Course Updates - phys.hawaii.eduvarner/PHYS272_Spr10/... · "Brute force" 2 ˆ dq dE k r r = r –...
Transcript of Course Updates - phys.hawaii.eduvarner/PHYS272_Spr10/... · "Brute force" 2 ˆ dq dE k r r = r –...
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Course Updateshttp://www.phys.hawaii.edu/~varner/PHYS272-Spr10/physics272.html
Reminders:
1) Assignment #8 will be able to do after today
2)Finish Chapter 28 today
3)Quiz next Friday
4)Review of 3 right-hand rules
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F qv B= ×rr r
B Force B Force (single charge):(single charge):
06
yz
x
BF
I
Lots of q:Lots of q:
F IL B= ×r rr
BvqFi
irrr
×∑=
BvqNF avgrrr
×=
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Straight wireStraight wire
Current in WireCurrent in Wire
Resulting B fieldResulting B field
II
:08
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Magnetic Dipole Moment
Area vectorMagnitude = AreaDirection uses R.H.R.
Magnetic Dipole moment
NIAμ ≡rr
19
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Current in WireCurrent in Wire
II
Consistent? Yes!
Resulting B fieldResulting B field
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Today is Ampere’s Law Day
"High symmetry"
IldB∫ =• 0μrv
´ I
Integral around a path …hopefully a simple one Current “enclosed” by that path
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Calculation of Electric Field
What are the analogous equations for theMagnetic Field?
• Two Ways to calculate
"Brute force"2
ˆdqdE k rr
=r
– Coulomb’s Law
"High symmetry"qSdE =•∫rr
0ε
– Gauss’ Law
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Calculation of Magnetic Field
• Two Ways to calculate
02
ˆ4μ I dl rdBπ r
×=r
r
´ I
– Biot-Savart Law (“Brute force”)
– Ampere’s Law (“High symmetry”)
These are the analogous equations
–AMPERIAN SURFACE/LOOP
IldB∫ =• 0μrv
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• Calculate field at distance Rfrom wire using Ampere's Law:
Ampere's Law simplifies the calculation thanks to symmetry of the current! (axial/cylindrical)
dl´ RI
• Choose loop to be circle of radius Rcentered on the wire in a plane ⊥ to wire. – Why?
• Magnitude of B is constant (function of Ronly)
• Direction of B is parallel to the path.
⇒πR
IμB2
0=
∫ =• )2( πRBldBrr
– Evaluate line integral in Ampere’s Law:– Current enclosed by path = I
IμπRB 02 =– Apply Ampere’s Law:
B-field of ∞ Straight Wire, revisited
X
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Arbitrary closed path about a straight current wire will satisfy
enclosedIldB 0μ=∫ ⋅rr
Arbitrary closed path outside a current wire will satisfy
0=∫ ⋅ ldBrr
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Two identical loops are placed in proximity to two identical current carrying wires.
A B
A B Same
For which loop is ∫ B • dl the greatest?→ →
Question 1:
A) B) C)
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Two identical loops are placed in proximity to two identical current carrying wires.
A B
A B Same
For which loop is ∫ B • dl the greatest?→ →
Question 1:
A) B) C)
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CB
B C Same
Now compare loops B and C. For which loop is ∫ B • dl the greatest?→ →
Question 2:
A) B) C)
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CB
B C Same
Now compare loops B and C. For which loop is ∫ B • dl the greatest?→ →
Question 2:
A) B) C)
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B Field Inside a Long Wire• Suppose a total current I flows
through the wire of radius a into the screen as shown.
• Calculate B field as a function ofr, the distance from the center of the wire.
x x x x xx x x x x x x x
x x x x x x x x xx x x x x x x x
x x x x x a
r
• B field is only a function of r ⇒ take path to be circle of radius r: ⇒ )π(2 rBldB =•∫
rr
• Current passing through circle: IarI 2
2
enclosed =
• Ampere's Law: enclosedoIμldB =•∫rr
⇒ 20
π2 arIμB =
B
• B field direction tangent to circles.
dl
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B Field of a Long Wire
• Inside the wire: (r < a)
• Outside the wire:( r > a )
r
B
a
0
2IBr
μπ
=
022
I rBa
μπ
=
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Question 3• Two cylindrical conductors each carry
current I into the screen as shown. The conductor on the left is solid and has radius R = 3a. The conductor on the right has a hole in the middle and carries current only between R = a andR = 3a. – What is the relation between the
magnetic field at R = 6a for the two cases (L = left, R = right)?(a) BL(6a)< BR(6a) (b) BL(6a)= BR(6a) (c) BL(6a)> BR(6a)
3aa
3a
I I
2a
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Question 3• Two cylindrical conductors each carry
current I into the screen as shown. The conductor on the left is solid and has radius R=3a. The conductor on the right has a hole in the middle and carries current only between R=a andR=3a.– What is the relation between the
magnetic field at R = 6a for the two cases (L=left, R=right)?
