exactly you got there if you must have a cgs answer, just find the magnetic susceptibility in SI, then divide by 4π, see the pf library cgs (emu) values:

Some books which give values of susceptibility use cgs (emu) units for electromagnetism.

Although susceptibility has no units, there is still a dimensionless difference between cgs and SI values, a constant, $4\pi$. To convert cgs values to SI, divide by $4\pi$ for electric susceptibility, and multiply by $4\pi$ for magnetic susceptibility.​
It is true that susceptibility has no units but the equation that I am using gives me the mass susceptibility, which has units of length3/mass. Does what you suggest change?
Also, even if you plug in the numbers, I get a value of magnitude of few 103 instead of something close to 10-6 which begs a question, is my equation even correct in the first place?

tiny-tim
Homework Helper
hi maqdah! … the equation that I am using gives me the mass susceptibility, which has units of length3/mass.
eugh! :yuck: i've never heard of mass susceptibility but to convert length3/mass from SI to cgs, you'd multiply by (m/cm)3/(kg/g), = 106/103 = 103

(i haven't followed your calculations, you seem to have mixed SI tesla with cgs grams )

mass susceptibility is just the volumetric susceptibility (the dimensionless quantity) divided by the density of the material. Molar susceptibility is the mass susceptibility multiplied by molar mass.
So now I understood how to convert to cgs units (thanks plenty for that), but I keep getting this feeling that the equation is wrong. Because if you do some dimensional analysis, it just doesn't add up.

tiny-tim
Homework Helper
… I keep getting this feeling that the equation is wrong. Because if you do some dimensional analysis, it just doesn't add up.
yup, i didn't understand your equation …
Fx= m/2 * μ0 * χm * dH^2/dx
… where does it come from? yup, i didn't understand your equation …

… where does it come from? The teacher gave it to me in the lab manual. I even found the same equation in other previous published papers that did the same experiment.

tiny-tim
Homework Helper
hmm …

well, i don't understand why it's H instead of B, i don't understand why H is squared, and i don't understand why there's no charge

can you scan or link to any of those papers? tiny-tim
Homework Helper
ah!

i] there's no m in there

ii] you're confusing the two different types of magnetic field

H is in amp-turns per metre, not tesla …

if you're using tesla, you need B2o, not µoH2

that's a lot of zeros! ah!

i] there's no m in there

ii] you're confusing the two different types of magnetic field

H is in amp-turns per metre, not tesla …

if you're using tesla, you need B2o, not µoH2

that's a lot of zeros! Im not sure I understand. So the equation become Fx = v/(2u0) * χm* dH2/dx?

tiny-tim
Homework Helper
Im not sure I understand. So the equation become Fx = v/(2u0) * χm* dH2/dx?
yes, but with B (in tesla) instead of H

so your original χm is multiplied by µ02 times 103/4π

= 4π 10-11 ~ 10-10 yes, but with B (in tesla) instead of H

so your original χm is multiplied by µ02 times 103/4π

= 4π 10-11 ~ 10-10 and the 103/4π is to convert it to cgs units? If yes I understand everything now :)

Thank you so much for your help. I really appreciate it.

tiny-tim
yes 