# Magnetization of the core of a long solenoid

• prodo123
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.

## 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?

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##.

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}##

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}##

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.

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.