University:
Santa Fe College
Course:
ETS 2527C | Electromechanical Components and Mechanisms
Academic year:

2024
Views:

435

Pages:

5
Author:

Jabin F.

For a cylindrical wire, if J in ∝ r^n: Bin ∝ r^n+1 Bout ∝ r^(-n) For a solenoid: If I ↑, B ↑ If I ↓, B ↓ Straight wire: I ẑ => B ϕ̂ ∮(∇ × ⃗B) d⃗s = ? ∮(∇ × ⃗B) d⃗s = ∮⃗B d⃗l ∮⃗B d⃗l = μ0 * Ienc Ienc = 10A Section 1 Reference: Question 40, Page-44 Heading: Magnetic field at the center of regular polygon Equation: B = (μ0 * n * I) / (2π * s) * sin(π/n) Section 2 Variables: n = number of sides s = distance from midpoint of each wire to center Condition: If m = 6 (hexagonal loop), then θ = 30°. Diagram: A hexagonal loop with sides of length a and radius R. Calculations: S = R * sin(θ) S = R * sin(30°) B = (μ0 * n * I) / (2π * R) * [sin(π/n) / cos(π/n)] Section 1 Reference: Question 45, Page-46 Equation: i = kt Conclusion: There is no steady current, but Bin is the same in a solenoid. Equation: Bin = μ0 * n * kt / 2 Section 2 Reference: Question 47, Page-46 Equation: Bin = (μ0 * μo * N * I) / (2πr) Toroidal: ⃗J = (I / A) * ẑ A = πr² ⃗J = (I / πr²) * ẑ ∇ × ⃗B = (μ0 / πr²) * ẑ for r < R ∇ × ⃗B = 0 for r > R Section 1 Reference: Question 49, Page-46 Equations: B = μ0 * I / (2πR) Bout ∝ 1/r Bin ∝ r Section 2 Reference: Question 51, Page-46 Heading: Magnetic field intensity Equation: H = (I0 / 2πr) * [sin(kr) - (γ / α) * cos(kr)] * ϕ̂ Equation: B = μ0 * H Method I for I: ∇ × ⃗B = μ0 * ⃗J I = ∫⃗J d⃗s Method II for I: ∮⃗B d⃗l = μ0 * Ienc Reference: Question 52, Page-46 Equation: Ienc = (μ0 / 4π) * ∮⃗B d⃗l Calculation: ∮⃗B d⃗l = ∫(Bϕ * ϕ̂) * a * dφ Substitution: Bϕ = (μ0 * I0) / (2πr) * [sin(kr) - (γ / α) * cos(kr)] Integration: ∮⃗B d⃗l = (μ0 * I0 * a) / (2π) * ∫[sin(kr) - (γ / α) * cos(kr)] dφ