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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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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Next Topic: Numerical Problems on Toroids

Science > Physics > Magnetic Effect of Electric Current > Ampere’s law

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