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Chapter 31 - Induction and Inductance
Problem Set #10 - due: Ch 31 - 1, 5, 11, 28, 35, 43, 51, 59, 70, 74, 83, 100
Lecture Outline
- Faraday's Law of Induction
- Motors and Generators
- The Meissner Effect and Superconductivity
- The Definition of Mutual Inductance
- The Definition of Self Inductance
- The LR Circuit
- Energy Storage in Inductors and B-fields
Magnet, wire, coil, & galvanometer (moving the coil vs. moving the magnet - it can't matter!)
1. Faraday's Law of Induction
Moving a wire through the field causes the charges within the wire to feel an upward magnetic force,
r F = q
r υ ×
r B ⇒ F = qυB. Positive charges accumulate at the top of the wire and negative charges at the bottom. This creates a downward electric field in the wire. The net force on the charges is given by the Second Law, ΣF = ma ⇒ qυB − qE = ma. Charges move upward until a=0. The electric field is then, E = υB.
If the wire was rolling along some rails connected to a voltmeter, the meter would give a reading due to the electric field in the wire. This voltage can be found from the E-field, ∆V = −
r ∫ E^ • dr^ s^ ⇒^ V^ =^ El^ ⇒^ V^ =^ Blυ^. Moving through a magnetic field creates a potential difference just like a battery would.
Example 1: Estimate the induced voltage across the 40.0m wingspread of an airplane traveling 800km/h (222m/s) perpendicular to the earth’s magnetic field of 50.0μT.
Using the Second Law when the charges in the wing have reached equilibrium, ΣF = ma ⇒ qυB − qE = 0 ⇒ E = υB. The induced voltage can be found from the E-field, ∆V = −
r ∫ E^ • dr^ s^ ⇒^ V^ =^ El^ ⇒^ V^ =^ Blυ^. V = Blυ ≈ (50.0μT)(40.0m)(222m/s) ≈ 0.444V
υ Fe
2
Fm
++
V (^) υ l
F
υ
Fe
2
Fm
++
The other way to look at this is to think about it in terms of the changing magnetic flux in the loop. The potential difference between the ends is, V = Blυ. Using the definition of
velocity, V = Bl dx dt
= B d dt
( lx) = B dA
dt
= d dt
( BA) = dΦB
dt
This explains why moving the magnet creates a voltage just like moving the wire does. If the wire has N turns then the voltage is N times bigger. Notice that the current flows in a direction such as to fight the change in the field. This is called "Lenz's Rule." Putting all this together gives Faraday's Law,
ε = −N dΦB dt
Faraday's Law
where we have defined the magnetic flux as,
ΦB ≡
r B • d
r ∫ A The Definition of Magnetic Flux
Beyond the Mechanical Universe (vol. 37 Ch 11,12,15,18,21,28)
Example 2: A small loop of N turns and area A is in the same plane as a long straight wire carrying a current i = io sinωt_. Find the induced voltage in the loop as a function of time and show that the peak-to-peak voltage is proportional to the peak magnetic field._
From the previous chapter we know the field due to a long straight wire is, B^ =
μo i 2 πr , where r is the distance from the wire to the center of the loop. The loop is small enough that at any given time the field is approximately constant. Therefor, the magnetic flux through the small loop is, (^) ΦB ≡
r B • d
r ∫ A=^ ∫ BdA^ =^ B dA∫ =^ BA^ = μo^ i 2 πr
A.
Applying Faraday's Law, ε = −N d dt
μo i 2 πr
A = − N μo 2 πr
A di dt
= − μoNA 2 πr
d dt
(i (^) o sinωt) = − μoNA 2 πr
ioω cosωt = − μo^ io 2 πr
NAωcosωt
The peak-to-peak voltage must be twice the maximum when cosωt=1, εpp = 2 μoio 2 πr
NAω = 2Bo NAω where Bo = μoio 2 πr
is the peak field due to the wire.
This is how the magnetic field probe you used in lab works! You can now see that the peak-to- peak voltage is proportional to the field assuming the frequency is kept constant.
V (^) υ
dx
l
A
B i(t)
ε^2 ≡
0 ε^2 dt
T
0 dt
T
εo^2 0 sin^2 2 π f tdt
T
0 dt
T
εo^2 0 sin^2 udu
2 π
2 π f T
= εo
(^2) π 2 π
= εo
2 2
εrms ≡ ε^2 = εo
2 2
⇒ εrms = 12 εo = 12 NBA2 π f
Example 4: A coil of 1000 turns and 12.0cm radius flips 180˚ about an axis that points northward. The coil has a resistance of 4.80 Ω. The vertical component of Earth's magnetic field is 46.0μT. Find the total charge that flows when the coils flips. Notice that the horizontal component of Earth's field contributes no flux. All the flux through the loop is due to the vertical component.
