Magnetization and Magnetic Intensity: Definition, Explanation and Formula

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Jasmine Grover

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Magnetization is a vector quantity that measures the density of a permanent or an induced dipole moment in any magnetic field. Magnetization is also known as magnetic polarization.

Growing up we have all played with magnets at some point or another. But have you ever wondered what causes these magnets to be magnetic or why certain materials are drawn towards these metals? And another point to be pondered upon is, why don’t all the materials possess their own magnetic field.

The magnetic behaviour of a magnet is marked by the alignment of the atoms within a substance. When a ferromagnetic substance is brought in contact with a strong external magnetic field, it experiences a torque in which the substance aligns itself in the direction of the magnetic field applied and thus gets strongly magnetized in the direction of the magnetic field. The materials can be classified very easily based on their magnetic properties with the help of magnetization.

Read More: Magnetic Pole

Key Terms: Magnetization, Magnetic fields, Dipole moment, Magnetic dipole, Electron, Atoms, Torque


What causes magnetization?

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Magnetization is caused as a result of a magnetic moment, which in turn is a result of the movement of electrons in an atom or by the spin of electrons. The net magnetization occurs as the response of that material to its external magnetic field combined with the unbalanced magnetic dipole moment that was innate in that material due to the movement of the electrons inside of it. Magnetization helps us in identifying different materials according to their magnetic properties. It is a vector quantity.

Causes of Magnetization

Causes of Magnetization

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What is Magnetization and its Mathematical Formula:

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Magnetization is denoted by M and is the net magnetic moment of any given material per unit volume.

Mathematically, Magnetization can be defined as:

M=mnet/V

Where, mnet is the magnetic moment of the material and V is the volume.

The concept of magnetization can be further explained by taking Solenoid as an example.

Integral Path for Ampere's Law

Integral Path for Ampere’s Law

Let's consider a solenoid with n number of turns per unit length. The current passing through it can be denoted by I. The magnetic field inside the solenoid can be calculated by:

B0= μ0nI

Let’s fill the interior of the solenoid with a material that has non-zero magnetization. Now the magnetic field inside the solenoid is greater than what it was before. The net magnetic field inside the solenoid can be now calculated by:

B=B0+Bm

Where Bm = magnetic field of the core material

Here, Bm is proportional to the magnetization of the material M and it can be mathematically calculated as:

Bm0M

Here, μ0 denotes the constant of permeability of a vacuum.

By introducing another vector field H, we can further this concept. The H represents the magnetic intensity of a given material. Magnetic intensity can be defined as a vector quantity that gives the strength of a magnetic field at any given point. We can calculate the magnetic intensity of a given material by:

H=B/μ0M

The total Magnetic Field of a given material can also be calculated as:

B0(H+M)

Where H= magnetic field due to external factors; (eg. current in the solenoid)

M= magnetic field due to the nature of the core

The nature of the core in the above formula is dependent upon the external forces and can be calculated by:

MH

Here, χ is known as the magnetic susceptibility of the material, which gives the response of any magnetic material to an external field. It is a dimensionless quantity.

The magnetic susceptibility is different for different materials. It is small and positive for paramagnetic materials and small and negative for diamagnetic materials. It can be calculated as:

B0(1+χ)H

0μrH

H

μr= 1 + χ; it is a dimensionless quantity called the relative magnetic permeability of any material. The Magnetic permeability of a substance (depicted by μ) can be defined as:

μ = μ0μr = μ0(1+χ)

