In this article, we shall study problems on current-carrying solenoid and current-carrying coil suspended in a uniform magnetic field.

Example – 01:

A solenoid has a core of material of relative permeability 4000. The number of turns is 1000 per metre. A current of 2 A flows through the solenoid. Find magnetic intensity, the magnetic field in core, magnetization, magnetic current and susceptibility.

Given: Relative permeability = μ_r = 4000, Number of turns per metre = n = 1000, Current flowing = i = 2A, μ_o = 4π x 10^-7 Wb/Am.

To Find:magnetic intensity = H = ?, the magnetic field in core = B = ?, magnetization = M_Z = ?, magnetic current = I_m =?, and susceptibility = χ = ?

Solution:

Magnetic intensity H = n i = 1000 x 2 = 2000 A/m

Intensity of magnetic field B = μ H = μ_r μ₀ H = 4000 x 4π x 10^-7 x 2000 = 10 T

Magnetization M_Z = (μ_r – 1)H = (4000 – 1) x 2000 = 3999 x 2000 = 7.998 x 10⁶ A/m

We have B = μ_r μ₀ (i + I_m)

10 = 4000 x 4π x 10^-7 x (2 + I_m)

(2 + I_m) = 10 / (4000 x 4 x 3.142 x 10^-7)

2 + I_m = 1989

Magnetic current I_m = 1987 A

We have μ_r = 1 + χ

∴ Susceptibility = χ = μ_r – 1 =  4000 – 1 = 3999

Example – 02:

A solenoid has 1000 turns and is 20 cm long. Find the magnetic induction produced at the centre of the solenoid by the current of 2 A. What is the flux at this point if the diameter of solenoid is 4 cm?

Given: Number of turns = N = 1000, Length  = l = 20 cm = 0.2 m, Current through solenoid = 2 A, diameter = 4 cm, radius of solenoid = 4/2 = 2 cm = 0.02 m, μ_o = 4π x 10^-7 Wb/Am.

To Find: magnetic induction = B = ?, magnetic flux = Φ = ?

Solution:

n = N/l = 1000/0.2 = 5000 turns per metre

B = μ H = μ n i = 4π x 10^-7 x 5000 x 2 = 4 x 3.142 x 10^-7 x 5000 x 2 = 0.01256 T

Now B = Φ/A

∴  Φ = B x A = B x πr² =  0.01256 x 3.142 x (0.02)²

∴  Φ = 0.01256 x 3.142 x 4 x 10^-4 =  1.58 x 10^-5 Wb

Ans: Magnetic induction produced = 0.01256 T and magnetic flux = 1.58 x 10^-5Wb

Example – 03:

A closely wound solenoid is 1 m long and has 5 layers of windings, each winding being of 500 turns. If the average diameter of the solenoid is 3 cm and it carries a current of 4 A, find the magnetic field at a point well within the solenoid.

Given: Number of turns = N =  500 x 5 = 2500, Length of solenoid = l = 1 m, Current through solenoid = 4 A, diameter = 3 cm, radius of solenoid = 3/2 = 1.5 cm = 0.015 m, μ_o = 4π x 10^-7 Wb/Am.

To Find: magnetic induction = B = ?

Solution:

n = N/l = 2500/1 = 2500 turns per metre

B = μ H = μ n i = 4π x 10^-7 x 2500 x 4 = 4 x 3.142 x 10^-7 x 2500 x 4 = 1.256 x 10^-2 T

B = 0.01256 T

Ans: Magnetic induction produced = 0.01256 T

Example – 04:

A solenoid 0.5 m long has a four layer winding of 300 turns each. What current must pass through it to produce a magnetic field of induction 2.1 x 10^-2 T at the centre.

Given: Number of turns = N =  300 x 4 = 1200, Length of solenoid = l = 0.5 m, Magnetic induction = B = 2.1 x 10^-2 T,  μ_o = 4π x 10^-7 Wb/Am.

