Numerical Problems on Magnetization, Susceptibility

Physics Chemistry  Biology  Mathematics
Science > Physics > Magnetism >You are Here

 

Example – 01:

  • Find the magnetization of bar magnet of length 10 cm and cross-sectional area 3 cm2. The magnetic moment of the magnet is 1 Am2.
  • Given: Length of magnet = l = 10 cm, cross-sectional area = A = 3 cm2, Magnetic moment = M =  1 Am2.
  • To Find: Magnetization = MZ = ?
  • Solution:

Volume of bar magnet = V = length x cross-sectional area = 10 x 3 = 30 cm3 = 30 x 10-6 m3.

MZ = M/V = 1/ (30 x 10-6) = 3.33 x 104 A/m.

Ans: Magnetization of the bar magnet is 3.33 x 104 A/m.2



Example – 02:

  • Find the magnetization of bar magnet of length 5 cm and cross-sectional area 2 cm2. The magnetic moment of the magnet is 1 Am2.
  • Given: Length of magnet = l = 5 cm, cross-sectional area = A = 2 cm2, Magnetic moment = M =  1 Am2.
  • To Find: Magnetization = MZ = ?
  • Solution:

Volume of bar magnet = V = length x cross-sectional area = 5 x 2 = 10 cm3 = 10 x 10-6 m2.

MZ = M/V = 1/ (10 x 10-6) = 1 x 105 A/m.

Ans: Magnetization of the bar magnet is 1 x 105 A/m.

Example – 03:

  • Find the magnetization of bar magnet of mass 180 g, the density of material 8 g/cm3 and the magnetic moment 3.5 Am2.
  • Given: Mass of magnet = m = 180 g, Density of material of magnet = 8 g/cm3, Magnetic moment = M =  3.5 Am2.
  • To Find: Magnetization = MZ = ?
  • Solution:

Density = Mass / Volume



∴  Volume = Mass/density = 180/8 = 22.5 cm3 = 22.5 x 10-6 m2.

MZ = M/V = 3.5/ (22.5 x 10-6) = 1.56 x 105 A/m.

Ans: Magnetization of the bar magnet is 1.56 x 105 A/m.

Example – 04:

  • A bar magnet made of steel has a magnetic moment of 2.5 Am2 and a mass of 6.6 x 10-3 kg. If the density of steel is 7.9 x 103 kg/m3, find the intensity of magnetization of the magnet.
  • Given: Mass of magnet = m = 6.6 x 10-3 kg, Density of material of magnet = 7.9 x 10 kg/m3, Magnetic moment = M =  2.5 Am2.
  • To Find: Magnetization = MZ = ?
  • Solution:

Density = Mass / Volume

∴  Volume = Mass/density = (6.6 x 10-3)/(7.9 x 103) = 8.354 x 10-7 m2.



MZ = M/V = 2.5/ (8.354 x 10-7) = 3.0 x 106 A/m.

Ans: Magnetization of the bar magnet is 3.0 x 106 A/m.

Example – 05:

  • Magnetic field and magnetic intensity are respectively 1.8 T and 1000 A/m. Find relative permeability and susceptibility.
  • Given: Magnetic field = B = 1.8 T, magnetic intensity = H = 1000 A/m, μ= 4π x 10-7 Wb/Am.
  • To Find: Relative permeability = μr = ?, susceptibility = χ = ?
  • Solution:

μr = B/(μoH) = 1.8/(4π x 10-7 x 1000) =1.8/(4 x 3.142 x 10-4) = 1.432 x 103  = 1432

We have μr = 1 + χ

∴  χ = μ– 1 =  1432 – 1 = 1431



Ans: Relative permeability is 1432 and susceptibility is 1431.

Example – 06:

  • Magnetic field and magnetic intensity are respectively 1.6 T and 1000 A/m. Find relative permeability and susceptibility.
  • Given: Magnetic field = B = 1.6 T, magnetic intensity = H = 1000 A/m, μ= 4π x 10-7 Wb/Am.
  • To Find: Relative permeability = μr = ?, susceptibility = χ = ?
  • Solution:

μr = B/(μoH) = 1.6/(4π x 10-7 x 1000) =1.6/(4 x 3.142 x 10-4) = 1.273 x 103  = 1273

We have μr = 1 + χ

∴  χ = μ– 1 =  1273 – 1 = 1272

Ans: Relative permeability is 1273 and susceptibility is 1272.



