Ampere’s Law

1. Ampere’s Law Alan Murray

2. dl B B I B dl dl B Try this … Create a contour for integration (a circle seems to make sense here!) Alan Murray – University of Edinburgh

3. Take a closed contour These currents are “enclosed” And these currents are not! òH.dl = Current Ienclosed • This is, as it turns out, Ampere’s Law and is the magnetic-field equivalent of Gauss’s law • If we define H=B÷μ, B=μH, thenòH.dl = Current “enclosed” = òòòJ.ds I4 I5 I1 I3 I2 I6 Alan Murray – University of Edinburgh

4. òH.dl = Current Ienclosed I4 I1 I3 I2 Alan Murray – University of Edinburgh

5. |J| = I/A B? I B? R Ampere’s Law – Worked Example • Calculate the magnetic field H both Outside (r>R) and Inside (r<R) A wire with uniformly-distributed current I,current density Alan Murray – University of Edinburgh

6. I, |J|=I/pR2 H, B R Outside … r>R, òH.dl = Ienclosed B r Alan Murray – University of Edinburgh

7. I, |J|=I/pR2 I =I pr2pR2 Inside … r<R, òH.dl = Ienclosed B? R r Alan Murray – University of Edinburgh

8. r B,H Inside … r<R, òH.dl = Ienclosed Alan Murray – University of Edinburgh

9. L B-field here? I Ampere – Worked example i.e. what is the magnetic fieldabove and close to a metal “track”on a printed circuit board or chip? Alan Murray – University of Edinburgh

10. I What does the B-field look like? B-field lines Alan Murray – University of Edinburgh

11. Width L Contour for integration b a Current enclosed = a´I/L I/L Amps/metre of width, out of the diagram òH.dl = Ienclosed? Alan Murray – University of Edinburgh

12. òH.dl = aH(H, dl parallel) òH.dl = 0(H, dl perpendicular) òH.dl = 0(H, dl perpendicular) H òH.dl = aH(H, dl parallel) òH.dl = Ienclosed? H a b b a Current enclosed = a´I/L òH.dl = aH + aH = 2aH òH.dl = Ienclosed® 2aH = aI/L |H| = _I_ 2L Alan Murray – University of Edinburgh

13. òH.dl = Ienclosed? |H| = -I_ 2L |H| = +I_ 2L Alan Murray – University of Edinburgh