To Find: Current through solenoid = i = ?

Solution:

n = N/l = 1200/0.5 = 2400 turns per metre

B = μ H = μ n i

∴  i = B/μ n = (2.1 x 10^-2)/(4π x 10^-7 x 2400) =  (2.1 x 10^-2)/(4 x 3.142 x 10^-7 x 2400)

∴  i = 6.96 A

Ans: Current through solenoid is 6.96 A.

Example – 05:

A solenoid (π/2) m long has a two-layer winding of 500 turns each and has radius 5 cm. What is the magnetic induction at the centre when it carries a current of 5 A.

Given: Number of turns = N =  500 x 2 = 1000, Length of solenoid = l = (π/2) m, Magnetic induction, Current through solenoid = i = 5 A.

To Find: Magnetic induction = B = ?

Solution:

n = N/l = 1000/(π/2) = (2000/π) turns per metre

B = μ H = μ n i = 4π x 10^-7 x (2000/π) x 5 = 4 x 10^-3 T

Ans: Magnetic induction produced = 4 x 10^-3 T

Example – 06:

A circular coil of 3000 turns per 0.6 m length carries a current of 1 A. What is the magnitude of magnetic induction (magnetic flux)?

Given: Number of turns = n = 3000 per 0.6 m = 3000/0.6 = 5000 per metre, current through the coil = 1 A.

To Find: Magnetic induction = B = ?

Solution:

B = μ H = μ n i = 4π x 10^-7 x 5000 x 1 = 4 x 3.142 x 10^-7 x 5000 x 1 = 6.284 x 10^-3 T

Ans: Magnetic induction produced = 6.284 x 10^-3 T or 6.284 x 10^-3 Wb/m²

Example – 07:

A solenoid of 20 turns per cm has a radius of 3 cm and is 40 cm long. Find the magnetic moment when it carries a current of 9.5 A.

Given: Number of turns = n = 20 per cm = 20/0.01 = 2000 per metre, radius of solenoid = 3 cm = 0.03 m, length of solenoid = l = 40 cm = 0.4 m, current through solenoid = 9.5 A.

To Find: Magnetic moment = M = ?

Solution:

Total turn of solenoid = N = n x l = 2000 x 0.4 = 800

M = N i A =  N i πr² = 800 x 9.5 x 3.142 x (0.03)² = 800 x 9.5 x 3.142 x 9 x 10^-4 = 21.5 Am²

Ans: Magnetic moment is 21.5 Am²

Example – 08:

A circular coil of 300 turns and diameter 14 cm carries a current of 15 A. What is the magnitude of magnetic moment associated with the coil?

Given: Number of turns = N = 300, diameter of coil = 14 cm, radius of coil = 14/2 = 7 cm = 0.07 m, current through the coil = 15 A.

To Find: Magnetic moment = M = ?

Solution:

M = N i A =  N i πr² = 300 x 15 x 3.142 x (0.07)² = 300 x 15 x 3.142 x 49 x 10^-4 = 69.28 Am²

Ans: the magnetic moment is 69.28 Am²

Example – 09:

A closely wound solenoid of 1000 turns and area 4.2 cm² carries a current of 3 A. It is suspended so as to move freely in horizontal plane in a horizontal magnetic field of  6 x 10^-2 T. Find the magnetic moment, torque acting on the solenoid when the axis of solenoid makes an angle of 30° with the external horizontal field.

Given: Number of turns = N = 1000, Area of cross-section of solenoid = A = 4.2 cm² = 4.2 x 10^-4 m², current through solenoid = 3 A, external magnetic field = B = 6 x 10^-2 T, Angle made by solenoid axis with field = θ =30° ,

To Find: Magnetic moment = M = ?

