# Magnetic susceptibility and relative permeability

1. Feb 21, 2014

### Marwa

1. The problem statement, all variables and given/known data
find the derivation of
χm=μr-1
where Xm is the magnetic susceptibility and μ is the relative permeability.

2. Relevant equations
M=Xm.H
M is the magnetization,H is the magnetic field
μr=μ/μo
where μo is the permeability of a vacuum,

3. The attempt at a solution
I don't know how to start ,I know some equations related to magnetic properties but i cannot connect them to each other to find the derivation.

2. Feb 21, 2014

### collinsmark

Hello Marwa,

Welcome to Physics Forums!

For convenience, I've retyped in your problem.

Derive the formula,
$$\chi_m = \mu_r - 1$$

Relevant equations:
$$\mathrm{\textbf{M}} = \chi_m \mathrm{\textbf{H}}$$
$$\mu = \mu_r \mu_0$$
I'm pretty sure you will need one or two more relevant equations.
Can you find any relevant equations that relate to the magnetic field, $\mathrm{\textbf{B}}$?
.
.
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3. Feb 22, 2014

### Marwa

Relevant equations:
β=μH

βo=μoH

μr=μ/μo

β = μoH+ μoM

M=Xm.H

4. Feb 22, 2014

### collinsmark

Okay, with these relevant equations,
$$\mathrm{\textbf{M}} = \chi_m \mathrm{\textbf{H}}$$
$$\mu = \mu_r \mu_0$$
$$\mathrm{\textbf{B}} = \mu \mathrm{\textbf{H}}$$
$$\mathrm{\textbf{B}} = \mu_0 \left( \mathrm{\textbf{H}} +\mathrm{\textbf{M}} \right)$$
you have enough to show that
$$\chi_m = \mu_r - 1.$$
What have you tried so far?

5. Feb 22, 2014

### Marwa

I have tried :

M=Xm.H
Xm=M/H
Xm=Mμ/B
as H=B/μ
and B=μo(H+M)
Xm=(M.μ)/μo(H+M)
so Xm=μr.M/(H+M)

6. Feb 22, 2014

### collinsmark

Try substituting
M = χmH
into
B=μ0(H+M)​
Then factor the H.

7. Feb 22, 2014

### Marwa

I got it:

B=μo(H+M)
and M=XmH
B=μo(H+XmH)
B=μoH(1+Xm)
so μH=μoH(1+Xm)
μ/μo=μr
Xm=1-μr

Thank you.

8. Jan 2, 2017

DERIVE THE RELATIONSHIP B = μ0(H + M)