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.

In this article, we shall study to derive an expression for magnetic induction and magnetic potential at any point in a magnetic field created by a bar magnet.

Magnetic Induction at Any Point Due to a Short Bar Magnet:

Consider a short magnetic dipole NS.  Let  be the magnetic moment of the dipole

M = m x 2l ………………(1)

The direction of magnetic induction is along the axis from S-pole to N-pole inside the magnet.

Consider a point ‘P’ near the dipole at distance ‘r’ from its centre O. i.e. OP = r Let ‘ θ’ be the angle between the line joining the point from the centre O and the axis of the dipole (angle between OP and SN). Resolving magnetic moment into two mutually perpendicular components, we have,  the component M Cosθ along OP and M Sinθ perpendicular to OP.

Now, the point P lies on the axis of M Cosθ. Hence, the magnetic induction at, the axis point of M Cos θ is given by

Also, the given point P lies on the equatorial-line of component M Sin θ. Hence, the magnetic induction at the equatorial point of M Sin θ is given by

Let B₁ and B₂ be represented by sides PQ and PT of completed parallelogram PQRT. The diagonal PR represents the resultant magnetic induction in magnitude and direction.

This is the magnitude of the resultant induction B at point P.

Let ∝ be the angle made by the resultant B with the direction of OP

This is the angle made by B with OP.  Hence, the total inclination of the resultant induction B with the axis of the dipole is  ( θ + ∝ )

Special cases:

Case 1: If P is a point on the axis of the dipole, then θ = 0° or θ = 180° and Cos θ = ± 1

Case – 2: If P is a point on the equator of the dipole, then θ = 90° and Cos θ = 0

Magnetic Potential at Any Point Due to a Short Bar Magnet:

The magnetic potential at a point in a magnetic field is defined as the work done in moving unit north pole from infinity to that point. It is denoted by ‘V’ and its S.I. unit is J/Am or Wb/m.

In free space, the magnetic potential at a point due to the magnetic pole of strength ‘m’ units and at a distance, r is given by

Consider a short magnetic dipole NS.  Let M be the magnetic moment of the dipole

M = m x 2l ………………(1)

The direction of M is along the axis from S-pole to N-pole.

Consider a point ‘P’ near the dipole at distance ‘r’ from its centre O. i.e. OP = r. Let ‘θ’ be the angle between the line joining the point from the centre O and the axis of the dipole (angle between OP and SN).

Now the magnetic potential due to the North Pole of a magnetic dipole is given by

The magnetic potential due to the North Pole of a magnetic dipole is given by

Since the magnetic potential is a scalar quantity, the resultant potential at a point P is given by

Case 1: If P is a point on the axis of the dipole, then θ = 0° or θ = 180° and Cos θ = ± 1

Case – 2: If P is a point on the equator of the dipole, then θ = 90° and Cos θ = 0

Hence V_equator = 0

Science > Physics > Magnetism > Magnetic Induction and Magnetic Potential at any Point

The post Magnetic Induction and Potential at any Point appeared first on The Fact Factor.

In this article, we shall study the force acting on poles and torque action on bar magnet suspended in a uniform magnetic field.

Force Between Two Magnetic Poles (Inverse Square law):

If two poles of strengths m₁ and m₂ separated by a distance ‘r’ from each other in a vacuum, then the force between the two magnetic poles is given by

Problems on Inverse Square Law

Example – 01:

The force between two magnetic poles in the air is 9.604 mN. If one pole is ten times stronger than other, calculate the pole strength of each magnet. The distance between the poles of two magnets is 0.1 m.

Given: Force = F = 9.604 mN = 9.604 x 10^-3 N, distance between poles = r = 0.1 m, Relation between pole strength m₂ = 10 m₁, μ_o/4π = 10^-7 Wb/Am.

To Find: pole strengths = ?

Solution:

m₂ = 10 m₁ = 10 x 9.8 = 98 Am

Ans: The pole strengths of magnets are 9.8 Am and 98 Am.

Example – 02:

What is the force of repulsion between two magnetic poles of strengths 1.6 Am and 7.2 Am separated by a distance 0.06 m in a vacuum?

