- #1

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## 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?