Lecture 23: Outline
Hour 1:
Concept Review / Overview
PRS Questions – possible exam questions
Hour 2:
Sample Exam
Yell if you have any questions
7:30-9 pm Tuesday
P23 - 1
Exam 2 Topics
• DC Circuits
• Current & Ohm’s Law (Macro- and Microscopic)
• Power
• Kirchhoff’s Loop Rules
• Charging/Discharging Capacitor (RC Circuits)
• Magnetic Fields
• Force due to Magnetic Field (Lorentz Force)
• Magnetic Dipoles
• Generating Magnetic Fields
• Biot-Savart Law & Ampere’s Law
P23 - 2
General Exam Suggestions
• You should be able to complete every problem
• If you are confused, ask
• If it seems too hard, you aren’t thinking enough
• Look for hints in other problems
• If you are doing math, you’re doing too much
• Read directions completely (before & after)
• Write down what you know before starting
• Draw pictures, define (label) variables
• Make sure that unknowns drop out of solution
• Don’t forget units!
P23 - 3
What You Should Study
•
•
•
•
•
•
Review Friday Problem Solving (& Solutions)
Review In Class Problems (& Solutions)
Review PRS Questions (& Solutions)
Review Problem Sets (& Solutions)
Review PowerPoint Presentations
Review Relevant Parts of Study Guide
(& Included Examples)
P23 - 4
Current & Ohm’s Law
dQ
I=
dt
Ohm’s Laws
G
G
E = ρJ = 1
G I
J ≡ Iˆ
A
G
J
( σ)
∆V = IR
R=
ρA
A
P23 - 5
Series vs. Parallel
Rs = R1 + R2
1
1
1
=
+
Cs C1 C2
1
1
1
= +
RP R1 R2
CP = C1 + C2
Series
• Current same
• Voltages add
Parallel
• Currents add
• Voltages same
P23 - 6
PRS Questions:
Light Bulbs
Class 10
P23 - 7
Current, Voltage & Power
Battery
Resistor
Capacitor
Psupplied = I ∆V = I ε
Pdissipated
Pabsorbed
∆V
= I ∆V = I R =
R
2
2
dQ Q
= I ∆V =
dt C
2
d Q
dU
=
=
dt 2C dt
P23 - 8
Kirchhoff’s Rules
I1 = I 2 + I 3
G G
∆V = − ∫ E ⋅ d s = 0
Closed
Path
P23 - 9
(Dis)Charging A Capacitor
dQ
I =±
dt
ε (1 − e
Q=C
Q
∑i ∆Vi = ε − C − IR = 0
Q final
τ
dQ
C − Q − RC
=0
dt
ε
ε
− t / RC
)
dQ
= e − t / RC
I=
P23 - 10
dt
R
General Comment: RC
All Quantities Either:
Value(t ) = Value Final (1 − e − t /τ )
Value(t ) = Value0 e − t /τ
τ can be obtained from differential equation
(prefactor on d/dt) e.g. τ = RC
P23 - 11
PRS Questions:
DC Circuits with Capacitors
Class 12
P23 - 12
Right Hand Rules
1. Torque: Thumb = torque,
Fingers show rotation
2. Feel:
Thumb = I,
Fingers = B,
Palm = F
3. Create: Thumb = I
Fingers (curl) = B
4. Moment: Fingers (curl) = I
Thumb = Moment (=B inside loop)
P23 - 13
Magnetic Force
G
G G
FB = qv × B
G
G G
dFB = Id s × B
G
G G
FB = I L × B
(
)
P23 - 14
PRS Questions:
Right Hand Rule
Class 14
P23 - 15
Magnetic Dipole Moments
G
G
µ ≡ IAnˆ ≡ IA
Generate:
Feel:
1) Torque to align with external field
2) Forces as for bar magnets
P23 - 16
Helmholtz Coil
Common Concept Question
Parallel (Helmholtz) makes
uniform field (torque, no force)
Anti-parallel makes zero, nonuniform field (force, no torque)
P23 - 17
PRS Questions:
Magnetic Dipole Moments
Class 17
P23 - 18
The Biot-Savart Law
Current element of length ds carrying current I
(or equivalently charge q with velocity v)
produces a magnetic field:
G
G µo q v x rˆ
B=
2
4π r
G
G µ 0 I d s × rˆ
dB =
2
4π r
P23 - 19
Biot-Savart: 2 Problem Types
I
I
P
I
I
P
Notice that r is the same
for every point on the
loop. You don’t really
need to integrate
(except to find path
length)
P23 - 20
G G
Ampere’s Law: ∫ B ⋅ d s = µ 0 I enc
.
