lecture10 - Lecture 10 Perfect Metals in Magnetism and...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
1 ECE 303 – Fall 2007 – Farhan Rana – Cornell University Lecture 10 Perfect Metals in Magnetism and Inductance In this lecture you will learn: • Some more about the vector potential • Magnetic field boundary conditions • The behavior of perfect metals towards time-varying magnetic fields • Image currents and magnetic diffusion • Inductance ECE 303 – Fall 2007 – Farhan Rana – Cornell University The Vector Potential - Review 0 . = A r In electroquasistatics we had: 0 = × E r Therefore we could represent the E-field by the scalar potential: φ −∇ = E r In magnetoquasistatics we have: Therefore we can represent the B-field by the vector potential: ( ) ( ) 0 . . = = H B o r r µ A H B o r r r × = = A vector field can be specified (up to a constant) by specifying its curl and its divergence Our definition of the vector potential is not yet unique – we have only specified its curl For simplicity we fix the divergence of the vector potential to be zero: A r A r
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
2 ECE 303 – Fall 2007 – Farhan Rana – Cornell University Magnetic Flux and Vector Potential Line Integral - Review The magnetic flux λ through a surface is the surface integral of the B-field through the surface ∫∫ = ∫∫ = a d H a d B o r r r r . . µ λ Since: A H B o r r r × = = () = ∫∫ × = ∫∫ = s d A a d A a d B r r r r r r . . . We get: B-field The magnetic flux through a surface is equal to the line-integral of the vector potential along a closed contour bounding that surface s d r A closed contour Stoke’s Theorem ECE 303 – Fall 2007 – Farhan Rana – Cornell University Vector Potential of a Line-Current z I ˆ x y Consider an infinitely long line-current with current I in the + z -direction The H-field has only a φ -component r Using Ampere’s Law: ( ) I r H = π 2 Work in cylindrical co-ordinates r I H 2 = But ( ) ( ) r I r r A r r A A H o z z o o 2 1 = = × = r () () = r r I r A r A o o o z z ln 2 If the current has only a z-component then the vector potential also only has a z -component which, by symmetry, is only a function of the distance from the line-current Integrating from r o to r : J A o r r = 2
Background image of page 2
3 ECE 303 – Fall 2007 – Farhan Rana – Cornell University Vector Potential of a Line-Current Dipole z I ˆ + x y d + r r Consider two infinitely long equal and opposite line-currents, as shown The vector potential can be written as a sum using superposition: () = = + + r r I r r I r r I r A o o o o o z ln 2 ln 2 ln 2 π µ r The final answer does not depend on the parameter r o r r Question: where is the zero of the vector potential?
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 02/02/2008 for the course ECE 3030 taught by Professor Rana during the Fall '06 term at Cornell University (Engineering School).

Page1 / 12

lecture10 - Lecture 10 Perfect Metals in Magnetism and...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online