Given: pole strengths m₁ = 1.6 Am and m₂ = 7.2 Am, distance between poles = r = 0.06 m, μ_o/4π = 10^-7 Wb/Am.

To Find: The force between poles = F = ?

Solution:

Ans: Force Between Poles is 3.2 x 10^-4 N

Force Experienced by a Pole of Magnet in Uniform Magnetic Field:

If we try to break the bar magnet at the centre and separate the poles, new poles are formed at the broken ends. It means it is not possible for us to separate the poles and study them individually hence the magnetic circuit is studied in analogy with the electrical circuit. Analogous to the quantity of electrical field intensity we have similar quantity magnetic field intensity. It is also referred as magnetic induction. The electrical field intensity at a point is defined as the force experienced by a unit charge kept at that point.

By analogy, we can define magnetic induction at a point as the force experienced by a pole of unit strength kept at that point. Let the strength of the magnetic field be B.

Thus the N-pole is acted upon by a force of magnitude mB and the south pole is acted upon by a force of magnitude mB, in the opposite direction to that on the north pole.

Torque Acting on Bar Magnet in a Uniform Magnetic Field:

Let us consider a bar magnet of pole strength, ‘m’ and magnetic length ‘2l’ suspended in a magnetic field of induction B such that it is free to rotate about a transverse axis passing through its centre. Let θ be the angle between the axis of the bar magnet and the direction of the magnetic field.

The magnetic dipole moment of a bar magnet is given by

M = m . 2l …………… (1)

Now, each pole of the bar magnet is acted upon by a force whose magnitude is given by

F = mB …………… (2)

The force acting on the north pole is equal to the force acting on the south pole but they act in the opposite direction. Similarly, the lines of action of these two forces are different, Hence they form what is called a couple.

The magnitude of the couple or moment of force is given by

Torque ( τ ) = Force (F) . Perpendicular distance between the forces

∴ τ  =    F  . SP

∴ τ  =    F  . 2l sin θ          ……………….   (3)

From equations (2) and (3)

τ =    m B . 2l sin θ

∴ τ  =    (m . 2l ) . B  sin θ   ………………..(4)

From equations (1) and (4)

τ =    M B sin θ

This is an expression for the torque acting on a magnet kept in the magnetic field. The direction of the torque is perpendicular to the plane passing through .i.e. perpendicular to the plane of the paper.

Tangent Law:

Statement:

If a bar magnet is free to rotate about an axis at right angles to two mutually perpendicular uniform magnetic fields of inductions B₁ and B₂, then it comes to rest in a direction making an angle q with the direction of , such that

B₂ = B₁ . tan θ

Proof:

It is found that when a bar magnet is suspended in two cross magnetic fields, it comes to rest with its axis along the direction of the resultant of the two magnetic fields. Consider a bar magnet suspended in two cross magnetic fields B₁ and B₂ .

Under the action of these two magnetic fields the needle rotates through angle θ and come to rest along the resultant of these two magnetic fields.

In the equilibrium condition, the needle is acted upon by two torques. one due to the magnetic field of induction given by m B₂ .2l. cosθ and second due to the horizontal component of the earth given by m B₁ .2l. sin θ. Under the action of these two torques, the needle remains in equilibrium. Hence the two couples should be equal.

∴  m B₂ . 2l . cos θ  = m B₁  . 2l. sin θ

∴    B₂ . cos θ   = B₁  . sin θ

B₂ = B₁ . (sin θ / cos θ)

B₂ = B₁ . tan θ

This relation is called as the tangent law.

Numerical Problems on Torque Acting on Magnet:

Example – 03:

A torque of moment 25 x 10^-2 Nm acts on a magnet suspended in a uniform magnetic field of induction 0.5 Wb/m² when making an angle of 30° with the field. Find the magnetic dipole moment of the magnet.

Given: Torque = τ = 25 x 10^-2 Nm, Magnetic induction = B = 0.5 Wb/m², angle with field = θ = 30°.

To find: Magnetic moment = M = ?

