Lecture 17: THU 18 Lecture 17: THU 18 MAR 10MAR 10

Ampere’s law Ampere’s law

Physics 2102

Jonathan Dowling

André Marie Ampère (1775 – 1836)

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

qFlux = 0!

q

∫ =⋅≡ΦSurface 0ε

qAdErr

∫ =⋅≡ΦSurface 0ε

qAdErr

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

Gauss’s Law for B-Fields!Gauss’s Law for B-Fields!

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∫ =• 0AdB

There are no SINKS or SOURCES of B-Fields!

What Goes IN Must Come OUT!

There are no SINKS or SOURCES of B-Fields!

What Goes IN Must Come OUT!

The circulation of B (the integral of B scalar ds) along an imaginary closed loop is proportional to the net amount of current traversing the loop.

The circulation of B (the integral of B scalar ds) along an imaginary closed loop is proportional to the net amount of current traversing the loop.

i1

i2i3

ds

i4

)( 3210loop

iiisdB −+=⋅∫ μrr

Thumb rule for sign; ignore i4

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

Ampere’s law: Closed Ampere’s law: Closed Loops Loops

rB⋅d

rs

LOOP—∫ =μ0ienclosed

The circulation of B (the integral of B scalar ds) along an imaginary closed loop is proportional to the net amount of current traversing the loop.

The circulation of B (the integral of B scalar ds) along an imaginary closed loop is proportional to the net amount of current traversing the loop.

€

rB ⋅d

r s

loop

∫ = μ0 (i1 − i2)

Thumb rule for sign; ignore i3

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

Ampere’s law: Closed Ampere’s law: Closed Loops Loops

rB⋅d

rs

LOOP—∫ =μ0ienclosed

rB⋅d

rs

LOOP—∫ =μ0ienclosed

enc Calculation of . We curl the fingers

of the right hand in the direction in which

the Amperian loop was traversed. We note

the

i

direction of the thumb.

All currents inside the loop to the thumb are counted as .

All currents inside the loop to the thumb are counted as .

All currents outside the loop are not count

pa

antiparal

rallel positive

lel negative

enc 1 2

ed.

In this example : .i i i= −

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 of the paths shown?

Path 1: ∫B∙ds = μ0•(–5.0A+3.0A)

Path 2: ∫B∙ds = μ0•(–5.0A–5.0A–3.0A)

Ampere’s Law: Example Ampere’s Law: Example 11

• Infinitely long straight wire with current i.

• Symmetry: magnetic field consists of circular loops centered around wire.

• So: choose a circular loop C so B is tangential to the loop everywhere!

• Angle between B and ds is 0. (Go around loop in same direction as B field lines!)

∫ =⋅C

isdB 0μrr

∫ ==C

iRBBds 0)2( μπ

R

iB

πμ2

0=R

iB

πμ2

0=

R

Much Easier Way to Get B-Field Around A Wire: No Calculus.

Ampere’s Law: Example Ampere’s Law: Example 22

• Infinitely long cylindrical wire of finite radius R carries a total current i with uniform current density

• Compute the magnetic field at a distance r from cylinder axis for:r < a (inside the wire)r > a (outside the wire)

• Infinitely long cylindrical wire of finite radius R carries a total current i with uniform current density

• Compute the magnetic field at a distance r from cylinder axis for:r < a (inside the wire)r > a (outside the wire)

iCurrent out of page,

circular field lines

∫ =⋅C

isdB 0μrr

∫ =⋅C

isdB 0μrr

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

∫ =⋅C

isdB 0μrr

Current out of page, field

tangent to the closed

amperian loop enclosedirB 0)2( μπ =

2

22

22 )(

R

rir

R

irJienclosed === π

ππ

r

iB enclosed

πμ2

0=

20

2 R

irB

πμ

= 20

2 R

irB

πμ

= For r < R For r>R, ienc=i, soB=μ0i/2πR = LONG WIRE!

For r>R, ienc=i, soB=μ0i/2πR = LONG WIRE!

Need Current Density J!

R

0

2

i

R

μπ

r

B

O

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

r < RB∝ rr < RB∝ r

r > RB∝1 / rr > RB∝1 / r

r < R

B=μ0ir2πR2

r < R

B=μ0ir2πR2

r > R

B=μ0i2πr

Outside Long Wire!

r > R

B=μ0i2πr

Outside Long Wire!

SolenoidsSolenoids

inBinhBhisdB

inhhLNiiNi

hBsdB

isdB

enc

henc

enc

000

0

)/(

000

μμμ

μ

=⇒=⇒=•

===

+++=•

=•

∫

∫∫

rr

rr

rr

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

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Magnetic Field in a Toroid “Doughnut” Solenoid

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Toriod Fusion Reactor: Power NYC For a Day on a Glass of H2O

€

B =μ0Ni

2πr

€

B =μ0Ni

2πr

Magnetic Field of a Magnetic Magnetic Field of a Magnetic DipoleDipole

30

2/3220

2)(2)(

zzRzB

μπμμ

πμ rrr

≈+

= 30

2/3220

2)(2)(

zzRzB

μπμμ

πμ rrr

≈+

=

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

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

Magnetic Dipole in a B-Field

rF =0rτ =

rμ ×

rB

U =−rμg

rB

Force is zero in homogenous fieldTorque is maximum when  = 90°Torque is minimum when  = 0° or 180°Energy is maximum when  = 180°Energy is minimium when  = 0°

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Neutron Star a Large Magnetic Dipole