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

2024
Views:

224

Pages:

7
Author:

Jabin F.

Ampere's Law (Integral Form): ∮B·dl = μ₀Ienc Ienc = I₁ - I₂ Ampere's Law (Integral Form) Applied to a Wire: ∮B·dl = μ₀(I₁ - I₂) ∮B·dl = μ₀I₂ ∫(∇×B)·dS = μ₀∫J·dS K = I/J⊥ ⇒ I = ∫Kdl⊥ J = I/A ⇒ I = ∫J·dS Differential Form of Ampere's Law: ∇×B = μ₀J Circuit Equations: I₁ + I₂ = 2 I₂ + I₃ = 4 I₁ + I₂ + I₃ = 1 Current Direction Convention: Negative current means into the page. Positive current means out of the page. Calculations: From equations (2) and (3), we get: -I₁ = 3A I₁ = -3A (into the page) Substituting I₁ into equation (1): -3A + I₂ = 2 I₂ = 5A (out of the page) Substituting I₂ into equation (3): 5A + I₃ = 4 I₃ = -1A (into the page) Conclusion: The correct option is (D). Diagram: A diagram showing three parallel current-carrying wires. Calculations: Magnetic Field at Point P: Bp = B₁ + B₂ + B₃ Bp = μ₀I/(2π(d+r)) + μ₀I/(2πr) - μ₀I/(2π(d-r)) Bp = μ₀I/(2π)[1/(d+r) + 1/r - 1/(d-r)] Simplifying the Expression: Multiplying the numerator and denominator of each term by the other denominators: Bp = μ₀I/(2π)[(d-r)(r) + (d+r)(d-r) - (d+r)(r)] / [(d+r)(r)(d-r)] Bp = μ₀I/(2π)[d² - r² + d² - r² - dr - r²] / [(d+r)(r)(d-r)] Bp = μ₀I/(2π)[2d² - 3r² - dr] / [(d+r)(r)(d-r)] Setting Bp to Zero: Bp = 0 when the numerator is zero: 2d² - 3r² - dr = 0 Rearranging terms: -3r² - dr + 2d² = 0 Using the quadratic formula: r = [-(-d) ± √((-d)² - 4(-3)(2d²))] / (2(-3)) r = (d ± √(d² + 24d²)) / (-6) r = (d ± 5d) / (-6) r = -d/3 or r = 2d/3 Conclusion: The possible values for r are -d/3 or 2d/3. Initial Equations: dJ_TS = dyŷ - dzẑ dF_TS = IB₀[dyŷ - dzẑ] × (-x̂) dF_TS = IB₀[dyẑ + dzŷ] Net Force: dF_qTS = 2IB₀dyẑ F_qTS = 2IB₀∫dyẑ Integration Limits: The limits of integration are from 0 to R/2. Final Result: F_qTS = 2IB₀(R/2)ẑ F_qTS = IB₀Rẑ