Solution:

M = N i A =  1000 x 3 x 4.2 x 10^-4 = 1.26 Am²

Now τ = MB sin θ = 1.26 x 6 x 10^-2 x sin 30° =  1.26 x 6 x 10^-2 x 0.5 = 0.0378 Nm

Ans: The magnetic moment is 1.26 Am²  and torque acting = 0.0378 Nm

Example – 10:

A closely wound solenoid of 1000 turns and area 2 x 10^-4 m² carries a current of 1 A. It is placed in horizontal axis at 30° with the direction of uniform magnetic field of 0.16 T. Calculate the magnetic moment of solenoid and torque experienced by it in the field.

Given: Number of turns = N = 1000, Area of cross-section of solenoid = A = 2 x 10^-4 m², current through solenoid = 1 A, external magnetic field = B = 0.16 T, Angle made by solenoid axis with field = θ =30° ,

To Find: Magnetic moment = M = ?

Solution:

M = N i A =  1000 x 1 x 2 x 10^-4 = 0.2 Am²

Now τ = MB sin θ = 0.2 x 0.16x sin 30° =  0.2 x 0.16x 0.5 = 0.016 Nm

Ans: The magnetic moment is 0.2 Am²  and torque acting = 0.016 Nm

Example – 11:

A closely wound solenoid of 2000 turns and area 1.6 x 10^-4 m² carries a current of 4 A. It is in equilibrium with horizontal axis at 30° with the direction of uniform magnetic field of 7.5 x 10^-2 T. Calculate the magnetic moment of the solenoid and also find the force and torque experienced by it in the field.

Given: Number of turns = N = 2000, Area of cross-section of solenoid = A = 1.6 x 10^-4 m², current through solenoid = 4 A, external magnetic field = B = 7.5 x 10^-2  T, Angle made by solenoid axis with field = θ =30° ,

To Find: Magnetic moment = M = ?

Solution:

M = N i A =  2000 x 4 x 1.6 x 10^-4 = 1.28 Am²

As the solenoid is in equilibrium position no force acts on it.

Now τ = MB sin θ = 1.28 x 7.5 x 10^-2  x sin 30° = 1.28 x 7.5 x 10^-2  x 0.5 = 0.048 Nm

Ans: The magnetic moment is 1.28 Am², force = 0, and torque acting = 0.048 Nm

Science > Physics > Magnetism > Numerical Problems on Current-Carrying Solenoids

The post Numerical Problems on Current-Carrying Solenoid appeared first on The Fact Factor.

In this article, we shall study problems to calculate magnetization, magnetic susceptibility, magnetic permeability, etc.

μ = B / H

μ_r  = B / B₀

Example – 01:

Find the magnetization of the 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 = M_Z = ?

Solution:

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

M_Z = 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 the 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 = M_Z = ?

Solution:

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

M_Z = 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 = M_Z = ?

Solution:

Density = Mass / Volume

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

M_Z = 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 the magnetization of the magnet.

Given: Mass of magnet = m = 6.6 x 10^-3 kg, Density of material of magnet = 7.9 x 10³  kg/m³, Magnetic moment = M =  2.5 Am².

To Find: Magnetization = M_Z = ?

Solution:

Density = Mass / Volume

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

M_Z = 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_o = 4π x 10^-7 Wb/Am.

To Find: Relative permeability = μ_r = ?, susceptibility = χ = ?

Solution:

μ_r = B/(μ_oH) = 1.8/(4π x 10^-7 x 1000) =1.8/(4 x 3.142 x 10^-4) = 1.432 x 10³  = 1432

We have μ_r = 1 + χ

∴  χ = μ_r – 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_o = 4π x 10^-7 Wb/Am.

To Find: Relative permeability = μ_r = ?, susceptibility = χ = ?

Solution:

μ_r = B/(μ_oH) = 1.6/(4π x 10^-7 x 1000) =1.6/(4 x 3.142 x 10^-4) = 1.273 x 10³  = 1273

We have μ_r = 1 + χ

∴  χ = μ_r – 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_o = 4π x 10^-7 Wb/Am.

