Chapter 33 & 34 Review
Chapter 33: The Magnetic Field
Cyclotron Motion
r r
F  q(v  B)
rcyc
Newton’s Law
r r
ma = F
Both a and F directed toward
center of circle
r
v2
a=
rcyc
r
F = qvB
Equate
v
qB
W=
=
= 2p fcyc
rcyc
m
v
v
rcyc = =
W 2p fcyc
What is the force on those N
electrons?
Force on a single electron
r
r r
F = - e v¥ B
(
)
Force on N electrons
Now use
r
eN v = I l
r
r
Lets make l a vector - eNv = I l
Force on wire segment:
r
r r
F = - eN v¥ B
(
)
points parallel to the wire in
the direction of the current
when I is positive
r r
r
Fwire = I l ¥ B
r r
r
Fwire = I l ¥ B
Comments:
Force is perpendicular to both B and l
Force is proportional to I, B, and length of line segment
Superposition: To find the total force on a wire you must break
it into segments and sum up the contributions from each
segment
r
Ftotal = I
Â
segments- i
r r r
li ¥ B(ri )
Electric Field
r
E(r) 
Magnetic Field
r
 0 qv  r̂
r
B(r) 
4 r 2
q
r̂
4 0 r 2
r̂
v
q
r
What are the magnitudes
and directions of the
electric and magnetic
fields at this point?
Assume q > 0
Comparisons: both go like r-2, are proportional to q, have 4 in
the denominator, have funny Greek letters
Differences: E along r, B perpendicular to r and v
Magnetic Field due to a current
r
 0 qv  r̂
r
B(r) 
4 r 2
Magnetic Field due to a single charge

r
B(r)  0
4
If many charges use superposition
r
qi v i  r̂i
 r2
ch arg esi
i
r̂1
r̂2
r3
r̂3
r2
#3 q3 , v3
#2 q2 , v2
#1 q1, v1
r1
Where I want to know what B is
For moving charges in a wire, first sum over charges in each
segment, then sum over segments


r
0
q
v

r̂
0
r
i i
i
 
B(r) 


4 sgments j  ch arg es in
ri 2  4
 each segment i


ch arg es in
each segment i
r
r
l j  r̂j
qi v i  r̂i

I
ri 2
rj 2

sgments j
I
r
l j  r̂j
rj 2
Summing over segments - integrating along curve
I
r
dl
r
r̂
r
r

r
B(r)  0
4
Biot Savart law

I
r
l j  r̂j
sgments j
rj 2
r

dl  r̂
 0—
I
4 
r2
Integral expression looks simple but…..you have to keep track
of two position vectors
r
r
which is where you want to know B
which is the location of the line segment that is contributing to
B. This is what you integrate over.
Magnetic field due to an infinitely long wire
z
Current I flows along z axis
r
0
dl  r̂
r
B(r) 
I
4 
r2
I
I want to find B at the point
r = xi  0 j  0k
r̂
x
r r
r = r - r¢ =
r
dl = dz¢k
r r
r - r¢
xi - z¢k
r̂ = r r =
r - r¢
x 2 + z¢2
x 2 + z¢2
r  = 0i  0 j  zk
I will sum over segments at points
r
0 I
B 
2 r
compare with E-field for a line charge

E 
2 0 r
Biot-Savart Law
implies Gauss’ Law
and Amperes Law
Gauss’ Law:
q r
r
B(r)  0 12 v  r̂
4 r
r
—
 B  dA  0
Ampere’s Law
r r
B(
r)
—
  ds =0 Ithrough
But also, Gauss’ law and
Ampere’s Law imply the Biot Savart law
 0 q1 r
r
B(r) 
v  r̂
2
4 r
Magnetic field due to a
single loop of current
Electric field due to a single charge
Guassian surfaces
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Graphics decompressor
are needed to see this picture.
r
—
 B  dA  0
r Qin
—
 E  dA 
0
r
B(r)
r
E(r)
r
ds
r
ds
r
ds
r r
—
 B(r)  ds  0 I
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are needed to see this picture.
r r r
0—
 E(r)  ds
r r
—
 B(r)  ds =0 Ithrough
r r
B(
r)
ds =Bl
—

