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.

Example 01:

A Rowland ring (toroid) of ferromagnetic material of a mean radius 15 cm has 3000 turns of wire wound it. The relative permeability of the material is 1000. What is the magnetic field in a core when a current of 2 A passes through the windings?

Given: mean radius = 15 cm = 0.15 m, Number of turns = 3000, the relative permeability = μ_r = 1000, current through coil = i = 2 A, μ_o = 4π x 10^-7 Wb/Am.

To Find: Magnetic field = B =?

Solution:

n = N/2πr = 3000 / (2π x 0.15)

∴ B = μ_r μ₀ n i = 1000 x 4π x 10^-7 x (3000 / (2π x 0.15) x 2

∴ B = 1000 x 2 x 10^-7 x (3000 /0.15) x 2 = 1000 x 2 x 10^-7 x (20000) x 2

∴ B = 8 T

Ans: Magnetic field = 8 T

Example 02:

A toroidal ring (toroid) of ferromagnetic material of mean radius 15 cm has 3500 turns of wire wound it. The relative permeability of the material is 800. What is the magnetic field in a core when a current of 1.2 A passes through the windings?

Given: mean radius = 15 cm = 0.15 m, Number of turns = 3500, the relative permeability = μ_r = 800, current through coil = i = 1.2 A, μ_o = 4π x 10^-7 Wb/Am.

To Find: Magnetic field = B =?

Solution:

n = N/2πr = 3500 / (2π x 0.15)

∴ B = μ_r μ₀ n i = 800 x 4π x 10^-7 x (3500 / (2π x 0.15) x 1.2

∴ B = 800 x 2 x 10^-7 x (3500 /0.15) x 1.2 = 4.48 T

Ans: Magnetic field = 4.48 T

Example 03:

A Rowland ring of ferromagnetic material has 3000 turns. The inner and outer radii of ring are 11 cm and 12 cm respectively. If a current of 0.7 A is passed through its coils, the magnetic field produced in the core is 2.5 T. What is relative permeability of the core?

Given: mean radius = (11 +12)/2 = 11.5 cm = 0.115 m, Number of turns = 3000, Magnetic field = B = 2.5 T, current through coil = i = 0.7 A, μ_o = 4π x 10^-7 Wb/Am.

To Find:  relative permeability = μ_r =?

Solution:

n = N/2πr = 3000 / (2π x 0.115)

∴ B = μ_r μ₀ n i

∴ 2.5   = μ_r x 4π x 10^-7 x (3000 / (2π x 0.115)) x 0.7

∴ 2.5   = μ_r x 2x 10^-7 x (3000 /0.115) x 0.7

∴ 2.5   = μ_r x 3.652 x 10^-3

∴ μ_r = 2.5/(3.652 x 10^-3) = 6.846 x 10² = 684.6

Ans: The relative permeability = 684.6

Example 04:

A soft iron ring of cross-sectional diameter 8 cm and mean circumference 200 cm has 400 turns of wire wound uniformly around it. Calculate the current necessary to produce magnetic flux of 5 x 10^-4 Wb. Take relative permeability of iron 1800.

Given: Cross-sectional diameter = 8 cm, cross-sectional radius = R = 4 cm = 0.04 m, mean circumference = 2πr = 200 cm = 2 m, Number of turns = 400, Magnetic flux = Φ = 5 x 10^-4 Wb, relative permeability = μ_r = 1800, μ_o = 4π x 10^-7 Wb/Am.

To Find:  current through coil = i =?

Solution:

n = N/2πr = 400 /2 = 200

B = Φ/A = Φ /πR² = (5 x 10^-4) / (3.142 x (0.04)²)

∴ B = ( 5 x 10^-4)/(3.142 x 16 x 10^-4) = 0.09946 T

B = μ_r μ₀ n i

∴ 0.09946   = 1800 x 4π x 10^-7 x 200 x i

∴ 0.09946   = 1800 x 4 x 3.142 x 10^-7 x 200 x i

∴ 0.07812   = 0. 4524 i

∴ i = 0.099462/0. 4524 = 0.22 A

Ans: The current through coil is 0.22 A

Example 05:

A soft iron ring of cross-sectional area 1.5 cm² and mean circumference 30 cm has 240 turns of wire wound uniformly around it. Calculate the relative magnetic permeability if the current necessary to produce magnetic flux of 7.5 x 10^-4 Wb is 2 A.