Solution:

τ = MB sin θ

∴ M = τ/ Bsin θ

∴ M = 25 x 10^-2/ (0.5 x sin 30°) = 25 x 10^-2/ (0.5 x 0.5)

∴ M = 25 x 10^-2/ (0.25) = 1 Am²

Ans: The magnetic dipole moment of the magnet is 1 Am²

Example – 04:

A magnet of magnetic dipole moment 2 Am² is deflected through 30° from the direction of a magnetic field of induction 2 Wb/m². Find the magnitude of the torque or couple.

Given: Magnetic moment = M = 2 Am², Magnetic induction = B = 2 Wb/m², angle with field = θ = 30°.

To find: Torque = τ =?

Solution:

τ = MB sin θ

∴ τ = 2 x 2 x sin 30° = 2 x 2 x 0.5 = 2 Nm

Ans: The torque acting on the magnet is 2 Nm

Example – 05:

A bar magnet of dipole moment 7.5 Am² experiences a torque of 1.5 x 10^-4 Nm, when placed inclined at 30° in a uniform magnetic field. Find the magnetic induction of the field.

Given: Magnetic moment = M = 7.5 Am², , angle with field = θ = 30°, Torque = τ = 1.5 x 10^-4 Nm

To find: Magnetic induction = B =?

Solution:

τ = MB sin θ

∴ B = τ/ Msin θ

∴ B = 1.5 x 10^-4/ (7.5 x sin 30°) = 1.5 x 10^-4/ (7.5 x 0.5)

∴ B = 4 x 10^-5 Wb/m²

Ans: The magnetic induction is  is 4 x 10^-5 Wb/m²  or 4 x 10^-5 T

Example – 06:

A magnet of dipole moment 0.05 Am² is suspended so to move freely in a horizontal plane. Find the couple required to hold it at right angles to the Earth’s horizontal magnetic induction of 0.32 x 10^-4 Wb/m².

Given: Magnetic moment = M = 0.05 Am², Magnetic induction = B = 0.32 x 10^-4 Wb/m², angle with field = θ = 90°.

To find: Torque = τ =?

Solution:

τ = MB sin θ

∴ τ = 0.05 x 0.32 x 10^-4 x sin 90°

∴ τ = 0.05 x 0.32 x 10^-4 x 1 = 1.6 x 10^-6 Nm

Ans: The torque acting on magnet is 1.6 x 10^-6 Nm

Example – 07:

Calculate the dipole moment of magnet which when placed at right angles to the earth’s horizontal magnetic induction 2 x 10^-5 Wb/m² experiences a couple of 2 x 10^-5 Nm.

Given: Torque = τ = 2 x 10^-5 Nm, Magnetic induction = B = 2 x 10^-5  Wb/m², angle with field = θ = 90°.

To find: Magnetic moment = M = ?

Solution:

τ = MB sin θ

∴ M = τ/ Bsin θ

∴ M = 2 x 10^-5/ (2 x 10^-5 x sin 90°) = 2 x 10^-5/ (2 x 10^-5 x 1)

∴ M = 1 Am²

Ans: The magnetic moment of the magnet is 1 Am²

Example – 08:

A magnet of moment 0.6 Am² makes an angle of 30° with the magnetic meridian. Calculate the torque that tends to bring  the magnet back to the meridian given that the earth’s horizontal field is 3.2 x 10^-5 Wb/m².

Given: Magnetic moment = M = 0.6 Am², Magnetic induction = B = 3.2 x 10^-5 Wb/m², angle with field = θ = 90°.

To find: Torque = τ =?

Solution:

τ = MB sin θ

∴ τ = 0.6 x 3.2 x 10^-5 x sin 30°

∴ τ = 0.6 x 3.2 x 10^-5 x 0.5 = 9.6 x 10^-6 Nm

Ans: The torque required is 9.6 x 10^-6 Nm

Example – 09:

A torque of 5 x 10^-3  Nm is required to hold a freely suspended horizontal bar magnet with the axis at an angle of 60° to uniform horizontal field of induction 3 x 10^-3 Wb/m². Find the magnetic moment of the magnet.

