Deriving the magnetic moment of a specimen in a given magnetic field

AI Thread Summary
The discussion revolves around deriving the relationship between the magnetization per unit volume and the magnetic field for a specimen in a magnetic field. The user is attempting to prove that \( \frac{M}{\mu N}= \tanh\left(\frac{1}{kT}\left(\mu \mu_0 H + k \theta \frac{M}{\mu N}\right)\right) \) using the given energy expression. They express confusion over how to equate the two sides and mention an unsuccessful attempt that led to an incorrect conclusion. Additionally, there is a request for guidance on formatting questions using LaTeX. The conversation highlights the challenges in solving the problem and the need for clarity in mathematical representation.
1v1Dota2RightMeow
Messages
75
Reaction score
7

Homework Statement


I'm working on a problem that says that
>If $\varepsilon_{\pm}=\mp (\mu \mu_0 H + k \theta \frac{M}{\mu N} )$ is the energy of the atom of a specimen that can orient itself either parallel or antiparallel in a magnetic field, show that $ \frac{M}{\mu N}= tanh(\frac{1}{kT}( \mu \mu_0 H + k \theta \frac{M}{\mu N}))$.

I'm not really sure how to make the 2 equal.

Homework Equations


$M=\mu Ntanh(x)$
$x=\frac{\mu B}{kT}$

The Attempt at a Solution


I tried setting $\frac{\mu B}{kT}= \frac{1}{kT}(\mu \mu_0 H + k\theta \frac{M}{\mu N})$ but it ended up with $\muB=0$ which can't be true.
 
Physics news on Phys.org
This is just reformatting for you.
1v1Dota2RightMeow said:

Homework Statement


I'm working on a problem that says that
> If ##\varepsilon_{\pm}=\mp (\mu \mu_0 H + k \theta \frac{M}{\mu N} )## is the energy of the atom of a specimen that can orient itself either parallel or antiparallel in a magnetic field, show that ## \frac{M}{\mu N}= tanh(\frac{1}{kT}( \mu \mu_0 H + k \theta \frac{M}{\mu N}))##.

I'm not really sure how to make the 2 equal.

Homework Equations


##M=\mu Ntanh(x)##
##x=\frac{\mu B}{kT}##

The Attempt at a Solution


I tried setting ##\frac{\mu B}{kT}= \frac{1}{kT}(\mu \mu_0 H + k\theta \frac{M}{\mu N})## but it ended up with ## \mu B=0## which can't be true.
 
MisterX said:
This is just reformatting for you.

How do I format my questions like that?
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top