0 I through  0 NI
B
0 NI
l
  0 (N / l)I
# turns per unit length
Biot-Savart Law
implies Gauss’ Law
and Amperes Law
Gauss’ Law:
q r
r
B(r)  0 12 v  r̂
4 r
r
—
 B  dA  0
Ampere’s Law
r r
B(
r)
—
  ds =0 Ithrough
But also, Gauss’ law and
Ampere’s Law imply the Biot Savart law
 0 q1 r
r
B(r) 
v  r̂
2
4 r
Chapter 34: Faraday’s Law of Induction
Faraday’s Law for Moving Loops
r r r
r
r r
d
d
E  v  B  dS  -   
B  dA

dt
dt Area
—
 
loop

Magnetic Flux
r r
   B  dA
S
Some surface
Remember for a closed surface   0
r
—
 B  dA  0
Closed surface
r
dA
B
Open surface
Magnetic flux measures how much magnetic field passes
through a given surface
Rectangular surface in a constant magnetic field. Flux depends on
orientation of surface relative to direction of B
r
Suppose the rectangle is oriented do that B and dA
are parallel
r r
r
r
   B  dA  B A  B ab
S
Lenz’s Law
In a loop through which there is a change in magnetic flux, and
EMF is induced that tends to resist the change in flux
What is the direction of the
magnetic field made by the
current I?
A. Into the page
B. Out of the page
Reasons Flux Through a Loop Can Change
r r
d
d

B  dA

dt
dt Area
A. Location of loop can change
B. Shape of loop can change
C. Orientation of loop can change
D. Magnetic field can change
Faraday’s Law for Moving Loops
r r r
r
r r
d
d
E  v  B  dS  -   
B  dA

dt
dt Area
—
 
loop

Faraday’s Law for Stationary Loops
r
r r
B r
—
 loop E  dS   Area
 t  dA
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I
VB
VR
R
I
VL
L
Now I have cleaned things up
making use of
IB=-IR, IR=IL=I.
Now use device laws:
VR = RI
VL = L dI/dt
KVL: VL + VR - VB = 0
dI
L + RI - VB = 0
dt
This is a differential
equation that determines
I(t). Need an initial
condition I(0)=0
dI(t)
L
+ RI(t)- VB = 0, I (0) = 0
dt
Solution:
VB
- t /t
I(t) =
1e
(
)
R
t = (L / R)
This is called the “L over R”
time.
Approaches a value VB/R
Current starts at zero
Let’s verify
What is the voltage across the resistor and the inductor?
VB
I(t) =
1- e- t /t )
(
R
VR = RI(t) = VB (1- e- t /t )
dI
VL = L = VB e- t /t
dt
I
VB
VR
R
I
VL
L
Initially I is small and VR is
small.
All of VB falls across the
inductor, VL=VB.
Inductor acts like an open circuit.
Time asymptotically I stops
changing and VL is small.
All of VB falls across the resistor,
VR=VB. I=VB/R
Inductor acts like an short
circuit.
Let’s take a special case of no current initially flowing
through the inductor
Initial
charge on
capacitor
d 2V (t) V (t)
+
= 0
2
dt
LC
Solution
A:
V (0) = VC (0)
dV
I L (0) = 0 = - C
dt
V (t) = VC (0)cos(wt)
w = 1 / LC
B:
t= 0
V (t) = VC (0)sin(wt)
Current through Inductor and Energy Stored
Energy
1
2
U C = CV
2
1 2
U L = LI L
2
t
+
Foolproof sign convention for two terminal devices
I
V2
1. Label current going in one
terminal (your choice).
2. Define voltage to be potential at
that terminal wrt the other
terminal
V= V2 -V1
V1
KVL Loop
Contribution to
voltage sum = +V
3. Then no minus signs
V = RI
dI
V= L
dt
dV
I= C
dt
Power to device
P = VI