1 exam 2 covers ch. 27-33, lecture, discussion, hw, lab chapter 27: the electric field chapter 29:...
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1
Exam 2 covers Ch. 27-33,Lecture, Discussion, HW, Lab
Chapter 27: The Electric Field Chapter 29: Electric potential & work Chapter 30: Electric potential & field
(exclude 30.7) Chapter 31: Current & Resistance Chapter 32: Fundamentals of Circuits
(exclude 32.8) Chapter 33: The Magnetic Field
(exclude 33.5-33.6, 33.9-10, & Hall effect)
Exam 2 is Tue. Oct. 27, 5:30-7 pm, 145 Birge
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Tue. Oct. 27, 2009 Physics 208 Lecture 16 2
Law of Biot-Savart
Each short length of current produces contribution to magnetic field.r
I in plane of pageds
€
dr B =
μo
4π
Idr s × ˆ r
r2
B out of page
ds
dB
r
€
μo = 4π ×10−7 N / A2= permeability of free space
r = distance from current element
Field from very short section of current
€
dr s
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Vector cross product
Tue. Oct. 27, 2009 Physics 208 Lecture 16 3
€
rC
€
rD
€
rC ×
r D
€
dr B =
μo
4π
Idr s × ˆ r
r2
€
dr B =
μoI
4π r2dr s × ˆ r
Short length of current
Unit vector toward point at which field is evaluated
Dist. to point at which field is evaluated
€
dr B
€
dr s
€
rr
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Tue. Oct. 27, 2009 Physics 208 Lecture 16 4
Field from a circular loop
Each current element produce dB All contributions add as vectors Along axis, all
components cancelexcept for x-comp
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Tue. Oct. 27, 2009 Physics 208 Lecture 16 5
Magnetic field from loop of current
Looks like magnetic dipole
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Tue. Oct. 27, 2009 Physics 208 Lecture 16 6
Building a solenoid
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Tue. Oct. 27, 2009 Physics 208 Lecture 16 7
Solenoid: many current loops
€
Bsolenoid =μoNI
L= μonI
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Tue. Oct. 27, 2009 Physics 208 Lecture 16 8
Magnetic Force on a Current
S
N
I
Current
Magnetic field
Magnetic force
€
rF =
r I ×
r B L
€
qr v ×
r B Force on each charge
Force on length of wire
€
dr s
€
Idr s ×
r B
Force on straight section of wire, length L
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Tue. Oct. 27, 2009 Physics 208 Lecture 16 9
Quick Quiz
A current I flows in a square loop of wire with side length L.
A constant B field points in the x-direction, perpendicular to the plane of the loop. What is the net force on the wire loop?
x
y
I
I
I
I
L
A. 4LB
B. 2LB
C. LB
D. 0
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No force, but torque
Torque is Net torque can be nonzero even when
net force is zero.
Tue. Oct. 27, 2009 Physics 208 Lecture 16 10
€
rr ×
r F
Lever armForce
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12/09/2002 U. Wisconsin, Physics 208, Fall 2006 11
Which of these loop orientations has the largest magnitude torque? Loops are identical apart from orientation.
(A) a (B) b (C) c
Question on torque
a bc
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Quick QuizWhich of these different sized current loops has
the greatest torque from a uniform magnetic field to the right? All have same current.
Tue. Oct. 27, 2009 Physics 208 Lecture 16 12
L
W
L/2
2W
2L
W/2
A. B.
C. D. All same
€
rB
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Tue. Oct. 27, 2009 Physics 208 Lecture 16 13
Torque on current loop
€
rτ =
rr ×
r F
€
rτ =2
l
2F sinθ
⎛
⎝ ⎜
⎞
⎠ ⎟
€
F = IBl ⇒ τ = AIBsinθ
€
A = l 2 =loop area
B
F
B
I
€
rr
I
F
Torque proportional to
• Loop area
• Current
• sinθ
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Tue. Oct. 27, 2009 Physics 208 Lecture 16 14
Current loops & magnetic dipoles Current loop produces magnetic dipole field. Magnetic dipole moment:
€
rμ
€
rμ =IA
currentArea of loop
€
rμ
magnitude direction
In a uniform magnetic field
Magnetic field exerts torqueTorque rotates loop to align with
€
rτ =
rμ ×
rB ,
€
rμ
€
rB
€
rτ =
rμ
rB sinθ
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Works for any shape planar loop
Tue. Oct. 27, 2009 Physics 208 Lecture 16 15
I
€
rμ =IA
€
rμ perpendicular to loop
Torque in uniform magnetic field
€
rτ =
rμ ×
rB ,
€
rτ =
rμ
rB sinθ
Potential energy of rotation:
€
U = −r μ ⋅
r B = −μBcosθ
Lowest energy aligned w/ magnetic field
Highest energy perpendicular to magnetic field
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Tue. Oct. 27, 2009 Physics 208 Lecture 16 16
Magnetic flux Magnetic flux is defined
exactly as electric flux (Component of B surface) x (Area element)
€
ΦB = B • dA∫
zero flux Maximum flux
SI unit of magnetic flux is the Weber ( = 1 T-m2 )
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Tue. Oct. 27, 2009 Physics 208 Lecture 16 17
Magnetic Flux Magnetic flux Φ through a surface:
(component of B-field surface) X (surface area) Proportional to
# B- field lines penetrating surface
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Tue. Oct. 27, 2009 Physics 208 Lecture 16 18
Why perpendicular component? Suppose surface make angle surface normal
ΦB = BA cos ΦB =0 if B parallel A ΦB = BA (max) if B A
Flux SI units are T·m2=Weber
€
rB = B||
ˆ s + B⊥ˆ n
€
ˆ n
€
ˆ s
€
rA = A ˆ n
Component || surface
Component surface
Only component‘goes through’ surface
€
ΦM =r B •
r A
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Tue. Oct. 27, 2009 Physics 208 Lecture 16 19
Total flux E not constant
add up small areas where it is constant
Surface not flat add up small areas
where it is ~ flat
€
δΦBi = BiδA icosθ =
r B i • δ
r A i
Add them all up:
€
ΦB =r B • d
r A
surface
∫
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Tue. Oct. 27, 2009 Physics 208 Lecture 16 20
Magnetic flux
What is that magnetic flux through this surface?
A. Positive
B. Negative
C. Zero
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Tue. Oct. 27, 2009 Physics 208 Lecture 16 21
Properties of flux lines Net magnetic flux through any closed
surface is always zero:
€
Φmagnetic = 0
No magnetic ‘charge’, so right-hand side=0 for mag.
Basic magnetic element is the dipole
€
Φelectric =Qenclosed
εo
For electric charges, and electric flux
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Tue. Oct. 27, 2009 Physics 208 Lecture 16 22
Time-dependent fields Up to this point, have discussed only magnetic
and electric fields constant in time. E-fields arise from charges B-fields arise from moving charges (currents)
Faraday’s discovery
Another source of electric field Time-varying magnetic field creates
electric field
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Tue. Oct. 27, 2009 Physics 208 Lecture 16 23
Measuring the induced field
A changing magnetic flux produces an EMF around the closed path.
How to measure this? Use a real loop of wire for the closed path.
The EMF corresponds to a current flow:
€
ε=IR