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A magnetic ring has a mean circumference of 20 cm and a cross section of 20 cm2 and has 800 numbers of turns of wire. When the exciting current is 5A, the flux is 2 mWb. The relative permeability of iron is nearly
- a)3.98
- b)398
- c)0.398
- d)39.8
Correct answer is option 'D'. Can you explain this answer?
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A magnetic ring has a mean circumference of 20 cm and a cross section ...
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A magnetic ring has a mean circumference of 20 cm and a cross section ...
Given data:
Mean circumference of magnetic ring, C = 20 cm
Cross-section of magnetic ring, A = 20 cm²
Number of turns of wire, N = 800
Exciting current, I = 5 A
Flux, Φ = 2 mWb
Formula used:
Magnetic flux, Φ = B × A, where B is the magnetic field in Tesla and A is the cross-sectional area of the magnetic ring.
Magnetic field, B = μ × H, where μ is the permeability of the magnetic material and H is the magnetic field intensity.
Magnetic field intensity, H = N × I / C, where N is the number of turns of wire, I is the exciting current and C is the mean circumference of the magnetic ring.
Calculation:
The magnetic field intensity, H = N × I / C = 800 × 5 / 20 = 200 A/m
The magnetic flux, Φ = B × A
2 × 10⁻³ = μ × H × A
μ = 2 × 10⁻³ / (H × A) = 2 × 10⁻³ / (200 × 10⁰ × 20 × 10⁻⁴) = 39.8
Therefore, the relative permeability of iron is nearly 39.8.
Hence, option (D) is the correct answer.
Mean circumference of magnetic ring, C = 20 cm
Cross-section of magnetic ring, A = 20 cm²
Number of turns of wire, N = 800
Exciting current, I = 5 A
Flux, Φ = 2 mWb
Formula used:
Magnetic flux, Φ = B × A, where B is the magnetic field in Tesla and A is the cross-sectional area of the magnetic ring.
Magnetic field, B = μ × H, where μ is the permeability of the magnetic material and H is the magnetic field intensity.
Magnetic field intensity, H = N × I / C, where N is the number of turns of wire, I is the exciting current and C is the mean circumference of the magnetic ring.
Calculation:
The magnetic field intensity, H = N × I / C = 800 × 5 / 20 = 200 A/m
The magnetic flux, Φ = B × A
2 × 10⁻³ = μ × H × A
μ = 2 × 10⁻³ / (H × A) = 2 × 10⁻³ / (200 × 10⁰ × 20 × 10⁻⁴) = 39.8
Therefore, the relative permeability of iron is nearly 39.8.
Hence, option (D) is the correct answer.
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A magnetic ring has a mean circumference of 20 cm and a cross section of 20 cm2 and has 800 numbers of turns of wire. When the exciting current is 5A, the flux is 2 mWb. The relative permeability of iron is nearlya)3.98b)398c)0.398d)39.8Correct answer is option 'D'. Can you explain this answer?
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