Given: Torque = τ = 5 x 10^-3 Nm, Magnetic induction = B = 3 x 10^-3  Wb/m², angle with field = θ = 60°.

To find: Magnetic moment = M = ?

Solution:

τ = MB sin θ

∴ M = τ/ Bsin θ

∴ M = 5 x 10^-3/ (3 x 10^-3 x sin 60°) 

∴ M = 5 x 10^-3/ (3 x 10^-3 x 0.8660)

∴ M = 1.925 Am²

Ans: The magnetic moment of the magnet is 1.925 Am²

Example – 10:

A short bar magnet placed with its axis at 30° to a uniform magnetic field of 0.2 T experiences a torque of 0.06 Nm. Calculate the magnetic moment of the magnet and find out what orientation of the magnet corresponds to stable equilibrium in magnetic field.

Given: Torque = τ = 0.06Nm, Magnetic induction =0.2 T, angle with field = θ = 30°.

To find: Magnetic moment = M = ? θ = ? for stable equilibrium.

Solution:

τ = MB sin θ

∴ M = τ/ Bsin θ

∴ M = 0.06/ (0.2 x sin 30°) = 0.06/ (0.2 x 0.5)

∴ M = 0.6 Am²

For stable equilibrium torque acting on magnet is zero.

τ = MB sin θ

∴  0 = MB sin θ

∴  sin θ = 0

∴  θ = 0°

Ans: The magnetic moment of the magnet is 0.6 Am² and for stable equilibrium θ = 0°

i.e. the axis of magnet should be parallel to the magnetic field.

Example – 11:

A bar magnet of moment  0.9 JT^-1 experiences a torque of 0.063 Nm, when placed inclined at 30° in a uniform magnetic field. Find the magnetic induction of the field.

Given: Magnetic moment = M = 0.9 JT^-1, angle with field = θ = 30°, Torque = τ = 0.063 Nm

To find: Magnetic induction = B =?

Solution:

τ = MB sin θ

∴ B = τ/ Msin θ

∴ B = 0.063/ (0.9 x sin 30°) = 0.063/ (0.9 x 0.5)

∴ B = 0.14 T

Ans: The magnetic induction is  is 0.14 T.

Example – 12:

A magnet of dipole moment 6.4 Am² and length 5 x 10^-3 m makes an angle of 60° with the magnetic field of 0.4 T. calculate the torque acting on it. Also find the pole strength of the magnet.

Given: Magnetic moment = M = 6.4 Am², Magnetic induction = B = 0.4 T, angle with field = θ = 60°, magnetic length = 5 x 10^-3 m

To find: Torque = τ =? pole strength = m = ?

Solution:

τ = MB sin θ

∴ τ = 6.4 x 0.4x sin 60° = 6.4 x 0.4x 0.8860 = 2.2 Nm

M = m x magnetic length

∴  m = M/magnetic length = 6.4/5 x 10^-3 = 1.28 x 10³ Am

Ans: The torque required is 1.28 x 10³ Nm

Example – 13:

Two magnets of magnetic moments M and √3M are joined together to form a cross. The combination is suspended freely in a uniform magnetic field. In the equilibrium position the magnet of magnetic moment M makes an angle θ with the field. Determine θ.

Solution:

The magnet of magnetic moment M makes an angle θ with the field.

As the two magnets are forming cross, the magnet of magnetic moment √3M makes an angle (90° – θ) with the field.

Let B be the magnetic strength of the external field.

For equilibrium the torque acting on the two magnets must be equal.

τ₁ = τ₂

∴  MB sin θ =  √3MB sin (90° – θ)

∴  sin θ =  √3 cos θ

∴  tan θ =  √3

∴  θ =  60°

Ans: Hence the value of θ is  60°

Science > Physics > Magnetism > Torque Acting on Bar Magnet in Uniform Magnetic Field

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In this article, we shall study the terminology of a bar magnet and the concept of the magnetic dipole moment of a magnet.

