Magnetization of the core of a long solenoid

  • Thread starter prodo123
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  • #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?
 

Answers and Replies

  • #2
TSny
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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##.
 
  • #3
17
4
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}##
 
  • #4
17
4
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}##
 
  • #5
TSny
Homework Helper
Gold Member
12,956
3,313
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.
 
  • #6
17
4
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.
 

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