Chapter 33 & 34 Review. Chapter 33: The Magnetic Field.
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Transcript of Chapter 33 & 34 Review. Chapter 33: The Magnetic Field.
![Page 1: Chapter 33 & 34 Review. Chapter 33: The Magnetic Field.](https://reader033.fdocuments.net/reader033/viewer/2022052312/5697c0251a28abf838cd55ae/html5/thumbnails/1.jpg)
Chapter 33 & 34 Review
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Chapter 33: The Magnetic Field
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Cyclotron Motion
Newton’s Law mra =
rF
Both a and F directed toward center of circle
ra =
v2
rcyc
rF = qvB
rF =q(
rv×
rB)
rcyc
Equate
W=
v
rcyc=qB
m= 2p fcyc
rcyc =
v
W=
v
2p fcyc
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What is the force on those N electrons?
Force on a single electron
rF = - e
rv¥
rB( )
Force on N electrons
rF = - eN
rv¥
rB( )
Now use eN
rv = I l
Lets make l a vector - eNrv = I
rl
points parallel to the wire in the direction of the current when I is positive
Force on wire segment:
rFwire = I
rl ¥
rB
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rFwire = I
rl ¥
rB
Comments:Force is perpendicular to both B and lForce 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
rFtotal = I
rli ¥
rB(
rri )
segments- iÂ
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rB(
rr) =
μ0
4πqrv×r̂r2
rE(
rr) =
q4πε0
r̂r2
Electric Field Magnetic Field
q
r̂
rv
What are the magnitudes and directions of the electric and magnetic fields at this point? Assume q > 0
r
Comparisons: both go like r-2, are proportional to q, have 4π in the denominator, have funny Greek letters
Differences: E along r, B perpendicular to r and v
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Magnetic Field due to a current
Magnetic Field due to a single charge
rB(
rr) =
μ0
4πqrv×r̂r2
If many charges use superposition
rB(
rr) =
μ0
4πqi
rvi ×r̂i
ri2
charges−i∑
#2 q2 , v2
#1 q1, v1
#3 q3 , v3
r3
r2
r1
Where I want to know what B is
r̂3
r̂2
r̂1
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For moving charges in a wire, first sum over charges in each segment, then sum over segments
rB(
rr) =
μ0
4πqi
rvi ×r̂i
ri2
chargesineachsegment−i
∑⎛
⎝
⎜⎜
⎞
⎠
⎟⎟sgments−j
∑ =μ0
4πI
rlj ×r̂j
rj2
sgments−j∑
qirv i ×r̂i
ri2
chargesineachsegment−i
∑ =I
rlj ×r̂j
rj2
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Summing over segments - integrating along curve
rB(
rr) ==
μ0
4πI
rlj ×r̂j
rj2
sgments−j∑ →
μ0
4πIdrl ×r̂r2—∫
drl
r̂
I r
Integral expression looks simple but…..you have to keep track of two position vectors
Biot Savart law
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.