(a) BL(6a)< BR(6a) (b) BL(6a)= BR(6a) (c) BL(6a)> BR(6a)
3aa
3a
I I
2a
• Use Ampere’s Law in both cases by drawing a loop in the plane ofthe screen at R=6a• Both fields have cylindrical symmetry, so they are tangent to theloop at all points, thus the field at R=6a only depends on currentenclosed• Ienclosed = I in both cases
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Question 4• Two cylindrical conductors each carry
current I into the screen as shown. The conductor on the left is solid and has radius R = 3a. The conductor on the right has a hole in the middle and carries current only between R = a andR = 3a.
(a) BL(2a)< BR(2a) (b) BL(2a)= BR(2a) (c) BL(2a)> BR(2a)
• What is the relation between the magnetic field at R = 2a for the two cases (L = left, R = right)?
3aa
3a
I I
2a
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Question 4• Two cylindrical conductors each carry
current I into the screen as shown. The conductor on the left is solid and has radius R=3a. The conductor on the right has a hole in the middle and carries current only between R=a andR=3a.– What is the relation between the
magnetic field at R = 2a for the two cases (L=left, R=right)?
(a) BL(2a)< BR(2a) (b) BL(2a)= BR(2a) (c) BL(2a)> BR(2a)
3aa
3a
I I
2a
Again, field only depends upon current enclosed...
LEFT cylinder: RIGHT cylinder:
IIaaIL 9
4)3()2(2
2
==ππ ( )
( ) IIaaaaIR 8
3)3()2(
22
22
=−−=
ππ
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B Field of ∞ Current Sheet• Consider an ∞ sheet of current described
by n wires/length each carrying current iinto the screen as shown. Calculate the Bfield.
• What is the direction of the field?• Symmetry ⇒ vertical direction
• Calculate using Ampere's law for a square ofside w:
• BwBwBwldB 200 =+++=•∫rr
• nwiI =therefore, IμldB 0=•∫
rr⇒
20niμB =
xxxxx
x
xx
xxxx
constant constant
w
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B Field of an ideal Solenoid• A constant magnetic field can (in principle) be
produced by an ∞ sheet of current. In practice, however, a constant magnetic field is often produced by a solenoid.
• If a << L, the B field is to first order contained within the solenoid, in the axial direction, and of constant magnitude. In this limit, we can calculate the field using Ampere's Law.
• A solenoid is defined by a current i flowing through a wire that is wrapped n turns per unit length on a cylinder of radius a and length L.
L
a
• To correctly calculate the B-field, we should use Biot-Savart, and add up the field from the different loops.
Ideal Solenoid
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B Field of an ∞ Solenoid• To calculate the B field of the solenoid using Ampere's
Law, we need to justify the claim that the B field is nearly 0 outside the solenoid (for an ∞ solenoid the B field is exactly 0 outside).
• The fields are in the same direction in the region between the sheets (inside the solenoid) and cancel outside the sheets (outside the solenoid).
• To do this, view the ∞ solenoid from theside as 2 ∞ current sheets.
x x x x x x x x x x x
• • • • • • • • • • • • • •
• Draw square path of side w:BwldB =•∫
rr
nwiI = ⇒ niμB 0=
x x x x x x x x
• • • • • • • • • • •Amp
LengthB ∝Note:
• B is uniform inside solenoid and zero outside.
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Which direction does the magnetic field point inside the iron core?
a) left b) right c) up d) down
e) out of the screen
A current carrying wire is wrapped around an iron core, forming an electro-magnet.
Question 5:
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Which direction does the magnetic field point inside the iron core?
a) left b) right c) up d) down
e) out of the screen
A current carrying wire is wrapped around an iron core, forming an electro-magnet.
