Magnetization of the core of a long solenoid

[prodo123]

Homework Statement

A long solenoid of 60 turns/cm carries a current of 0.15 A. It wraps a steel core with relative permeability $\mu_r=5200$. Find the magnitude of the magnetization of the core.

Homework Equations

$N=\lambda L$
$\chi = \mu_r-1$
$\mu = \mu_r\mu_0$
$\vec{M}=\chi\vec{H}$
where $\vec{H}$ is the external magnetic field applied to the core and $\vec{M}$ is the magnetization of the core. This equation never showed up on the textbook for some reason.
$\vec{B_m}=\mu_0\vec{M}$
where $\vec{B_m}$ is the additional magnetic field in the core induced by the external field $\vec{H}$.Constants for the problem:
$\lambda=6000$ (60 turns/cm = 6000 turns/m)
$I=0.15$

The Attempt at a Solution

The magnetic field induced by the solenoid is the external magnetic field $\vec{H}$ applied to the core.
Ampere's Law finds the induced magnetic field of the solenoid along the center axis as:
$\int \vec{H}\cdot d\vec{l} = \mu_0 N I$
$HL=\mu_0 \lambda L I$
$H=\mu_0 \lambda I$
$H = 1.13 \text{ mT}$

$\chi=\mu_r-1=\frac{M}{H}$
$M = H(\mu_r-1)$
$M = 5.88\text{ A/m}$

The textbook has a completely different answer:
$M = \mu_r \lambda I$
$M = 4.68\text{ MA/m}$

which I assume took the following steps:
$H=\mu_0 \lambda I$
$B_m=\mu\lambda I = \mu_0 M$
$M = \frac{\mu}{\mu_0}\lambda I$
$M = \mu_r \lambda I$

Which is the right approach and why?

 

[TSny]

Ampere's Law finds the induced magnetic field of the solenoid along the center axis as:
$\int \vec{H}\cdot d\vec{l} = \mu_0 N I$

Hello. There should not be a factor of $\mu_0$ on the right side of the above equation.

Also, note $M = H(\mu_r - 1) \approx H \mu_r$.

 

[prodo123]

Hello. There should not be a factor of $\mu_0$ on the right side of the above equation.

Sorry for the confusion, all the fields are B-fields, let me revise...maybe that's the issue?

$\int \vec{B}\cdot d\vec{l}=\mu_0 N I$
$BL=\mu_0 \lambda L I$
$B=\mu_0 L I$
$B=1.13\text {mT}$

 

[prodo123]

Hello. There should not be a factor of $\mu_0$ on the right side of the above equation.

Also, note $M = H(\mu_r - 1) \approx H \mu_r$.

The textbook doesn't discuss at all H-fields, so if I read what's online correctly,
Para- and diamagnetism have B- and H-fields proportional such that $\vec{B} = \mu \vec{H}$. The external field $\vec{B}$ is therefore equal to $\mu_0 \vec{H}$.
$M=\chi H = \chi\frac{B}{\mu_0}$
$M = (\mu_r-1)\frac{B}{\mu_0}\approx\mu_r\frac{B}{\mu_0}$
$M=4.68\text{ MA/m}$

 

[TSny]

The textbook doesn't discuss at all H-fields, so if I read what's online correctly,
Para- and diamagnetism have B- and H-fields proportional such that $\vec{B} = \mu \vec{H}$. The external field $\vec{B}$ is therefore equal to $\mu_0 \vec{H}$.

OK. By "external field" you mean just the part of $\vec{B}$ that is due to the current in the winding of the solenoid.

$M=\chi H = \chi\frac{B}{\mu_0}$
$M = (\mu_r-1)\frac{B}{\mu_0}\approx\mu_r\frac{B}{\mu_0}$
$M=4.68\text{ MA/m}$

OK. In the first two lines here, $\vec{B}$ is the "external" field.

 

[prodo123]

OK. By "external field" you mean just the part of $\vec{B}$ that is due to the current in the winding of the solenoid.
OK. In the above three lines, $\vec{B}$ is the "external" field.

Yes, the external field $\vec{B}$ in the equations is the B-field due to the current in the coils only.