# Numerical Problems on Current Through Toroids or Rowland Ring

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#### Example – 01:

• A Rowland ring (toroid) of ferromagnetic material of 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 passe through the windings?
• Given: mean radius = 15 cm = 0.15 m, Number of turns = 3000, The relative permeability = μr = 1000, current through coil = i = 2 A, μ= 4π x 10-7 Wb/Am.
• To Find: Magnetic field = B = ?
• Solution:

n = N/2πr = 3000 /(2π x 0.15)

∴  B = μr 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 passe through the windings?
• Given: mean radius = 15 cm = 0.15 m, Number of turns = 3500, The relative permeability = μr = 800, current through coil = i = 1.2 A, μ= 4π x 10-7 Wb/Am.
• To Find: Magnetic field = B = ?
• Solution:

n = N/2πr = 3500 /(2π x 0.15)

∴  B = μr 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= 4π x 10-7 Wb/Am.
• To Find:  relative permeability = μr = ?
• Solution:

n = N/2πr = 3000 /(2π x 0.115)

∴  B = μr μ0 n i

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

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

∴ 2.5   =  μr x 3.652 x 10-3

∴ μr = 2.5/(3.652 x 10-3) = 6.846 x 10= 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: Crossectional 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 = μr = 1800, μ= 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 = μr 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: Crossectional 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= 4π x 10-7 Wb/Am.
• To Find:   relative permeability = μr =?
• Solution:

n = N/2πr = 240 /0.3= 800

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

B = μr μ0 n i

∴ 5   = μr x 4π x 10-7 x  800 x 2

∴ 5   = μr x 4 x 3.142 x 10-7 x  800 x 2

∴ 5   = μr x 2.01 x 10-3

∴ μr = 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= 4π x 10-7 Wb/Am.
• To Find:  current through coil = i = ?
• Solution:

n = N/2πr = 2000 /(2π x 0.1)

∴  B = μr 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= 4π x 10-7 Wb/Am.
• To Find:   Magnetic field = B = ?
• Solution:

∴  B = μr μ0 n i

∴  B = μr 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= 4π x 10-7 Wb/Am.
• To Find:   Magnetic field = B = ?
• Solution:

∴  B = μr μ0 n i

∴  B = μr 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 μr = 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 μr = 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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