Magnetization of the core of a long solenoid

In summary: The M-field is the magnetic field produced by the magnetization of the core due to the external field ##\vec{B}##.In summary, the problem involves finding the magnitude of the magnetization of a steel core wrapped by a solenoid with 60 turns per cm and carrying a current of 0.15 A. Using Ampere's Law, the external magnetic field ##\vec{B}## is found to be 1.13 mT. The textbook provides two different methods for calculating the magnetization, one using the equation ##M = H(\mu_r - 1)## and the other using ##M = \mu_r \lambda I##. Both methods result in different values, 5.
  • #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?
 
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  • #2
prodo123 said:
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
TSny said:
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
TSny said:
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
prodo123 said:
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
TSny said:
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.
 

1. What is a solenoid?

A solenoid is a type of electromagnet that is formed by wrapping a wire into a cylindrical shape with a coil. When an electric current is passed through the wire, a magnetic field is created.

2. How does a solenoid magnetize the core?

A solenoid magnetizes the core by creating a magnetic field within the coil. This field aligns the magnetic domains in the core material in the same direction, causing the core to become magnetized.

3. What is the purpose of magnetizing the core of a solenoid?

The purpose of magnetizing the core of a solenoid is to increase the strength of the magnetic field produced by the solenoid. This allows for more precise control and manipulation of the magnetic field for various applications.

4. How does the length of a solenoid affect its magnetization?

The length of a solenoid affects its magnetization by increasing the strength of the magnetic field as the length increases. This is because there are more turns of wire in a longer solenoid, resulting in a stronger magnetic field.

5. Can the magnetization of a solenoid's core be reversed?

Yes, the magnetization of a solenoid's core can be reversed by changing the direction of the electric current flowing through the wire. This will cause the magnetic field to change direction, resulting in a reversal of the magnetization of the core.

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