Bar Magnet:

A bar magnet is a rectangular parallelepiped body which exhibits magnetic properties. When a bar magnet is suspended in the air such that it is free to rotate about the transverse axis passing through its centre, then it is found that the bar magnet always aligns itself in the north-south direction. The end of the magnet which is pointing towards the geographical north is called north-seeking pole or simply north pole, while the end of the magnet pointing towards the geographical south is called south seeking pole or simply south pole. From the behaviour of a bar magnet, we can say that earth itself is behaving like a magnet. The magnetic north pole of the earth is at geographical south pole while the magnetic south pole of the earth is at the geographical north pole.

 A Magnetic Monopole Does Not Exist:

A bar magnet is said to have two poles located at the two ends of the magnet. If we try to break the bar magnet at the centre and separate the poles, new poles are formed at the broken ends. Thus two new magnets are formed each having two opposite poles at their ends. If the magnet is broken down into very small pieces further each piece will be a magnet with two poles. It means it is not possible for us to separate the poles. Thus we can say that a magnetic monopole does not exist.

Fictitious Poles:

It is not possible to separate the two poles (the south pole and the north pole) by breaking the magnet into two parts. Similarly, it is not possible to locate the position of the poles of the magnet. Hence a magnetic dipole is supposed to be made up of two fictitious or imaginary poles. The position of these fictitious poles is found using a compass needle. And it is found that they are not located exactly at the ends of the magnet but slightly inside.

• Geometric Length of magnet: The length of the edge parallel to the magnetic axis is called the geometric length of the bar magnet.
• Magnetic Length of Magnet: The distance between the poles of a bar magnet is called magnetic length.

Magnetic length of bar magnet × 1.2 = Geometric length of the bar magnet.

• Axis of Magnet: The line joining the poles of the bar magnet to called an axis of the magnet.

• Equator of Magnet: The perpendicular drawn to the magnetic axis through the centre of magnet is called the equator of the magnet.

Need of Analogy of Magnetic Circuit with Electric Circuit:

If we try to break the bar magnet at the centre and separate the poles, new poles are formed at the broken ends. It means it is not possible for us to separate the poles and study them individually hence the magnetic circuit is studied in analogy with the electrical circuit. Formulae & concepts are derived from an electrical circuit and by analogy formula for a magnetic circuit is written.

Magnetic Induction at a Point in Magnetic Field:

The magnetic intensity or magnetic induction at any point in the magnetic field is equal to the number of tubes of force passing through the unit area of a small surface element drawn at that point.

B = ∅ / A

Where B = magnetic intensity or intensity of the magnetic field

or magnetic induction.

∅ = Magnetic flux.  Its unit as weber (Wb)

A = Area through which magnetic flux pass.

Unit of Magnetic induction B is Wb/m² or tesla (T)

 Magnetic Dipole:

A magnetic dipole can be defined as two equal and opposite magnetic poles separated by a finite distance. A magnetic dipole consists of two equal and opposite magnetic charges having pole strength +m & -m separated by finite distance ‘2l’.

Magnetic Dipole Moment:

The magnetic dipole moment is defined as the product of the pole strength and the magnetic length of a magnet.

M = m × 2l

The magnetic dipole moment is a vector quantity. Its direction is from -m to + m i.e. from the south pole to the north pole. SI unit of Magnetic dipole moment is ampere. metre²  i.e. Am²

A current-carrying coil behaves like a magnet.

Magnetic dipole moment of current-carrying coil = NiA

Where N = Number of turns of coil

I= Current through the coil

A = Area of the coil

Numerical Problems on Bar Magnet:

Example – 01:

A bar magnet of geometric length 18 cm has pole strength 100 Am. Find the magnetic dipole moment of the bar magnet.

Given: Geometric length of bar magnet = 18 cm = 0.18 m, Pole strength = m = 100 Am

To Find: Magnetic dipole moment = M = ?

Solution:

Magnetic length = (5/6) x Geometric length = (5/6) x 0.18 = 0.15 cm

M = m x magnetic length

∴  M = 100 x 0.15 = 15Am²

Ans: The magnetic dipole moment of bar magnet is 15 Am².