Given: Cross-sectional area = 1.5 cm² = 1.5 x 10^-4 m², mean circumference = 2πr = 30 cm = 0.3 m, Number of turns = 240, Magnetic flux = Φ = 7.5 x 10^-4 Wb, current through coil = i = 2 A, μ_o = 4π x 10^-7 Wb/Am.

To Find:   relative permeability = μ_r =?

Solution:

n = N/2πr = 240 /0.3= 800

B = Φ/A = (7.5 x 10^-4) / (1.5 x 10^-4) = 5 T

B = μ_r μ₀ n i

∴ 5   = μ_r x 4π x 10^-7 x 800 x 2

∴ 5   = μ_r x 4 x 3.142 x 10^-7 x 800 x 2

∴ 5   = μ_r x 2.01 x 10^-3

∴ μ_r = 5/( 2.01 x 10^-3) = 2488

Ans:  relative permeability is 2488.

Example 06:

The radius of an Amperian loop in a toroid of 2000 turns is 10 cm. If the magnetic induction inside the toroidal space is 0.08 T. What is the magnitude of the current flowing?

Given: mean radius = 10 cm = 0.1 m, Number of turns = 2000, Magnetic field = B = 0.08 T, μ_o = 4π x 10^-7 Wb/Am.

To Find:  current through coil = i =?

Solution:

n = N/2πr = 2000 / (2π x 0.1)

∴ B = μ_r μ₀ n i

∴ 0.08   = 1 x 4π x 10^-7 x (2000 / (2π x 0.1)) x i

∴ 0.08   = 2 x 10^-7 x (20000) x i

∴ 0.08   = 4 x 10^-3 i

∴ i = 0.08 / (4 x 10^-3) = 20 A

Ans: The current through coil is 20 A

Example 07:

A toroid has 5000 turns wound upoin it and carries a current of 15 A. What is the magnetic flux density inside the torroid at point 25 cm from the centre of toroidal circle?

Given: distance from centre = r = 25 cm = 0.25 m, Number of turns = 5000, current through coil = i = 15 A, μ_o = 4π x 10^-7 Wb/Am.

To Find:   Magnetic field = B =?

Solution:

∴ B = μ_r μ₀ n i

∴ B = μ_r μ₀ (N/2πr) i

∴ B   = 1 x 4π x 10^-7 x (5000 / (2π x 0.25)) x 15

∴ B   = 2 x 10^-7 x (5000 /0.25) x 15

∴ B   = 2 x 10^-7 x 20000 x 15

∴ B   = 0.06 T

Ans: Magnetic flux density is 0.06 T

Example – 08:

A closed wound narrow toroid has 500 turns wound upoin it and carries a current of 5 A. What is the magnetic flux density inside the torroid at point 5 cm from the centre of toroidal coil?

Given: distance from centre = r = 5 cm = 0.05 m, Number of turns = 500, current through coil = i = 5 A, μ_o = 4π x 10^-7 Wb/Am.

To Find:   Magnetic field = B =?

Solution:

∴ B = μ_r μ₀ n i

∴ B = μ_r μ₀ (N/2πr) i

∴ B   = 1 x 4π x 10^-7 x (500 / (2π x 0.05)) x 5

∴ B   = 2 x 10^-7 x (500/0.05) x 5

∴ B   = 2 x 10^-7 x 10000 x 5

∴ B   = 0.1 T

Ans: Magnetic flux density is 0.1 T

Example 09:

A toroid is wound on a paramagnetic substance of susceptibility 3 x 10^-4. What will be the percentage increase in magnetic field of toroid?

Given: susceptibility = χ =3 x 10^-4,

To Find:   % Change in magnetic field =?

Solution:

We have μ_r = χ + 1= 3 x 10^-4 + 1 = 0.0003 + 1 = 1.0003

Ans: Percentage increase in the magnetic field is 0.03 %

Example – 10:

A toroid is wound on a paramagnetic substance of susceptibility 6.8 x 10^-5. What will be the percentage increase in the magnetic field of the toroid?

Given: susceptibility = χ = 6.8 x 10^-5,

To Find:   % Change in the magnetic field =?