B
Long
Circular
Symmetry
I
B
(Infinite) Current Sheet
X
X
X
X
X
X
X
X
X
Solenoid
=
2 Current
Sheets
B
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
Torus/Coax
P23 - 21
PRS Questions:
Making B Fields
Classes 14-19
P23 - 22
SAMPLE EXAM:
P23 - 23
Problem 1: Wire Loop
P
D
2D
A current flowing in the circuit pictured produces a magnetic field
at point P pointing out of the page with magnitude B.
a) What direction is the current flowing in the circuit?
b) What is the magnitude of the current flow?
P23 - 24
Solution 1: Wire Loop
P
D
2D
I
a) The current is flowing counter-clockwise, as shown above
b) There are three segments of the wire: the semi-circle, the
two horizontal leads, and the two vertical leads.
The two vertical leads do not contribute to the B field (ds || r)
The two horizontal leads make an infinite wire a distance D
from the field point.
P23 - 25
Solution 1: Wire Loop
For infinite wire use Ampere’s Law:
P
D
2D
I
For the semi-circle
use Biot-Savart:
G G
v∫ B ⋅ d s = µ0 I enc ⇒ B ⋅ 2π D = µ0 I
µ0 I
B=
2π D
G µ 0 I d sG × rˆ
dB =
4π r 2
G
D
r = and d s ⊥ rˆ
2
G
µ 0 I d s × rˆ
B = ∫ dB = ∫
4π r 2
µ0 I
µ0 I µ0 I
π r) =
=
=
2 (
4π r
4r
2D
P23 - 26
Solution 1: Wire Loop
Adding together the two parts:
P
D
µ0 I µ0 I µ0 I ⎛ 1 ⎞
B=
+
=
⎜1 + ⎟
2π D 2 D 2 D ⎝ π ⎠
2D
I
They gave us B and want I to make that B:
2 DB
I=
⎛ 1⎞
µ0 ⎜1 + ⎟
⎝ π⎠
P23 - 27
Problem 2: RC Circuit
R1
R2
i1
i3
ε
R3
i2
C
Initially C is uncharged.
1. When the switch is first closed,
what is the current i3?
2. After a very long time, how much
charge is stored on the capacitor?
3. Obtain a differential equation for
the charge on the capacitor
(Here only, let R1=R2=R3=R)
Now the switch is opened
4. Immediately after opening the
switch, what is i1? i2? i3?
5. How long before i2 falls to 1/e of
this initial value?
P23 - 28
Solution 2: RC Circuit
Initially C is uncharged → Looks like short
R1
R2
i1
i3
ε
→
i2
C
R1
R2
i1
i3
ε
R3
→
i2
ε
R3
Req = R3 +
1
1
1
+
R1 R2
R1 || R2
i3
R3
⇒ i3 =
ε
Req
P23 - 29
Solution 2: RC Circuit
After a long time, C is full → i2 = 0
R1
R2
i1
i3
ε
i2
C
R3
i1 = i3 =
→
R1
i1
ε
i3
i2=0
C
R3
ε
R1 + R3
R1
Q = CVC = C ( i1 R1 ) = Cε
R1 + R3
P23 - 30
Solution 2: RC Circuit
R
R
Kirchhoff’s Loop Rules
Left: − i3 R + ε − i1 R = 0
Right: − i R + ε − i R − q
i2
i1
ε
i3
C
+q
-q
3
Current: i3 = i1 + i2
R
2
c
=0
Want to have i2 and q only ( L − 2 R) :
0 = − ( i1 + i2 ) R + ε − i1 R + 2 ( i1 + i2 ) R − 2ε + 2i2 R + 2q
= 3i2 R − ε + 2q
dq
i2 = +
dt
c
c
→
dq ε
2q
=
−
dt 3R 3RC
P23 - 31
Solution 2: RC Circuit
Now open the switch.
R1
i1
ε
R2
i3
i3 = 0
i2
→
C
R1
i1
R2
i2
C
+q
-q
R3
Capacitor now like a battery, with:
R1
Q
VC = = ε
C
R1 + R3
VC
R1
1
i1 = −i2 =
=ε
R1 + R2
R1 + R3 R1 + R2
P23 - 32
Solution 2: RC Circuit
How long to fall to 1/e of initial current? The time constant!
