- #1
prodo123
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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?