Solution:

We have μ_r = χ + 1= 6.8 x 10^-4 + 1 = 0.000068 + 1 = 1.000068

Ans: Percentage increase in the magnetic field is 6.8 x10^-3 %

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The device which uses the electric or magnetic field to guide and accelerate a beam of charged particles to high speed is called a particle accelerator. Charged particles used may be protons or electrons. These high-velocity particles are used in nuclear physics and high energy physics. Depending upon the direction of motion of charged particles, they have classified into two types a) Linear accelerator and b) Circular accelerator or Cyclotron.

In linear accelerator charged particles move in a straight line. To achieve velocity in kilometres, the length of linear accelerators is also in kilometres, thus the length of linear accelerators is very large. In Cyclotron the same distance as in linear accelerators is covered in concentric circles. Hence the size of the cyclotron is greatly reduced. The first cyclotron was developed by Lawrence and Livingston in 1931 at the University of California.

Cyclotron:

Principle:

An electric field can accelerate a charged particle and the magnetic field can throw the particle in a circular orbit such that its frequency of the revolution does not depend on its speed. Thus the radius of circular path increases, the speed of the particles goes on increasing.

Diagram:

Construction:

It consists of a pair of hollow metal chambers labelled D₁ and D₂ which are flat and semicircular in the shape of the letter ‘D’ back to back. Hence these chambers are called ‘Dees’. A source of a charged particle is located at the midpoint of the gap between the two dees.

The dees are connected to the radio frequency oscillator to apply high frequency between the dees. Dees act as electrodes and the potential between the two dees is made to alter very rapidly. The dees are then fixed in a large metal box which is evacuated. The whole apparatus is put between a strong electromagnet which makes the particle to move in a circular path.

Working:

Let us assume a positive charge particle (proton) having a charge ‘q’ is emitted from the source. Let us assume at this instant D₁ is negative. It is accelerated towards D₁. Let v₁ be the velocity of the particle when it reaches D₁. Due to the magnetic field of induction, it starts moving in a circular path in D₁. Let r₁ be the radius of the circular path and ‘m’ be the mass of the charged particles.

After completing the semicircular path, the particle reaches gap again. At this instant, D₂ is made negative.  Let v₂ be the velocity of the particle when it reaches D₂. Due to the magnetic field of induction, it starts moving in a circular path in D₂. Let r₂ be the radius of the circular path.

Dividing equation (1) by (2) we get

Thus the angular velocity and hence the period is constant or it is independent of the velocity of the particle. Thus at every crossing of the gap the velocity of the charged particle increases. It traces a flat spiral of increasing radius. Then at the periphery point, it is deflected on target with very high velocity using a deflecting plate.

Uses:

• The fast-moving charged particles obtained using cyclotron can be used in artificial transmutation.
• The fast-moving charged particles obtained using cyclotron can be further used to obtain high energy fast-moving particles like neutrons, which can be used in nuclear fission.
• The fast-moving charged particles obtained using cyclotron can be used in inducing artificial radioactivity.

Previous Topic: Tangent Galvanometer

Science > Physics > Magnetic Effect of Electric Current > Cyclotron

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In this article, we shall study, the principle, construction, working, sensitivity, and accuracy of the tangent galvanometer.

Principle:

The tangent galvanometer works on the principle of tangent law. The magnetic needle is subjected to two magnetic fields which are perpendicular to each other. One field is due to the horizontal component of the earth’s magnetic field and the other is created by passing the current through the coil of the tangent galvanometer. Under the action of two magnetic fields the needle comes to rest making angle θ with B_H, such that B =  B_H tan θ

Diagram:

Construction:

The tangent galvanometer consists of the following parts.

A coil having a large number of turns:

The coil consists of an insulated copper wire wound on a vertical circular frame. The frame is made up of nonmagnetic material like ebonite.

The frame is fixed vertically on a horizontal base with leveling screws. The coil is provided with two or more coils of a different number of turns. Thus the number of turns can be changed.

A small magnetic needle:

A magnetometer box with a magnetic needle is kept at the centre of the coil in a horizontal plane. The magnetic needle is pivoted at the centre of the coil and can rotate in a horizontal plane.

A light aluminium pointer is pivoted at a right angle to the magnetic needle.  When the needle turns through a certain angle the aluminum pointer also turns through the same angle.

The deflection can be read on a circular degree scale. A mirror is placed below the pointer to avoid error due to parallax.