To Find: permeability = μ = ?, relative permeability = μ_r = ?, susceptibility = χ = ?

Solution:

μ = B/H = 1./900 = 1.44 x 10^-3

μ_r = B/(μ_oH) = 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 μ_r = 1 + χ

∴  χ = μ_r – 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_o = 4π x 10^-7 Wb/Am.

To Find: permeability at saturation = μ = ?

Solution:

μ = μ_o ( 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_o = 4π x 10^-7 Wb/Am.

To Find: permeability at saturation = μ = ?

Solution:

μ = μ_o ( 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_o = 4π x 10^-7 Wb/Am.

To Find: permeability = μ = ?, Magnetic flux per unit area = B = ?

Solution:

μ = μ_o ( 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 =  0.904T

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

∴  χ₁/χ₂ = T₂/T₁

∴  T₂ = (χ₁/χ₂) x T₁

∴  T₂ = (1.2 x 10^-5/1.8 x 10^-5) x 300 = = (2/3) x 300 = 200 K

Ans: At temperature 200 K, magnetic 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

∴  χ₁/χ₂ = T₂/T₁

∴  T₂ = (χ₁/χ₂) x T₁

∴  T₂ = (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

∴  χ₁/χ₂ = T₂/T₁

∴  χ₂ = (T₁/T₂) x χ₁

∴  χ₂ = (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 magnetic susceptibility is 1.2 x 10^-5.

Science > Physics > Magnetism > Numerical Problems on Magnetic Susceptibility

The post Numerical Problems on Magnetic Susceptibility appeared first on The Fact Factor.

In this article, we shall study the origin of magnetism, magnetic intensity, magnetization, and magnetic susceptibility.

Magnetic Field Due to Current-Carrying Coil:

The magnetic induction at a point on the axis at a distance of ‘x’ from the centre of a circular coil of radius ‘a’ is given by

It is directed along the axis of the coil and away from it and perpendicular to the plane of the coil. For x >>a, we can neglect a² from denominator in the expression.

This is an expression for magnetic induction due to current carrying loop

Now, the electric intensity due to electric dipole on its axis is given by

From the two equations (3) and (4) we can say that the magnetic dipole moment is analogous to electrostatic dipole moment P and the magnetic field is analogous to the electric field. Thus the planar current loop is analogous to a magnetic dipole. i.e. current-carrying loop produces a magnetic field and behaves like a magnetic dipole.

Dipole Moment of Revolving Electron:

The origin of magnetism in substances can be explained by considering the circular motion of electrons. The negatively charged electrons in atoms move in circular orbits around the positively charged nucleus which are equivalent to a circular coil carrying current.

The period of revolution of electron

The direction of this magnetic moment is into the plane of the paper.

If m_e is the mass of the electron, multiplying and dividing numerator and denominator of R.H.S.

L_o is the angular momentum of the electron which is coming out of the plane of the paper. The quantities e, m_e are constant, hence the magnetic dipole moment of the electron is directly proportional to its angular momentum. The angular momentum of electron and magnetic dipole moment are opposite to each other. In vector form

The ratio of magnetic dipole moment to the angular momentum of the revolving electron is constant and is called the gyromagnetic ratio.

Gyromagnetic ratio is also called the magnetogyric ratio.

The orbital motion of electrons gives rise to an orbital magnetic moment. In addition, the electrons spin about its own axis constituting a spin magnetic moment.  The resultant magnetic moment of an atom is the vector sum of the orbital and spin magnetic moment.

Origin of Magnetism on the Basis of Circulating Charges:

The origin of magnetism in substances can be explained by considering the circular motion of electrons. The electrons in atoms move in circular orbits around the nucleus which are equivalent to a circular coil carrying current.

