Coil of N turns is wound on a cast iron ring which has mean length of 50 cm and its crosssection is of 4 cm diameter. The current flowing through the coil is 2 A which produces a flux of 6 mWb in the air-gap of 2mm length. If the relative permeability of iron is 1000, calculate the number of turns N.?

Question

Coil of N turns is wound on a cast iron ring which has mean length of 50 cm and its cross­section is of 4 cm diameter. The current flowing through the coil is 2 A which produces a flux of 6 mWb in the air-gap of 2mm length. If the relative permeability of iron is 1000, calculate the number of turns N.?

Answer

Calculation of Number of Turns N in a Coil Wound on a Cast Iron Ring

Given Parameters:

Mean length of the cast iron ring = 50 cm

Diameter of the cross-section of the ring = 4 cm

Current flowing through the coil = 2 A

Flux produced in the air-gap = 6 mWb

Length of the air-gap = 2 mm

Relative permeability of iron = 1000

Formulae used:

Flux density (B) = Flux (Φ) / Area (A)

Area (A) = πr²

Reluctance (S) = l / (μA)

Flux (Φ) = NI / (S)

Solution:

First we need to calculate the area of the cross-section of the ring:

Radius (r) = Diameter / 2 = 4 / 2 = 2 cm

Area (A) = πr² = π(2)² = 12.57 cm²

Next, we need to calculate the reluctance of the magnetic circuit:

Length of the air-gap (l) = 2 mm = 0.2 cm

Relative permeability of iron (μ) = 1000

Reluctance (S) = l / (μA) = 0.2 / (1000 x 12.57 x 10^-4) = 1.59 x 10^(-3) AT/Wb

Now, we can calculate the number of turns in the coil:

Flux (Φ) = NI / (S)

Flux density (B) = Flux (Φ) / Area (A)

Flux density (B) = μH

H = NI / l

B = μNI / (lA)

6 x 10^(-3) = 1000 x 2 x N / (1.59 x 10^(-3) x 12.57 x 10^(-4))

Number of turns N = 188.5

Answer:

The number of turns in the coil wound on the cast iron ring is 188.5.

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