R1
i1
R2
This is an easy circuit since it just
looks like a resistor and capacitor in
series, so:
i2
C
+q
-q
τ = ( R1 + R2 ) C
Notice that this is different than the charging time
constant, because there was another resistor in
the circuit during the charging
P23 - 33
Problem 3: Non-Uniform Slab
y
Consider the slab at left with
non-uniform current density:
G
x ˆ
J = Jo k
d
Find B everywhere
P23 - 34
Solution 3: Non-Uniform Slab
y
G
x ˆ Direction: Up on right, down on left
J = Jo k
d
G G
Inside: (at 0<x<d): B ⋅ d s = µ 0 I enc
B
v∫
B
A
x
G G
v∫ B ⋅ d s = BA + 0 + 0 + 0 x
G G
J0 x
µ 0 I enc = µ 0 ∫∫ J ⋅ dA = µ 0 ∫
Adx
d
0
2
J 0A x
= µ0
d 2
J0 x2
B = µ0
up
d 2
P23 - 35
Solution 3: Non-Uniform Slab
y
G
x ˆ Direction: Up on right, down on left
J = Jo k
d
G G
B ⋅ d s = µ 0 I enc
Outside: (x > d):
B
v∫
B
A
x
G G
v∫ B ⋅ d s = BA + 0 + 0 + 0 d
G G
J0 x
µ 0 I enc = µ 0 ∫∫ J ⋅ dA = µ 0 ∫
Adx
d
0
2
J 0A d
= µ0
d 2
B = 12 µ 0 J 0 d up
P23 - 36
Problem 4: Solenoid
R
X
X
X
X
X
X
X
X
X
X
X
X
A current I flows up a very long solenoid
and then back down a wire lying along its
axis, as pictured. The wires are negligibly
small (i.e. their radius is 0) and are
wrapped at n turns per meter.
a) What is the force per unit length
(magnitude and direction) on the straight
wire due to the current in the solenoid?
b) A positive particle (mass m, charge q)
is launched inside of the solenoid, at a
distance r = a to the right of the center.
What velocity (direction and non-zero
magnitude) must it have so that the field
created by the wire along the axis never
exerts a force on it?
P23 - 37
Solution 4: Solenoid
R
X
X
X
X
X
X
X
SUPERPOSITION
You can just add the two fields from each
part individually
a) Force on wire down axis
Since the current is anti-parallel to the field
produced by the solenoid, there is no force
(F=0) on this wire
X
X
X
X
X
b) Launching Charge q
The central wire produces a field that wraps
in circles around it. To not feel a force due to
this field, the particle must always move
parallel to it – it must move in a circle of
radius a (since that is the radius it was
launched from).
P23 - 38
Solution 4: Solenoid
R
X
X
X
X
X
X
X
X
X
X
X
X
b) Launching Charge q
So first we should use Ampere’s law to
calculate the field due to the solenoid:
G G
v∫ B ⋅ d s = Bl = µ0 NI
µ 0 NI
B=
= µ 0 nI up the solenoid
l
Now we just need to make a charge q move
in a circular orbit with r = a:
G
2
G G
FB = q v × B = qvB = m v
r
= mv
2
a
qBa q µ 0 nIa
v=
out of the page
=
m
m
P23 - 39
Problem 5: Coaxial Cable
X
a
b
c
Consider a coaxial cable of with inner
conductor of radius a and outer
conductor of inner radius b and outer
radius c. A current I flows into the
page on the inner conductor and out
of the page on the outer conductor.
What is the magnetic field everywhere
(magnitude and direction) as a
function of distance r from the center
of the wire?
P23 - 40
Solution 5: Coaxial Cable
b c
r
a
X
Everywhere the magnetic field is
clockwise. To figure out the
magnitude use Ampere’s Law:
G G
v∫ B ⋅ d s = µ0 I enc ⇒ B ⋅ 2π r = µ0 I enc
µ 0 I enc
⇒B=
2π r
Drawn for a < r < b
The amount of current penetrating our Amperian loop
depends on the radius r:
r ≤ a: I enc
2
r
=I 2
a
µ 0 Ir
clockwise
⇒ B=
2
2π a
P23 - 41
Solution 5: Coaxial Cable
b c
r
Remember: Everywhere
µ 0 I enc
B=
clockwise
2π r
a
X
a ≤ r ≤ b: I Encl
b ≤ r ≤ c: I Encl
⇒
µ0 I
=I ⇒ B=
clockwise
2π r
⎛ r 2 − b2 ⎞
= I ⎜1 − 2 2 ⎟
⎝ c −b ⎠
µ0 I ⎛ r 2 − b2 ⎞
B=
⎜1 − 2 2 ⎟ clockwise
2π r ⎝ c − b ⎠
P23 - 42
Solution 5: Coaxial Cable
b c
r
a
X
Remember: Everywhere
µ 0 I enc
B=
clockwise
2π r
r ≥ c: I Encl = 0
⇒
B=0
P23 - 43