The orbital motion of electrons gives rise to an orbital magnetic moment. In addition, the electrons spin about its own axis constituting a spin magnetic moment.  The resultant magnetic moment of an atom is the vector sum of the orbital and spin magnetic moment. These small magnets are called elementary or atomic magnets. When the material is not magnetized, these elementary magnets form closed chains thereby annualizing each other’s effect. When the material is magnetized, these elementary magnets are aligned in the same direction.

On the basis of magnetic properties, substances are classified into three groups namely diamagnetic, paramagnetic and ferromagnetic.

The Atomic theory of magnetism explains following facts.

• Single poles cannot exist. Poles always exist in a pair.
• The magnetic poles of a magnet are of equal strength.
• When a magnet is heated, the thermal energy of atomic magnet increases. Due to which they again form closed chain and
  magnetism is lost.

Magnetization of Ferromagnetic Material using Rowland Ring (Toroid With Iron Core):

The magnetization of a ferromagnetic material such as iron can be studied with an arrangement called Toroid with an iron core.

Let the toroid have ‘n’ number of turns per unit length and ‘I’ be the current through it. The magnitude of magnetic field inside the coil when iron core is not present is given by

B₀ = μ₀ n I

When an iron core is present in the toroid, the magnetic field increases, which is given by

B = B₀  +  B_M    ……… (1)

Where B_M is magnetic field contributed by the iron core.

It is found that BM is directly proportional to magnetization of iron and is given by

B_M = μ₀ M_z        ……. (2)

The strength of magnetic field at a point can be given in terms of vector quantity called magnetic intensity (H).

Thus  B₀ = μ₀ H  ……… (3)

Where, H = nI. Unit of magnetic intensity is A/m and its dimensions are [L^-1M⁰T⁰I¹].

Substituting values of equations (2) and (3) in (1)

B = μ₀ H + μ₀ M_z 

B = μ₀ (H + M_z )  ………….. (4)

Magnetization can be expressed in terms of magnetic intensity as

M_z   = χ H

Where χ (chi) is called the magnetic susceptibility.

Substituting in equation (4)

B = μ₀ (H + χ H)

∴   B = μ₀ (1 + χ ) H

The quantity (1 + χ )  is called relative magnetic permeability and is denoted by μ_r. It is a dimensionless quantity

∴   B = μ₀ μ_r H  = μ H

Note: 

μ_r is called relative magnetic permeability of the substance. μis called absolute magnetic permeability of the substance. μ₀ is called magnetic permeability of free space.

Terminology:

Magnetization:

The net magnetic dipole moment per unit volume is called as the magnetization of a sample. It is denoted by M_z. It is a vector quantity. S.I. unit of magnetization is A/m and its dimensions are [AL^-1]. By definition, magnetization.

Magnetic Intensity:

Magnetic intensity is a quantity used in describing the magnetic phenomenon in terms of their magnetic fields. The strength of magnetic field at a point can be given in terms of vector quantity called magnetic intensity (H). S.I. unit of magnetization is A/m and its dimensions are [AL^-1]. By definition, magnetic intensity

Magnetic susceptibility:

The ratio of the intensity of magnetization to magnetic intensity is called magnetic susceptibility. Magnetic susceptibility is dimensionless and unitless quantity. By definition,

Magnetic Permeability:

The ratio of the magnitude of the total field inside the material to that of the magnetic intensity of the magnetizing field is called magnetic permeability. S.I. unit of magnetic permeability is H/m. Its dimensions are [L¹M¹T^-2I^-2]. By definition  

μ = B / H

Relative Permeability:

The ratio of the magnitude of the total field inside the material to that of magnetizing field is called relative permeability. Relative permeability is dimensionless and unitless quantity. By definition, 

μ_r  = B / B₀

Curie’s Law of Magnetization:

The magnetization of a paramagnetic sample is directly proportional to the external magnetic field and inversely proportional to the absolute temperature.

Science > Physics > Magnetism > Magnetization and Magnetic Intensity

The post Magnetization and Magnetic Intensity appeared first on The Fact Factor.