Adjustment:

• All nearby magnets and magnetic materials are removed away from the instrument.
• The instrument is first leveled using spirit level by adjusting leveling screws at the base so that the needle is exactly horizontal and the coil is exactly vertical.
• The coil is then rotated about its vertical axis so that its plane is parallel to the needle.  Thus the coil remains in the magnetic meridian.
• The magnetometer box is rotated so that the pointer reads 0° – 0°.

Working:

When the current is not passed through the coil the magnetic needle is acted upon by only a horizontal component of the earth and thus the magnetic needle remains in the north-south direction. When the current is passed through the coil a magnetic field is created by the coil. Thus the magnetic needle is now acted upon by two magnetic fields which are at a right angle to each other. Hence the needle will turn through an angle say θ.

Let B_H be the horizontal component of the earth’s magnetic induction and B be the magnetic induction at the centre of the coil due to the current in the coil.  The direction of B will be perpendicular to the plane of the coil and hence perpendicular to B_H. By the tangent law

B = B_H tan θ   …………. (1)

The magnetic induction at the centre of the coil due to current in the coil is given by

Where k is a constant, called the reduction factor of the tangent galvanometer.

∴  i   ∝  tan θ

Thus in a tangent galvanometer, the current through the coil is directly proportional to the tangent of the angle of deflection of the needle.  Due to this characteristic of the galvanometer is known as the tangent galvanometer. Knowing k and θ we can calculate the value of the current through the coil.

The Sensitivity of a Tangent Galvanometer:

The sensitivity of the tangent galvanometer is defined as the ratio of the change in deflection of the galvanometer to the current producing this deflection.

Sensitivity = dθ / di

A galvanometer is said to be sensitive if it gives larger deflection for a small current.

Thus the sensitivity of tangent galvanometer can be increased by

• Increasing the number turns (n) of the coil,
• Keeping the value of cos²θ  small, i.e. deflection should be small,
• Decreasing the magnetic induction (B_H).  It can be done by putting auxiliary bar magnet below the magnetic needle such that it opposes the magnetic induction B_H
• Decreasing the radius (r) of the coil.

Limitations in Increasing Sensitivity:

• If the number of turns of the coil is increased the radius of the coil will not be the same for all turns and the magnetic induction at the centre is not uniform.
• The increase in the number of turns increases the resistance of the coil of the galvanometer.
• If the radius of the coil is made small the magnetic field at the centre will be non-uniform.
• If θ is kept small accuracy decreases. Accuracy is maximum when θ = 45°

Accuracy of Tangent Galvanometer:

Let dθ be the small error in measuring the deflection θ so that the corresponding error in the current calculated is di when measuring a current ‘i’. Then

Now it is clear that the accuracy is maximum when the error is minimum i.e. the quantity di/i  is minimum. i.e. the quantity (sin 2θ)  is maximum. i.e.

sin 2θ = 1

∴   2θ = sin^-1(1)

∴   2θ = 90°

∴   θ = 45°

Thus the accuracy of a T.G. is maximum at a deflection of 45°.

The accuracy can be increased by

• Using light and long pointer.
• Increasing the radius of each turn in the coil. Larger is the radius greater will be the accuracy.  Because due to the larger radius the magnetic induction produced near the centre of the needle will be more and nearly uniform.
• Decreasing the number of turns in the coil. For greater accuracy, the number of turns in the coil must be lesser. Because if the number of turns is large there can be the difference in the radii of different turns and hence current cannot be calculated accurately.

Note:

If we study the conditions of accuracy and sensitivity, we can note that they are contradictory to each other thus accuracy and sensitivity cannot be combined in a single instrument.

Distinguishing Between MCG and a TG:

Previous Topic: Ammeters and Voltmeters

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Science > Physics > Magnetic Effect of Electric Current > Tangent Galvanometer

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Ammeter:

An ammeter is an electrical measurement device (apparatus) which is used to measure the electric current in the electrical circuit.

Requirements of Good Ammeter:

• The resistance of an ammeter must be as small as possible.  The ideal resistance of an ammeter should be zero.
• The ammeter should have a range sufficient for measuring a given current.
• It should be sufficiently sensitive (i.e. for a small change in current there should be a reasonable change in deflection).
• It should be direct reading.

