Lecture 23: WED 11 MARLecture 23: WED 11 MARAmpereAmpere’’s laws law

Physics 2102

Jonathan Dowling

André Marie Ampère (1775 – 1836)

Exam 02 and Midterm Grade

Q1&P1, Dr. Schafer, Office hours: MW 1:30-2:30 pm, 222B Nicholson

Q2&P2, Dr. Gaarde, Office hours: TTh 2:30-3:30 pm, 215B Nicholson

Q3&P3, Dr. Lee, Office hours: WF 2:30-3:30 pm, 451 Nicholson

Q4&P4, Dr. Buth, Office hours: MF 2:30-3:30 pm, 222A Nicholson

Exam 02 Average: 63/100

You can calculate your midterm letter grade, as posted on paws, via this handy formula used by allsections:

[(MT01+MT02)*500/200+HWT/761]*30]/530*100 = Midterm Score

Here MT01 and MT02 are your scores on the first and second midterms and HWT is your totalhomework points for HW01-07 only.

Once you have this Midterm Score you then can get your letter grade via:

A: 90-100 B: 80-89 C: 60-79 D: 50-59 F: below 50.

Details on the grading policy are at: http://www.phys.lsu.edu/classes/spring2009/phys2102/

I have posted both midterm exam grades and your midterm class score on WebAssign.

a

IB

!

µ

2

10

1=

Magnetic field due to wire 1 where the wire 2 is,

1221BILF =

a

I2I1L

a

IIL

!

µ

2

210=

Force on wire 2 due to this field,F

Forces Between WiresForces Between Wires

Neil’s Rule: Same Currents – Wires Attract!Opposite Currents – Wires Repel!

SummarySummary• Magnetic fields exert forces on moving charges:

• The force is perpendicular to the field and the velocity.• A current loop is a magnetic dipole moment.• Uniform magnetic fields exert torques on dipole moments.

• Electric currents produce magnetic fields:•To compute magnetic fields produced by currents, use Biot-Savart’s law for each element of current, and then integrate.• Straight currents produce circular magnetic field lines, withamplitude B=µ0i/2πr (use right hand rule for direction).• Circular currents produce a magnetic field at the center (givenby another right hand rule) equal to B=µ0iΦ/4πr

• Wires currying currents produce forces on each other: Neil’s Rule:parallel currents attract, anti-parallel currents repel.

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

qFlux = 0!

q

! ="#$Surface 0

%

qAdE!!

Remember Gauss Law for E-Fields?Remember Gauss Law for E-Fields?

New Gauss Law for B-Fields!New Gauss Law for B-Fields!

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

! =• 0AdB

There are no SINKS or SOURCES of B-Fields!

What Goes IN Must Come OUT!

The circulation of B(the integral of Bscalar ds) along animaginary closed loop isproportional to the netamount of currenttraversing the loop.

i1

i2i3

ds

i4

)(3210

loop

iiisdB !+="# µ!!

Thumb rule for sign; ignore i4

If you have a lot of symmetry, knowing the circulation of Ballows you to know B.

AmpereAmpere’’s law: Closed Loopss law: Closed Loops

!B !d!s

LOOP

"" = µ0ienclosed

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?

Path 1: ∫B·ds = µ0•(–5.0A+3.0A)

Path 2: ∫B·ds = µ0•(–5.0A–5.0A–3.0A)

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 loopC -- B is tangential to theloop everywhere!

• Angle between B and ds =0. (Go around loop in samedirection as B field lines!)

! ="C

isdB0

µ!!

! ==C

iRBBds 0)2( µ"

R

iB

!

µ

2

0=

R

Much Easier Way to Get B-FieldAround A Wire: No Cow-Culus.

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

• Infinitely longcylindrical wire of finiteradius R carries a totalcurrent i with uniformcurrent density

• Compute the magneticfield at a distance rfrom cylinder axis for:– r < a (inside the wire)– r > a (outside the wire)

iCurrent out

of page,circular field

lines

! ="C

isdB0

µ!!

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

! ="C

isdB0

µ!!

Current out ofpage, field

tangent to theclosed

amperian loopenclosedirB 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 = LONG WIRE!

Need Current Density J!

R

0

2

i

R

µ

!

r

B

O

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

r < R

B ! r

r > R

B !1 / r

r < R

B =µ0ir

2!R2

r > R

B =µ0i

2!r

Outside Long!Wire!

SolenoidsSolenoids

inBinhBhisdB

inhhLNiiNi

hBsdB

isdB

enc

henc

enc

000

0

)/(

000

µµµ

µ

=!=!=•

===

+++=•

=•

"

"

"

!!

!!

!!

The n = N/L is turns per unit length.