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

2024
Views:

323

Pages:

6
Author:

Jabin F.

Title: "Boundary Conditions on B and A" Subtitle: "In Electrostatics" Equations: ∇ ⋅ E = ϵ 0 ρ →E above −E below = ϵ 0 σ ∇ ⋅ B =0→B above −B below =0 Diagram: A diagram illustrating the boundary conditions for electric and magnetic fields at an interface between two media. Z-axis: The vertical axis representing the direction perpendicular to the interface. Interface: The plane at z=0 separating the "above" and "below" regions. Electric Fields: E above and E below are the electric fields in the regions above and below the interface, respectively. Magnetic Fields: B above and B below are the magnetic fields in the regions above and below the interface, respectively. Surface Charge Density: σ represents the surface charge density at the interface. Equations: ∇× A = B ∇⋅ A =0⇒Coulomb Gauge Condition A above −A below =0 ∮ A ⋅d l =∫ B ⋅d S Diagram: A diagram illustrating a surface with a normal vector pointing upwards. The magnetic field B is shown as perpendicular to the surface. Text: "Normal Component of A is same at above and below of Surface" "Continuous" "Φ net =∮ S B ⋅d S =0" Equation: ∮c⃗A⋅d⃗U=0=A′′above(l)+0+A′′below(−l)+0 Conclusion: Aabove+Abelow=0 Continuity Statement: A′′ and A′′ are continuous at below and above. Section 2 Reference: Question 20, page-61 Equation: I=I0Sinωt Equations: Ein∝(−∂I∂T), Eout∝(−∂I∂T) Condition: If I is steady (time-independent), then Ein=0 and Eout=0. Section 1 Reference: Question 21, page-62 Diagram: A cylindrical conductor with a current flowing through it, surrounded by a circular loop. Equation: ∮E⋅d⃗l=−∫S∂B∂t⋅d⃗S Calculation: E⋅2πr=−B0πr2 Conclusion: E=−B0R22πr Section 2 Reference: Question 29, page-63 Equation: ∮E⋅d⃗l=−∫S∂B∂t⋅d⃗S Calculation: E⋅2πr=−(0.4)πr2 Result: |E|=4mV/m Direction: ⃗E is clockwise