Physics 2102Physics 2102Lecture 16Lecture 16AmpereAmpere’’s laws law

Physics 2102

Jonathan Dowling

André Marie Ampère (1775 – 1836)

Given an arbitrary closed surface, the electric flux through it isproportional to the charge enclosed by the surface.

qFlux=0!

q

! ="#$Surface 0

%

qAdErr

AmpereAmpere’’s law:s law:Remember GaussRemember Gauss’’ law? law?

GaussGauss’’ law for Magnetism: law for Magnetism:

No isolated magnetic poles! The magnetic flux through any closed“Gaussian surface” will be ZERO. This is one of the four“Maxwell’s equations”.

! =• 0AdB

! ="0

#

qAdErr

0

loop

B d s iµ! ="r r

�

The circulation of B (the integral of B scalards) along an imaginary closed loop is proportionalto the net amount of currenttraversing the loop.

i1

i2i3

ds

i4

)(3210

loop

iiisdB !+="# µrr

Thumb rule for sign; ignore i4

As was the case for Gauss’ law, if you have a lot of symmetry,knowing the circulation of B allows you to know B.

AmpereAmpere’’s law:s law:a Second Gaussa Second Gauss’’ law. law.

Sample ProblemSample Problem• Two square conducting loops carry currents

of 5.0 and 3.0 A as shown in Fig. 30-60.What’s the value of ∫B·ds through each ofthe paths shown?

AmpereAmpere’’s Law: Example 1s Law: Example 1• Infinitely long straight wire

with current i.• Symmetry: magnetic field

consists of circular loopscentered around wire.

• So: choose a circular loop C-- B is tangential to the loopeverywhere!

• Angle between B and ds = 0.(Go around loop in same

direction as B field lines!)

! ="C

isdB0

µrr

! ==C

iRBBds 0)2( µ"

R

iB

!

µ

2

0=

R

AmpereAmpere’’s Law: Example 2s Law: Example 2

• Infinitely long cylindricalwire of finite radius Rcarries a total current i withuniform current density

• Compute the magnetic fieldat a distance r fromcylinder axis for:– r < a (inside the wire)– r > a (outside the wire)

i

! ="C

isdB0

µrr

Current intopage, circular

field linesr

R

AmpereAmpere’’s Law: Example 2 (cont)s Law: Example 2 (cont)

! ="C

isdB0

µrr

Current intopage, field

tangent to theclosed

amperian loop

r

enclosedirB 0)2( µ! =

2

22

2

2 )(R

rir

R

irJi

enclosed=== !

!!

r

iB

enclosed

!

µ

2

0=

2

0

2 R

irB

!

µ= For r < R For r>R, ienc=i, so

B=µ0i/2πR

SolenoidsSolenoids

inBinhBhisdB

inhhLNiiNi

hBsdB

isdB

enc

henc

enc

000

0

)/(

000

µµµ

µ

=!=!=•

===

+++=•

=•

"

"

"

rr

rr

rr

Magnetic Field of a Magnetic DipoleMagnetic Field of a Magnetic DipoleA circular loop or a coil currying electrical current is a magneticdipole, with magnetic dipole moment of magnitude µ=NiA.Since the coil curries a current, it produces a magnetic field, that canbe calculated using Biot-Savart’s law:

3

0

2/322

0

2)(2)(

zzRzB

µ

!

µµ

!

µrr

r"

+=

All loops in the figure have radius r or 2r. Which of thesearrangements produce the largest magnetic field at thepoint indicated?

Sample ProblemSample Problem

Calculate the magnitude and direction of theresultant force acting on the loop.