lecture 17: thu 18 mar 10 ampere’s law physics 2102 jonathan dowling andré marie ampère (1775...
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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