Necessity of Shunting for an Ammeter:

We know that an ammeter is used for measuring current.  It is to be connected in series in the circuit for this purpose. If the resistance of ammeter is not very low but considerable, then, when it is connected in a circuit the effective resistance of the circuit will considerably increase. Due to the increase in the value of the effective resistance of the circuit the current in the circuit will decrease compared to the original current in the circuit which was to be measured.  Thus the decreased value of the current will be measured by the ammeter. In order to avoid this, the resistance of the ammeter should be very low. i.e. then the ammeter will be able to measure the current more accurately.

The effective resistance of the parallel combination is always smaller than the least value in the parallel combination. Hence a low-value resistance is connected in parallel across the galvanometer. This process is known as shunting.

Uses of a Shunt in an Ammeter:

• Due to the shunt, the effective resistance of the ammeter will be very low.
• Shunt increases the range of measurement of the current by the galvanometer and hence the range of ammeter is increased.
• Shunt protects the galvanometer coil from being damaged due to the excess flow of current.

Conversion of a Moving Coil Galvanometer into an Ammeter:

A sensitive moving coil galvanometer (pivoted type) is taken. To measure given the range of current a low resistance of suitable value is connected in parallel with it.

The scale is calibrated by comparison with a standard instrument.

Expression for the Value of a Shunt in an Ammeter:

Let G be the resistance of the galvanometer coil.  Let I_g be the current required to give full-scale deflection in the galvanometer.  Let S be the resistance to be connected in parallel with the galvanometer coil so as to measure a maximum current I. Let I_s be the current through the shunt.

This is the expression for the low-value resistance which is to be connected in parallel with the galvanometer to convert it into an ammeter to measure the given current.

Voltmeter:

A voltmeter is an electrical measurement device (apparatus) which is used to measure the potential difference in the electrical circuit.

Requirements of Good Voltmeter :

• The resistance of a voltmeter must be as large as possible.  The ideal resistance of an ammeter should be infinity.
• The voltmeter should have a range sufficient for measuring a given potential.
• It should be sufficiently sensitive (i.e. for a small change in voltage there should be a reasonable change in deflection).
• It should be direct reading.

Necessity of Connecting High-Value Resistance in a Voltmeter:

We know that a voltmeter is used to measure the potential difference between two points. (or across a resistance).  It is to be connected in parallel for this purpose. If the resistance of the voltmeter is not very high it will divert a considerable current from the circuit through it, as a result, there will be a fall in potential difference. Instead of measuring the actual potential difference this decreased potential difference will be measured by the voltmeter. In order to avoid this the resistance of the voltmeter should be very high i.e. then the voltmeter will be able to measure the voltage more accurately.

We know that the effective resistance of the series combination is always larger than the largest value in the series combination. Hence a large value resistance connected in series with the galvanometer to convert it into the voltmeter.

Use of a High-Value Resistance in a Voltmeter:

• Due to the high-value resistance, the effective resistance of the voltmeter will be very high.
• High-value resistance increases the range of measurement of the potential difference by the galvanometer and hence the range of voltmeter is increased.
• High-value resistance protects the galvanometer coil from being damaged due to the excess voltage applied.

Conversion of a Moving Coil Galvanometer into a Voltmeter:

A sensitive moving coil galvanometer (pivoted type) is taken. To measure given the range of potential difference a high resistance of suitable value is connected in series with it.

The value of high-value resistance is given by

The scale is calibrated by comparison with a standard instrument.

Expression for Value of a Resistance in a Voltmeter:

Let G be the resistance of the galvanometer coil.  Let I_g be the current required to give full-scale deflection in the galvanometer.  Let R be the resistance to be connected in series with the galvanometer coil so as to measure a maximum potential difference V. Let Is be the current through the shunt.

Diagram:

This is the expression for the high-value resistance which is to be connected in series with galvanometer to convert it into a voltmeter to measure the given potential difference.

Distinguishing Between Ammeter and Voltmeter:

Distinguishing Between Voltmeter and Potentiometer:

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In this article, we shall study principle, construction, working, sensitivity and accuracy of the moving coil galvanometer

Principle:

When a current-carrying coil is suspended in a uniform magnetic field it is acted upon by a torque. Under the action of this torque, the coil rotates and the deflection in the coil in a moving coil galvanometer is directly proportional to the current flowing through the coil.