Example – 02:

A bar magnet of magnetic length 0.1 m has pole strength of 10 Am. What is the dipole moment of this magnet?

Given: Magnetic length of bar magnet = 0.1 m, Pole strength = m = 10 Am

To Find: Magnetic dipole moment = M = ?

Solution:

M = m x magnetic length

∴  M = 10 x 0.1 = 1Am²

Ans: The magnetic dipole moment of bar magnet is 1 Am².

Example – 03:

A bar magnet of the magnetic moment 5 Am² has poles 0.2 m apart. Calculate the pole strength.

Given: Magnetic moment = 5 Am², magnetic length of bar magnet = 0.2 m,

To Find: pole strength = m = ?

Solution:

M = m x magnetic length

∴ m = M/ magnetic length = 5/0.2 = 25 Am

Ans: The pole strength of bar magnet is 25 Am.

Example – 04:

A bar magnet has pole strength 48 Am which are 0.25 m apart. What is the magnetic moment of the magnet?

Given: Magnetic length of bar magnet = 0.25 m, Pole strength = m = 48 Am

To Find: Magnetic dipole moment = M = ?

Solution:

M = m x magnetic length

∴  M = 48 x 0.25 = 12Am²

Ans: The magnetic dipole moment of bar magnet is 12 Am².

Problems on Magnetic Dipole Moment of Current-Carrying Coil:

Example – 05:

A circular coil of 20 turns and radius 10 cm has a magnetic dipole moment of 3.142 Am². What is the current flowing through the coil.

Given: Number of turns = n = 20, radius of coil = 10 cm = 0. 1 m, magnetic dipole moment = M = 3.142 Am².

To Find: Current through the coil = i = ?

Solution:

M = n I A

∴   i = M/nA = M/ (n x π r² )

∴   i = 3.142/ (20 x 3.142 x (0.1)² ) = 5 A

Ans: The current through the coil is 5 A.

Example – 06:

A circular coil of 50 turns and 10 cm radius carries a current of 2 A. Find the magnetic moment of the coil.

Given: Number of turns = n = 50, radius of coil = 10 cm = 0. 1 m, Current through coil = i = 2 A.

To Find: Magnetic dipole moment = M = ?

Solution:

M = n i A = n i (π r²)

∴  M =  50 x 2 x 3.142 x (0.1)² = 3.142 Am²

Ans: The magnetic dipole moment is 3.142 Am².

Example – 07:

A circular coil has 1000 turns each of area 2 m². If a current of 3 mA flows through the coil, find the magnetic moment of the coil.

Given: Number of turns = n = 1000, Area of coil = A = 2 m², Current through coil = i = 3 mA = 3 x 10^-3 A

To Find: Magnetic dipole moment = M = ?

Solution:

M = n i A

∴  M =  1000 x 3 x 10^-3 x 2 = 6Am²

Ans: The magnetic dipole moment is 6 Am².

Example – 08:

A rectangular coil of length 8 cm and breadth 5 cm has 200 turns of insulated wire. Find the magnetic dipole moment of the coil. When a current of 2 A flows through it.

Given: Number of turns = n = 200, Area of coil = A = 8 cm x 5 cm = 40 cm² = 40 x 10^-4 m², Current through coil = i = 2 A

To Find: Magnetic dipole moment = M = ?

Solution:

M = n i A

∴  M =  200 x 40 x 10^-4 x 2 = 1.6Am²

Ans: The magnetic dipole moment is 1.6 Am².

Example – 09:

A steel wire of length ‘l‘ has a magnetic moment M. If the wire is bent in a semicircular arc, what would its moment be?

Solution:

Let m be the pole strength

Now, M = m x magnetic length

∴  m = M/l

Curved length of semicircle = half circumference

∴   l = π r

∴ r = l /π

Now new magnetic dipole moment

M = m x new magnetic length

∴ M = (M/l)x 2r = (M/l)x 2 x (l /π) = 2M/π

Ans: New magnetic moment is 2M/π

Previous Subject: Types of Magnetic Materials

Science > Physics > Magnetism > Magnetic Dipole Moment of a Magnet

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