Lect09Magnetism - ECE 3030 Electromagnetic Fields and Waves...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
1 ECE 3030 Electromagnetic Fields and Waves Fall 2009 Lecture 9 2009/9/16 More Magnetoquasistatics Boundary Conditions CdtE f ft 1 Rana and Swartz 06/5/25 Electromagnetic Fields and Waves – Fall 2009 Lecture 9: Instructor: Dr. Wesley E. Swartz Conductor Effects Image Currents Inductance H I ( t ) The Vector Potential – Review In electrostatics we had Therefore we could write the E-field as the gradient of a potential: In magnetostatics we have: Therefore we can represent the B-field by the curl of a vector potential: 0 = × E ( ) ( ) 0 = = H B o μ A H B o × = = φ −∇ = E 2 Rana and Swartz 06/5/25 Electromagnetic Fields and Waves – Fall 2009 Lecture 9: A vector field can be specified (up to a constant) by specifying its curl and its divergence. Since we had so far only specified the curl of the vector potential, it was not yet unique. For simplicity we fixed the divergence of the vector potential to be zero: This choice is known as the “Coulomb Gage” . (There are other possible choices.) A 0 = A Magnetic Flux and the Vector Potential Line Integral – Review The magnetic flux λ through a surface is the surface integral of the B- field through the surface. Since   = = a d H a d B o A H B o × = = B-field 3 Rana and Swartz 06/5/25 Electromagnetic Fields and Waves – Fall 2009 Lecture 9: We get This says 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 A a d A a d B s d A closed contour Stoke’s Theorem Vector Potential of a Line Current Consider an infinitely-long line current with current I in the + z -direction. The H-field has only a Φ -component. Using Ampere’s Law: By symmetry, the vector potential only has a z - component which is only a function of the z I ˆ x y r I r H = π 2 Work in cylindrical co-ordinates r I H 2 = 4 Rana and Swartz 06/5/25 Electromagnetic Fields and Waves – Fall 2009 Lecture 9: distance from the line-current. Now , or Integrating from r 0 to r yields: r r A 1 A H z 0 o = × = r 2 I r ) r ( A 0 z = = = r r ln 2 I r A r A 0 0 0 z z Vector Potential of a Line-Current Dipole Consider two infinitely long equal and opposite line-currents, as shown. Using superposition: z I ˆ + x y d + r r = + r r ln 2 I r r ln 2 I r A o o o o z r z I ˆ ( ) = r ln I r A o 5 Rana and Swartz 06/5/25 Electromagnetic Fields and Waves – Fall 2009 Lecture 9: This final answer does not depend on r o .
Background image of page 1

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

View Full DocumentRight Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 4

Lect09Magnetism - ECE 3030 Electromagnetic Fields and Waves...

This preview shows document pages 1 - 2. Sign up to view the full document.

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