Construction of Suspended Type Moving Coil Galvanometer:

The suspended type consists of a rectangular coil of thin insulated copper wires having a large number of turns. The coil is suspended between the poles of a powerful horseshoe magnet by a suspension fibre of phosphor-bronze.  A spring is attached to the other end of the coil.  The current enters the coil through the fibre and leaves the coil through the spring. The upper end of the suspension fibre is connected to a rotating screw head so that the plane of the coil can be adjusted in any desired position. The horseshoe magnet has cylindrically concave pole-pieces. Due to this shape, the magnet produces radial magnetic field so that when coil rotates in any position its plane is always parallel to the direction of the magnetic field. When current flows through the coil it gets deflected.

A soft iron cylinder is fixed inside the coil such that the coil can rotate freely between the poles and around the cylinder. Due to the high permeability, the soft iron core increases the strength of the radial magnetic field. A small plane mirror M is fixed to the suspension fibre. This along with lamp and scale arrangement is used to measure the deflection of the coil.

Theory:

Consider a rectangular coil PQRS of single turn having length ‘l’ and breadth ‘b’ suspended in a uniform magnetic field of induction B such that the plane of the coil is parallel to the magnetic field. Let ‘I’ be the current through the coil.

The sides PS and QR being parallel to the magnetic field do not experience any force, but the sides PQ and RS being perpendicular to the magnetic field experience force. The force experienced by each side is given by

F = B I l

By Fleming’s left-hand rule these forces are opposite in direction. As these two forces are equal and opposite they form what is called as a couple and due to which a torque acts on the coil which tries to deflect the coil. The deflection torque is given by,

Torque = Force x Perpendicular distance between the forces.

τ =   F   x   b
∴ τ   =   B I l × b

But l τ b = A, the area of the coil

∴   τ =   B I A

If the coil has ‘n’ turns, then the deflecting torque is given by

∴   τ   = n BIA

Under the action of this torque, the plane of the coil rotates through an angle θ before coming to rest. Due to the radial magnetic field, the plane of the coil is always parallel to the direction of the magnetic field.  Thus at any position, the deflecting torque has constant magnitude. The rotation of the coil produces a twist in the fibre which produces a restoring torque which is directly proportional to the angle of deflection θ.

τ ∝  θ
 ∴    τ  = k  θ

Where k is the torque per unit twist (or torsional constant) of the suspension fibre.

When the coil comes to rest i.e. when it attains equilibrium, the restoring torque will balance the deflecting torque. So in equilibrium position of the coil,

Deflecting torque    =     Restoring torque.
n B I A  = k θ

The quantities in bracket are constant, therefore
∴   I ∝ θ

Thus in a moving coil galvanometer current in the coil is directly proportional to the angle of deflection of the coil.

Advantages of Moving Coil Galvanometers:

• They are not affected by a strong magnetic field.
• They have a high torque to weight ratio.
• They are very accurate and reliable.
• Their scales are uniform.

Disadvantages of Moving Coil Galvanometers:

• The change in temperature causes a change in restoring torque.
• Restoring torque cannot be easily changed.
• There is a possibility of damage to the phosphor bronze fibre or helical restoring spring due to severe stresses.
• Such instruments can only be used for measurement of direct current quantities and can not be used for measurement of alternating current quantities.

Pivoted Type Moving Coil Galvanometer:

Construction :

The rectangular coil of thin insulated copper wires having a large number of turns is pivoted between the poles of a powerful horseshoe magnet. The coil is mounted on a pivot between two supports. The supports are bearings with almost no friction. Two hairsprings are attached one above the coil and other below the coil which controls the rotation of the coil. The two coils are spiralled in opposite directions. Current enters through one coil and leaves through the other. A long pointer is attached to the coil which directly moves over a graduated scale. The whole assembly is fitted in a box with a window through which deflection can be observed.

Note:

The principle, working and theory of pivoted type moving coil galvanometer is the same as suspended type moving coil galvanometer.

Sensitivity of Moving Coil Galvanometer:

The sensitivity of moving coil galvanometer is defined as the ratio of the change in deflection of the galvanometer to the change in the current.

Sensitivity = dθ / di

A galvanometer is said to be sensitive if it gives larger deflection for a small current. The current in moving coil galvanometer is given by

Thus the sensitivity of moving coil galvanometer can be increased by

• Increasing the number turns (n) of the coil,
• Increasing the area (A) of the coil,
• increasing the magnetic induction (B) and
• Decreasing the couple per unit twist (k) of the suspension fibre.

