A rectangular shaped core is made up of Steel Plate 15mm and 20mm cross section. The mean length Path of magnetic path is 20cm. The existing coil has 300 turns and current 0.7A. Calculate: A) Magnetizing force B) Flux density C) Reluctance D) Flux of magnetic circuits
The correct answer and explanation is:
To solve this magnetic circuit problem, we will go through each part step-by-step.
Given Data:
- Cross-sectional dimensions: 15 mm × 20 mm
- Mean magnetic path length: 20 cm = 0.2 m
- Number of turns (N): 300
- Current (I): 0.7 A
- Material: Steel (assume relative permeability μr\mu_r ≈ 1000)
- Permeability of free space μ0\mu_0 = 4π×10−7 H/m4\pi \times 10^{-7} \, \text{H/m}
A) Magnetizing Force (H):
The magnetizing force is given by: H=N⋅IlH = \frac{N \cdot I}{l} H=300×0.70.2=2100.2=1050 A/mH = \frac{300 \times 0.7}{0.2} = \frac{210}{0.2} = 1050 \, \text{A/m}
B) Flux Density (B):
First, calculate the magnetic field strength BB: B=μ⋅H=μ0⋅μr⋅HB = \mu \cdot H = \mu_0 \cdot \mu_r \cdot H μ=4π×10−7×1000=1.2566×10−3 H/m\mu = 4\pi \times 10^{-7} \times 1000 = 1.2566 \times 10^{-3} \, \text{H/m} B=1.2566×10−3×1050=1.3194 TB = 1.2566 \times 10^{-3} \times 1050 = 1.3194 \, \text{T}
C) Reluctance (ℜ):
Reluctance is given by: ℜ=lμ⋅A\mathcal{ℜ} = \frac{l}{\mu \cdot A}
Cross-sectional area: A=15×10−3×20×10−3=3×10−4 m2A = 15 \times 10^{-3} \times 20 \times 10^{-3} = 3 \times 10^{-4} \, \text{m}^2 ℜ=0.21.2566×10−3⋅3×10−4=0.23.7698×10−7≈530,576 A/Wb\mathcal{ℜ} = \frac{0.2}{1.2566 \times 10^{-3} \cdot 3 \times 10^{-4}} = \frac{0.2}{3.7698 \times 10^{-7}} \approx 530,576 \, \text{A/Wb}
D) Magnetic Flux (Φ):
Flux is given by: Φ=MMFℜ=N⋅Iℜ\Phi = \frac{MMF}{\mathcal{ℜ}} = \frac{N \cdot I}{\mathcal{ℜ}} Φ=300×0.7530576≈210530576=3.96×10−4 Wb\Phi = \frac{300 \times 0.7}{530576} \approx \frac{210}{530576} = 3.96 \times 10^{-4} \, \text{Wb}
Summary:
- A) Magnetizing Force (H) = 1050 A/m
- B) Flux Density (B) = 1.32 T
- C) Reluctance (ℜ) = 530,576 A/Wb
- D) Magnetic Flux (Φ) = 3.96 × 10⁻⁴ Wb
Explanation:
A magnetic circuit is similar to an electric circuit, but it deals with magnetic flux instead of electric current. The coil on the core creates magnetomotive force (MMF), which drives magnetic flux through the steel. The magnetizing force (H) is the intensity of the magnetic field created along the core’s length. The flux density (B) shows how concentrated the magnetic field is in the core, depending on the material’s permeability. Reluctance acts like resistance, opposing the magnetic flux. Magnetic flux (Φ) is the total magnetic field passing through the core’s area. By knowing the number of turns, current, core dimensions, and material properties, all these quantities can be calculated using magnetic circuit laws.