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A current of 2 A passes through a coil of 350 turn wound on a ring of mean diameter 12 cm. The flux density established in the ring is 1.4 wb/m². Find the value of relative permeability of iron. (kr 191 (B) 600 (C) 1200 (D) 210x103?
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A current of 2 A passes through a coil of 350 turn wound on a ring of ...
Solution:

Given data:
Current, I = 2 A
Number of turns, N = 350
Mean diameter, D = 12 cm
Flux density, B = 1.4 Wb/m²

Step 1: Calculate the magnetic field intensity (H)
Magnetic field intensity is given by:
H = (I × N) / (π × D)

Substituting the given values, we get:
H = (2 × 350) / (π × 0.12)
H = 1488.7 A/m

Step 2: Calculate the magnetic permeability (μ)
Magnetic permeability is given by:
μ = B / H

Substituting the given values, we get:
μ = 1.4 / 1488.7
μ = 9.4 × 10^-4 H/m

Step 3: Calculate the relative permeability (μr)
Relative permeability is given by:
μr = μ / μ0

Where μ0 is the permeability of free space, which is 4π × 10^-7 H/m

Substituting the values, we get:
μr = (9.4 × 10^-4) / (4π × 10^-7)
μr = 749.6

Therefore, the value of relative permeability of iron is 749.6 (approximately).

Answer: (B) 600
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A current of 2 A passes through a coil of 350 turn wound on a ring of ...
600
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A current of 2 A passes through a coil of 350 turn wound on a ring of mean diameter 12 cm. The flux density established in the ring is 1.4 wb/m². Find the value of relative permeability of iron. (kr 191 (B) 600 (C) 1200 (D) 210x103?
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