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

2024
Views:

416

Pages:

5
Author:

Jabin F.

Magnetic Field of a Solenoid: n = number of turns per unit length Bz = magnetic field along the z-axis Diagram 1: A solenoid with current-carrying loops. Diagram 2: A cross-section of the solenoid showing a single loop. Diagram 3: A close-up of a single loop with dimensions labeled. Calculations: For one ring: Bz = μ₀ * I * a² / (2(a² + z²)^(3/2)) Total turns in a coil: N = n * dz dBz = μ₀ * I * a² * n * dz / (2(a² + z²)^(3/2)) Calculations: From figure 2: a/z = tanθ => z = a cotθ dz = -a csc²θ dθ dBz = -μ₀I * a³ * m * csc²θ dθ / (2a³ csc³θ) dBz = -μ₀I * m * sinθ dθ / (2a) Bz = -μ₀I * m / (2a) * [cosθ₂ - cosθ₁] Case: If infinite solenoid, θ₁ = 180°, θ₂ = 0° B_in = μ₀ * I * m / (2a) Problem: Question 14 Page 41 Calculations: B = μ₀nI/2 * (cosθ₂ - cosθ₁) OP = 8D/15 + 2D/3 = 18D/15 = 6D/5 pq = √((36/25)D² + D²/4) = √(169/100)D² = 13D/10 cosθ₂ = 6D/5 / (13D/10) = 12/13 cosθ₁ = 4/5 B = μ₀nI/2 * (12/13 - 4/5) = 4/65 * μ₀nI Case 2: At the End of a Semi-Infinite Solenoid: B_end = μ₀nI / 2 Case 3: Outside the Solenoid: B_out = 0 Magnetostatics: Current I = I₀ is possible. Current I = αt² is not possible in magnetostatics. Question 7/Page 49: Two long coaxial solenoids carrying current I in opposite directions. Inner solenoid (radius a) has n turns per unit length. Outer solenoid (radius b) has m turns per unit length. Find B in each region. Three Regions: (i) Inside the inner solenoid (ii) Between the solenoids (iii) Outside both Calculations: B_in = μ₀nI B_out = 0 B₁ = μ₀n₁I + μ₀n₂I(-z) = μ₀(n₁ - n₂)Iz B₂ = 0 + μ₀n₂I(-z) = μ₀n₂I(-z) B₃ = 0 + 0 = 0