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Ampere’s Law. Alan Murray. dl. B. B. I. B. dl. dl. B. Try this …. Create a contour for integration (a circle seems to make sense here!). Take a closed contour. These currents are “enclosed”. And these currents are not!. ò H . dl = Current I enclosed.

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ampere s law

Ampere’s Law

Alan Murray

try this

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

h dl current i enclosed

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

h dl current i enclosed4
òH.dl = Current Ienclosed

I4

I1

I3

I2

Alan Murray – University of Edinburgh

ampere s law worked example

|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

outside r r h dl i enclosed

I, |J|=I/pR2

H, B

R

Outside … r>R, òH.dl = Ienclosed

B

r

Alan Murray – University of Edinburgh

inside r r h dl i enclosed

I, |J|=I/pR2

I =I pr2pR2

Inside … r<R, òH.dl = Ienclosed

B?

R

r

Alan Murray – University of Edinburgh

inside r r h dl i enclosed8

r

B,H

Inside … r<R, òH.dl = Ienclosed

Alan Murray – University of Edinburgh

ampere worked example

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

what does the b field look like

I

What does the B-field look like?

B-field lines

Alan Murray – University of Edinburgh

h dl i enclosed

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

h dl i enclosed12

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

h dl i enclosed13
òH.dl = Ienclosed?

|H| = -I_ 2L

|H| = +I_ 2L

Alan Murray – University of Edinburgh