Limitations to Increase in Sensitivity of Moving Coil Galvanometer:

• If the turns of the coil are increased the length of wire and hence the resistance of the coil increases.
• Increasing the area of the coil beyond limit makes the instrument bulky.
• Increase in the number of turns and area of the coil increases the load on suspension fibre. Hence spring higher value of k should be used which decreases the sensitivity of the galvanometer.
• Increasing the strength of magnetic induction leads to an increase in the weight of the apparatus.
• Decreasing the couple per unit twist of the spring leads to a decrease in the strength of the spring.

Accuracy of Moving Coil Galvanometer:

The relative error in the measurement of current is given by di/i. For moving coil galvanometer, the current through it is given by

Thus the error in the measurement of current depends only on the measurement of the deflection in the galvanometer dθ.

For greater accuracy of the galvanometer, the ratio di / i should be small. It is small when the deflection is large. Thus for greater accuracy, the deflection in the galvanometer should be large for small current in it. As the expression of accuracy does not contain the terms n, A, B and k the accuracy is independent of the number of turns of the coil, the area of the coil, the magnetic induction and constant for the spring.

Previous Topic: Numerical Problems on Toroids

Next Topic: Ammeters and Voltmeters

Science > Physics > Magnetic Effect of Electric Current > Moving Coil Galvanometer

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In this article, we shall study the Ampere’s law and its application in finding magnetic induction due to long straight conductor, solenoid and toroid.

Statement:

The line integral of the magnetic field B around any closed path is equal to μ₀ times the net steady current enclosed by this path.

Mathematically,

Proof:

The magnetic field produced by a long straight conductor is in the form of concentric circles. These circles are in the plane perpendicular to the length of the conductor. The direction of the field is given by right-hand grip rule.

The magnitude of the magnetic field due to a long straight conductor at a point P on the closed-loop of radius ‘a’ is given by

Let us consider a field line through point P and of a very small length ‘dl’ of this field element. Now both B and dl are tangential to the field line at P.

Applications of Ampere’s Law:

Expression for Magnetic Field Due to Solenoid and Toroid:

A cylindrical coil of a large number of turns is called a solenoid. The magnetic field inside the solenoid is given by 

B = μ₀nI.

Where n = number of turns per unit length and

I = Current through the wire of a solenoid

A toroid is a coil which is wound on a torus or a doughnut-shaped structure. For a toroid, 

B = μ₀nI. 

Where n = number of turns per unit circumference and

I = Current through the wire of a toroid

Expression for Magnetic Induction at a Point Due to a Long Straight Conductor Carrying Current:

Let us consider a very long straight conductor carrying electric current ‘I’ as shown. Let us consider a point P at a distance of ‘a’ from the conductor. Let us consider an Amperian loop as an imaginary circle, having radius ‘a’ and passing through point P. Let us consider a very small element of length ‘dl’ of this Amperian loop. Now both  B and dl are tangential to the field line at P.

This is an expression for magnetic induction at a point due to a long straight conductor carrying current.

Expression for Magnetic Induction at a Point  on the Axis of a Long Straight Solenoid and Well Inside it:  

Let us consider a very long straight solenoid having ’n’ turns per unit length and carrying electric current ‘I’ as shown. Let us consider a point well inside the solenoid at which the magnetic induction is to be found.

Consider a rectangular path ABCD of the line of induction such that AB = L = length of the rectangular path. The number of turns enclosed by the rectangle is nL. Hence the total electric current flowing through the rectangular path is nLI. According to  Ampere’s law

Near the ends of the solenoid, the lines of the field are crowded. While for the rest of the space the lines are so widely spaced that the magnetic field is negligible.

This is an expression for magnetic induction at a point on the axis of long straight solenoid and well inside it.

The expression for magnetic induction at a point near and at the end of long straight solenoid is

Expression for Magnetic Induction at a Point  on the Axis of a Toroid:

A toroid is a solenoid bent into a shape of a hollow doughnut. Let us consider a toroidal solenoid of average radius ‘r’ having centre O and carrying current ‘I’. Let us consider an Amperian loop of radius r and traverse it in a clockwise direction. Let N be the number of turns of the toroid. Then the total current flowing through the toroid is ‘NI’. According to Ampere’s law

This is an expression for magnetic induction at a point on the axis of a toroid.

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