You have seen that materials like iron, steel etc are strongly attracted by external magnetic field, while wood, aluminium etc. do not show any attraction. What is the reason for these types of behaviour? If we observe a piece of magnetic material on an atomic scale we can see that the electrons orbiting around the nuclei and spinning about their own axes cause tiny currents. These current loops are so small that we can consider them as magnetic dipoles for practical purposes. Ordinarily they cancel each other because of the random orientation of atoms. But when a magnetic field is applied, a net alignment of these magnetic dipoles occurs, and the medium becomes magnetically polarized or magnetized. The net magnetic dipole moment developed per unit volume of a material is called intensity of magnetization or simply magnetization M (Its unit is A/m).
Different types of materials show different types of magnetization. Some materials acquire a magnetization parallel to the external applied magnetic field while some others opposite to it. There are substances which retain their magnetization ever after the external field has been removed (permanent magnets)
Consider a long solenoid of n turns per unit length and carrying current I. The magnetic field in the interior of the solenoid is
B0=mu0nI
The magnetic intensity H is a quantity related to currents in coils and conductors. In this case it is defined as
H= B0/mu0
H=nI
The magnetic intensity is a vector with dimensions of L-1A. Its unit is Am-1.
If we use another material such as iron as the core of the solenoid, keeping the current I constant, the magnetic field inside will be different from B0. This is because the core gets magnetized in the magnetic field produced by the current carrying coil, and this magnetized core in turn produces its own magnetic field. Let this magnetic material possess a dipole moment m. We define a relevant quantity called the magnetisation M which is equal to the magnetic moment per unit volume (V).
M=m/V
M is a vector with dimensions L-1A and unit Am-1. Thus M and H have the same units.
We can write the resultant field as B=B0+Bm
where Bm is the magnetic field produced due to the magnetization of the core. But Bm=mu0M.
Hence the total magnetic field in the material is B=mu0H+mu0M
B=mu0(H+M)
We have partitioned the contribution to the total magnetic field inside the solenoid into two parts: one, due to external factors such as the current in the solenoid, represented by H and the other is due to the specific nature of the magnetic material represented by M. The M can be influenced by external factors, which can be mathematically expressed as
M directly proportional to H or M=psi H
where psi is a dimensionless constant called the magnetic susceptibility. It is a measure of how a magnetic material responds to an external field. It is large and positive for materials which are called ferromagnetic. It is small and positive for materials which are called paramagnetic. It is small and negative for materials which are termed diamagnetic.
We obtain B=mu0(1+psi)H
B=mu0murH
where mur =(1+psi) is a dimensionless quantity called the relative permeability of the material inside the solenoid. For vaccum mur=1, psi=0
We can write mu0mur=mu, the absolute permeability of the medium
Therefore B=muH.
Different types of materials show different types of magnetization. Some materials acquire a magnetization parallel to the external applied magnetic field while some others opposite to it. There are substances which retain their magnetization ever after the external field has been removed (permanent magnets)
Consider a long solenoid of n turns per unit length and carrying current I. The magnetic field in the interior of the solenoid is
B0=mu0nI
The magnetic intensity H is a quantity related to currents in coils and conductors. In this case it is defined as
H= B0/mu0
H=nI
The magnetic intensity is a vector with dimensions of L-1A. Its unit is Am-1.
If we use another material such as iron as the core of the solenoid, keeping the current I constant, the magnetic field inside will be different from B0. This is because the core gets magnetized in the magnetic field produced by the current carrying coil, and this magnetized core in turn produces its own magnetic field. Let this magnetic material possess a dipole moment m. We define a relevant quantity called the magnetisation M which is equal to the magnetic moment per unit volume (V).
M=m/V
M is a vector with dimensions L-1A and unit Am-1. Thus M and H have the same units.
We can write the resultant field as B=B0+Bm
where Bm is the magnetic field produced due to the magnetization of the core. But Bm=mu0M.
Hence the total magnetic field in the material is B=mu0H+mu0M
B=mu0(H+M)
We have partitioned the contribution to the total magnetic field inside the solenoid into two parts: one, due to external factors such as the current in the solenoid, represented by H and the other is due to the specific nature of the magnetic material represented by M. The M can be influenced by external factors, which can be mathematically expressed as
M directly proportional to H or M=psi H
where psi is a dimensionless constant called the magnetic susceptibility. It is a measure of how a magnetic material responds to an external field. It is large and positive for materials which are called ferromagnetic. It is small and positive for materials which are called paramagnetic. It is small and negative for materials which are termed diamagnetic.
We obtain B=mu0(1+psi)H
B=mu0murH
where mur =(1+psi) is a dimensionless quantity called the relative permeability of the material inside the solenoid. For vaccum mur=1, psi=0
We can write mu0mur=mu, the absolute permeability of the medium
Therefore B=muH.
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