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Previous Year Questions 

  1. An electron in motion is associated with… [ KCET 1996]
  2. Two parallel wires carry electric current in same direction. The wires... [ KCET 1996]
  3. Resistance of an ideal ammeter is...[ KCET 1996]
  4. If a current of 0.1 A is passed through the coil, what is the couple acting?...[CBSE Class 12 ]
  5. The acceleration of the falling magnet is..[NEET]
  6. A bar magnet is equivalent to ............[KCET 2004]
  7. A metal ring is held horizontally and bar magnet is dropped through….[NEET]
  8. The total charge, induced in a conducting loop when it is moved in magnetic field depend on...[NEET 1992]
  9. If a charge particle enters perpendicular in the uniform magnetic field, then...[JIPMER 2016]
  10. Who invented the cyclotron?
  11. The ferromagnetic substance is converted into paramagnetic substances … [JIPMER 1996]
  12. The relative permeability is represented by μr … [KCET 2001]
  13. A device which converts electrical energy into mechanical energy … [JKCET 1999]
  14. A circular loop of radius R, carrying current I, lies in x-y plane … [JEE Advance 1999]
  15. When a material is placed in a magnetic field B … [COMEDK UGET]
  16. Susceptibility of a magnetic substance is found to depend on temperature … [JEE Advance 1998]
  17. If a magnetic dipole of moment M situated in the direction of …
  18. If the susceptibility of dia, para and ferro magnetic materials are ….
  19. Materials suitable for permanent magnet, must have which of the following properties ?
  20. At what temperature, the ferromagnetic substances become paramagnetic ?
  21. a magnetic induction of strength at its centre is...

Sample Questions

Ques.  What is magnetic permeability? (1 mark)

Ans. Magnetic permeability is the ratio of magnitude of the total field that is inside of the material to the magnetic intensity of the magnetizing field. The S.I. unit of magnetic permeability is H/m.

Ques. What is magnetic susceptibility? (1 mark)

Ans. Magnetic susceptibility is the ratio of the intensity or the degree of magnetization to the applied magnetic field. It is a dimensionless proportionality constant that has no unit. 400. 

Ques.  A solenoid has a core of a material with relative permeability 400. The windings of the solenoid are insulated from the core and carry a current of 2A. If the number of turns is 1000 per metre, calculate (a) H, (b) M, (c) B and (d) the magnetising current Im . (3 marks)

Ans.(a) The field H is dependent of the material of the core, and is 

H = nI = 1000 × 2.0 = 2 ×103 A/m. 

(b) The magnetic field B is given by 

B = µr µ0 H

 = 400 × 4π ×10–7 (N/A2 ) × 2 × 103 (A/m)

 = 1.0 T 

(c) Magnetisation is given by

 M = (B– µ0 H)/ µ0 

= (µr µ0 H–µ0 H)/µ0 = (µr – 1)H = 399 × H

 ≅ 8 × 105 A/m 

(d) The magnetising current IM is the additional current that needs to be passed through the windings of the solenoid in the absence of the core which would give a B value as in the presence of the core. Thus B = µr n (I + IM ). Using I = 2A, B = 1 T, we get IM = 794 A

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CBSE CLASS XII Related Questions

1.
(a) A circular coil of 30 turns and radius 8.0 cm carrying a current of 6.0 A is suspended vertically in a uniform horizontal magnetic field of magnitude 1.0 T. The field lines make an angle of 60° with the normal of the coil. Calculate the magnitude of the counter torque that must be applied to prevent the coil from turning. 
(b) Would your answer change, if the circular coil in (a) were replaced by a planar coil of some irregular shape that encloses the same area? (All other particulars are also unaltered.)

      2.
      Figure shows tracks of three charged particles in a uniform electrostatic field. Give the signs of the three charges. Which particle has the highest charge to mass ratio?

          3.
          Two charges 5 × 10–8 C and –3 × 10–8 C are located 16 cm apart. At what point(s) on the line joining the to charges is the electric potential zero? Take the potential at infinity to be zero.

              4.

              A series LCR circuit connected to a variable frequency 230 V source. L = 5.0 H, C = 80mF, R = 40 Ω.

              (a) Determine the source frequency which drives the circuit in resonance.

              (b) Obtain the impedance of the circuit and the amplitude of current at the resonating frequency.

              (c) Determine the rms potential drops across the three elements of the circuit. Show that the potential drop across the LC combination is zero at the resonating frequency

                  5.
                  Check that the ratio \(\frac{ke^2}{ Gm_e m_p}\) is dimensionless. Look up a Table of Physical Constants and determine the value of this ratio. What does the ratio signify?

                      6.
                      A series LCR circuit with R = 20 W, L = 1.5 H and C = 35 μF is connected to a variable-frequency 200 V ac supply. When the frequency of the supply equals the natural frequency of the circuit, what is the average power transferred to the circuit in one complete cycle?

                          CBSE CLASS XII Previous Year Papers

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