Example – 07:

  • Magnetic field and magnetic intensity are respectively 1.3 T and 900 A/m. Find permeability, relative permeability, and susceptibility.
  • Given: Magnetic field = B = 1.3 T, magnetic intensity = H = 900 A/m, μ= 4π x 10-7 Wb/Am.
  • To Find: permeability = μ = ?, relative permeability = μr = ?, susceptibility = χ = ?
  • Solution:

μ = B/H = 1./900 = 1.44 x 10-3

μr = B/(μoH) = 1.3/(4π x 10-7 x 900) =1.3/(4 x 3.142 x 10-7 x 900) = 1.149 x 103  = 1149

We have μr = 1 + χ

∴  χ = μ– 1 =  1149 – 1 = 1148



Ans: Permeability is 1.44 x 10-3, relative permeability is 1149 and susceptibility is 1148.

Example – 08:

  • The susceptibility of annealed iron at saturation is 5500. Find the permeability of annealed iron at saturation.
  • Given: susceptibility = χ = 5500,  μ= 4π x 10-7 Wb/Am.
  • To Find: permeability at saturation = μ = ?
  • Solution:

μ = μo ( 1 + χ)

∴  μ = 4π x 10-7  ( 1 + 5500) =  4 x 3.142 x 10-7 x 5501 = 6.9 x 10-3 H/m.

Ans: Permeability at saturation is  6.9 x 10-3 H/m.

Example – 09:

  • The susceptibility of magnetic material at saturation is 4000. Find its permeability at saturation.
  • Given: susceptibility = χ = 4000,  μ= 4π x 10-7 Wb/Am.
  • To Find: permeability at saturation = μ = ?
  • Solution:

μ = μo ( 1 + χ)

∴  μ = 4π x 10-7  ( 1 + 4000) =  4 x 3.142 x 10-7 x 4001 = 5.028 x 10-3 H/m.



Ans: Permeability at saturation is  5.028 x 10-3 H/m.

Example – 10:

  • An iron rod is subjected to a magnetizing field of 1200 A/m. The susceptibility of iron is 599. Find the permeability and the magnetic flux per unit area produced.
  • Given: Magnetizing field = H = 1200 A/m, susceptibility = χ = 599,  μ= 4π x 10-7 Wb/Am.
  • To Find: permeability = μ = ?, Magnetic flux per unit area = B = ?
  • Solution:

μ = μo ( 1 + χ)

∴  μ = 4π x 10-7  ( 1 + 599) =  4 x 3.142 x 10-7 x 600 = 7.536 x 10-4 H/m.

Now μ = B/H



∴ B = μ H =  7.536 x 10-4 x 1200 =  0.904 T

Ans: Permeability is  7.536 x 10-4 H/m and magnetic flux per unit area = 0.904 T

Example – 11:

  • The susceptibility of magnesium at 300 K is 1.2 x 10-5. At what temperature will the susceptibility increases to 1.8 x 10-5.
  • Given: Initial temperature = T1 = 300K, Initial susceptibility = χ1 = 1.2 x 10-5, Final susceptibility = χ2 = 1.8 x 10-5.
  • To Find: Final temperature = T2 = ?

We have χ ∝ 1/T

∴  χ12 = T2/T1

∴  T= (χ12) x T1

∴  T= (1.2 x 10-5/1.8 x 10-5) x 300 = = (2/3) x 300 = 200 K



Ans: At temperature 200 K susceptibility increases to 1.8 x 10-5.

Example – 12:

  • The susceptibility of magnesium at 400 K is 1.5 x 10-5. At what temperature will the susceptibility increases to 1.8 x 10-5.
  • Given: Initial temperature = T1 = 400K, Initial susceptibility = χ1 = 1.5 x 10-5, Final susceptibility = χ2 = 1.8 x 10-5.
  • To Find: Final temperature = T2 = ?

We have χ ∝ 1/T

∴  χ12 = T2/T1

∴  T= (χ12) x T1

∴  T= (1.5 x 10-5/1.8 x 10-5) x 400 = = (5/6) x 400 = 333.33 K

Ans: At temperature 333.33 K susceptibility increases to 1.8 x 10-5.

Example – 13:

  • The susceptibility of magnetic material at 250 K is 1.44 x 10-5. At what will the value of susceptibility at 300 K.
  • Given: Initial temperature = T1 = 250 K, Initial susceptibility = χ1 = 1.44 x 10-5,  1.8 x 10-5, Final temperature = T2 = 300 K
  • To Find: Final susceptibility = χ2 = ?
  • Solution:

We have χ ∝ 1/T

∴  χ12 = T2/T1

∴  χ= (T1/T2) x χ1

∴  χ= (250/300) x 1.44 x 10-5 = = (5/6) x 1.44 x 10-5 = 1.2 x 10-5

Ans: At temperature 300 K susceptibility is 1.2 x 10-5.

 

Science > Physics > Magnetism >You are Here
Physics Chemistry  Biology  Mathematics

Leave a Comment

Your email address will not be published. Required fields are marked *