Question 5:
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Which side of the solenoid should be labeled as the magnetic north pole?
A current carrying wire is wrapped around an iron core, forming an electro-magnet.
a) right
b) left
Question 6:
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Which side of the solenoid should be labeled as the magnetic north pole?
A current carrying wire is wrapped around an iron core, forming an electro-magnet.
a) right
b) left
Question 6:
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Use the wrap rule to find the B-field: wrap your fingers in the direction ofthe current, the B field points in the direction of the thumb (to the left).Since the field lines leave the left end of solenoid, the left end is the north pole.
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SolenoidsThe magnetic field of a solenoid is essentially
identical to that of a bar magnet.
The big difference is that we can turn the solenoid on and off ! It attracts/repels other permanent magnets; it attracts ferromagnets, etc.
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Solenoid ApplicationsDigital [on/off]:
– Doorbells
– Power door locks– Magnetic cranes– Electronic Switch “relay”
Close switchcurrentmagnetic field pulls in plungercloses larger circuit
Advantage:A small current can be used to switch a much larger one– Starter in washer/dryer, car ignition, …
Magnet off plunger held in place by springMagnet on plunger expelled strikes bell
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Solenoid ApplicationsAnalog (deflection ∝ I
):– Variable A/C valves– Speakers
In fact, a typical car has over 20 solenoids!
Solenoids are everywhere!
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Toroid• Toroid defined by N total turns
with current i.• B=0 outside toroid! (Consider
integrating B on circle outside toroid)
• To find B inside, consider circle ofradius r, centered at the center of the toroid.
x
x
x
xx
xx
x
x x
xx x
x
x
x
•
••
•
• •
•
••
•
• •
•
•
••
r
B
)π(2 rBldB =•∫rr
NiI =
IμldB 0=•∫rr
πrNiμB
20=⇒Apply Ampere’s Law:
• Direction? tangent to circle.
• Magnitude depends on? r (not theta or z).
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ITER Tokamak; giant toroid for fusion
A joint US-Europe-Japan project to be built in southern Franceby 2016. A toroidalfield contains the hotplasma. Fusion shouldprovide clean power.ITER is prototype forfuture machines and issuppose to produce 500MW of power.
person
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Example B-field Calculations:
– Inside a Long Straight Wire
– Infinite Current Sheet
– Solenoid
– Toroid
– Circular Loop
Summary
B = μ0 I2 π
ra2
20niμB =
niμB 0=
πrNiμB
20=
Bz ≈μ0 iR2
2z 3
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RI
II
III
P
2
ˆ4
oIdl rdBr
μπ
×=r
r
dl dl
dl
On segments I and III, dl x r^ is zero
Make sure you understand why
2 24 4 4o o o
II
dl RdBR R R
μ μ π μπ π
= = =∫
What is the direction of the B field ?
By the right hand rule, it points into the paper
28.30
Is the field from a straight wire always zero ? No !!! See 28.76
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Simplified coaxial cable
bc
a
Current I flowing out of the page through inner conductor
Current I flowing into the page through outer conductor
28.37, 28.38
o CB dl Iμ• =∫rr
2 o cB r Iπ μ= For a<r<b
2oIBr
μπ
= For r>c, IC=+I-I =0hence B=0 for r>c
For a<r<b
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202
2 ˆ1 ( )I rJ ka aπ
⎡ ⎤= −⎢ ⎥⎣ ⎦
r
Find B for r>a
o cB dl Iμ• =∫rr
J is current density, so integrate dA to obtain current passing through
( ) ( )2a a
o o
I J r dA J r rdrπ= =∫ ∫
202
2 1 ( ) 2a
o
I rI rdra a
ππ
⎡ ⎤= −⎢ ⎥⎣ ⎦∫
28.77
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2 2 4 21 1(4 / ) /2 4
a
o oo
I I a r r a I⎡ ⎤= − =⎢ ⎥⎣ ⎦
0(2 ) o C oB r I Iπ μ μ= =
/ 2oB I rμ π= Outside the cylinder
How should this calculation be modified inside the cylinder r<a ?
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More weekend fun?• HW #8 get cracking – due Monday
• Office Hours immediately after this class (9:30 –10:00) in WAT214 (1:30-2/1-1:30 M/WF)
• Next week: Chap 29 [Midterm 2 is Chap 25 – 29]
• Quiz next Friday