rr
r′r
rr
r′r
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Magnetic field due to an infinitely long wire
x
z
I
drl = d ¢z k
r̂
Current I flows along z axis
I want to find B at the point
rB(
rr) ==
μ0
4πIdrl ×r̂r2∫
rr = xi + 0 j + 0k
I will sum over segments at points r′r = 0i + 0 j + ′z k
r̂ =
rr -
r¢r
rr -
r¢r
=xi - ¢z k
x2 + ¢z 2
r =
rr -
r¢r = x2 + ¢z 2
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rB =
μ0 I2πr
r
compare with E-field for a line charge
rE =
λ2πε0r
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rB(
rr) =
μ0q1
4πr2
rv×r̂
Gauss’ Law:
rB⋅d
rA—∫ =0
rB(
rr)⋅d
rs=—∫ μ0 I through
Biot-Savart Law implies Gauss’ Law and Amperes Law
But also, Gauss’ law and Ampere’s Law imply the Biot -Savart law
Ampere’s Law
rB(
rr) =
μ0q1
4πr2
rv×r̂
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Electric field due to a single chargeMagnetic field due to a single loop of current
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Guassian surfaces
rB⋅d
rA—∫ =0
rE⋅d
rA—∫ =
Qin
ε0
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drs
rE(
rr)
drs
drs
rB(
rr)
0 =
rE(
rr)⋅d
rs—∫
rB(
rr)⋅d
rs—∫ =μ0 I
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rB(
rr)⋅d
rs=—∫ μ0 I through
rB(
rr)⋅d
rs=—∫ Bl
μ0Ithrough = μ 0NI
B=μ0NI
l=μ0 (N / l)I
# turns per unit length
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rB(
rr) =
μ0q1
4πr2
rv×r̂
Gauss’ Law:
rB⋅d
rA—∫ =0
rB(
rr)⋅d
rs=—∫ μ0 I through
Biot-Savart Law implies Gauss’ Law and Amperes Law
But also, Gauss’ law and Ampere’s Law imply the Biot -Savart law
Ampere’s Law
rB(
rr) =
μ0q1
4πr2
rv×r̂
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Chapter 34: Faraday’s Law of Induction
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loop
rE +
rv×
rB( )—∫ ⋅d
rS=-
ddt
Φ =−ddt
rB⋅d
rA
Area∫
Faraday’s Law for Moving Loops
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Magnetic Flux
Φ =
rB ⋅d
rA
S∫
Some surface
Remember for a closed surface Φ =0
rB d
rA
Magnetic flux measures how much magnetic field passes through a given surface
Open surface
rB⋅d
rA—∫ =0
Closed surface
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Suppose the rectangle is oriented do that are parallel
Φ =
rB ⋅d
rA
S∫ =
rB A =
rB ab
Rectangular surface in a constant magnetic field. Flux depends on orientation of surface relative to direction of B
rB and d
rA
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Lenz’s LawIn 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 pageB. Out of the page
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Reasons Flux Through a Loop Can Change
A. Location of loop can change
B. Shape of loop can change
C. Orientation of loop can change
D. Magnetic field can change
d
dtΦ =
ddt
rB⋅d
rA
Area∫
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loop
rE +
rv×
rB( )—∫ ⋅d
rS=-
ddt
Φ =−ddt
rB⋅d
rA
Area∫
Faraday’s Law for Moving Loops
Faraday’s Law for Stationary Loops
loop
rE⋅d
rS—∫ ==−
∂rB∂t
⋅drA
Area∫
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L
VB
R
VRVL
I I Now I have cleaned things up making use of IB=-IR, IR=IL=I.
Now use device laws:VR = RIVL = L dI/dt
KVL: VL + VR - VB = 0
LdI
dt+ RI - VB = 0
This is a differential equation that determines I(t). Need an initial condition I(0)=0
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LdI (t)
dt+ RI (t)- VB = 0, I (0) = 0
Solution:
I(t) =VBR
1- e- t /t( )
t = (L / R)
Let’s verify
This is called the “L over R” time.
Current starts at zero
Approaches a value VB/R
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What is the voltage across the resistor and the inductor?
I(t) =VBR
1- e- t /t( )
VR = RI(t) = VB 1- e- t /t( )
VL = LdI
dt= VBe
- t /t
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L
VB
R
VRVL
I I 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/RInductor acts like an short circuit.
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Let’s take a special case of no current initially flowing through the inductor
d 2V (t)
dt 2+V (t)
LC= 0
V (0) = VC (0)
IL (0) = 0 = - CdV
dt t= 0
Solution
V (t) = VC (0)cos(wt)
w= 1 / LC
A:
B: V (t) = VC (0)sin(wt)
Initial charge on capacitor
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Current through Inductor and Energy Stored
Energy
UC =1
2CV 2
UL =1
2LIL
2
t
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+
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IV2
V1
Foolproof sign convention for two terminal devices
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
3. Then no minus signs
V = RI
V = L
dI
dt
I = C
dV
dt
P = VI
Power to deviceKVL Loop
Contribution to voltage sum = +V