Faraday's Law requires, ε = −N dΦB
dt
The charge that flows is related to Ohm's Rule, ε = IR = dq
dt
R.
Setting the voltages equal, dq dt
R = −N dΦB dt
⇒ Rdq = −NdΦB ⇒ R 0 dq
Q
∫ = −^ N^ Φ odΦB
Φ
∫ ⇒^ RQ^ = −N^ (^ Φ− Φo)
The initial flux is just the product of the vertical component of the field and the area. The final flux
in just the opposite of the initial flux, RQ = −N ( −Bvπr^2 − Bvπr^2 ) = 2NBvπr^2.
⇒ Q = 2NBvπr
2 R
= 2(1000)(46.0x
− (^6) )π(0.120) 2
= 8.67x10−^4 C.
3. The Meissner Effect and Superconductivity
picture, floating magnet toy, coil & power supply, high-TC superconductor
Why does a floating magnet imply superconductivity? Two magnets repel each other, but the lower magnet can be replaced with a current carrying coil. The floating magnet implies a current in the superconductor. Since there is no power source, and the current lasts indefinitely, the material is conducting without resistance and is therefor a superconductor.
How does the current get there to begin with? If you bring the magnet toward the superconductor, Faraday's Law explains the induced current just like moving the magnet around near a coil. In ordinary conductors these currents die away due the resistance. The currents in the superconductor continue.
The Meissner Effect Set the magnet on a warm chunk of superconducting material and nothing happens because the material is only superconducting at low temperatures. When you cool the sample, the magnet lifts off and hovers above the superconductor. This is not explained by Faraday's Law, but it is called "The Meissner Effect." It is caused by the fact that superconductors exclude magnetic fields just like ordinary conductors exclude electric fields. In order for the superconductor to keep the field inside zero, it must set up a current that cancels out the field due to the magnet. These opposing fields cause the magnet to hover.
B
Bh
Bv
N
Superconductor
current
S
N
4. The Definition of Mutual Inductance
If a coil carrying a current I 1 is near a coil with N 2 turns, the magnetic field caused by the first coil will create a flux through the second coil. If the current in the first coil changes, the flux through the second coil will change and according to Faraday's Law a voltage will be induced in the second coil. The relationship between the voltage induced in the second coil and the rate of change of current in the first coil is called the "Mutual Inductance."
To find voltage induced in the second coil use Faraday's Law, ε 2 = −N 2 dΦ^21 dt
The flux through coil 2 caused by coil 1 can be found from the definition of flux, Φ 21 =
r B 1 • d
r A 2 coil 2
∫.
The field caused by coil 1 can be found by the Biot-Savart Rule,
r B 1 = μoI^1 4 π
dr s × ˆr coil 1^ ∫^ r^2
The flux can now be written, Φ 21 = μoI^1 4 π
dr s × ˆr coil 1^ ∫^ r^2
r A 2 coil 2
∫ =^
I 1
N 2
M 21
where M 21 = μoN^2 4 π
dr s × ˆr coil 1^ ∫^ r^2
r A 2 coil 2
∫. Notice that M^21 only depends on the shapes of the coils and
the relative positions not on the current in coil 1 or on time.
Putting the flux into Faraday's Law, ε 2 = −N 2 d dt
I 1
N 2
M 21
⇒ ε 2 ≡ −M 21 dI^1 dt
. This equation is
usually called the definition of mutual inductance,
ε 2 ≡ −M 21 dI^1 dt
The Definition of Mutual Inductance
5. The Definition of Self Inductance
Beyond the Mechanical Universe (vol. 37 Ch 25)
There is no reason that coil 1 and coil 2 have to be separate. They could be part of a solenoid for instance. In this case we call the quantity that relates the induced voltage to the rate of change of current the "Self Inductance" and we use the symbol L instead of M.
ε ≡ −L dI The Definition of Self Inductance dt
I 1 N^2
The current as a function of time for a "charging" LR circuit is shown at the left. The word charging is used, but keep in mind that the inductor is storing flux (or field, or energy) not charge like a capacitor would.
For a "discharging" LR circuit the initial current would be large enough for all the voltage to be dropped across the resistor. As the resistor removes energy from the circuit the current begins to drop toward zero. The loop theorem can again be used to find the current as a function of time. The result is,
I = Ioe
− RL t "Discharging" LR Circuit
The graph of current versus time is shown at the left.
Example 6: The inductor from example 5 is made of copper. Find (a)its resistance, (b)the equilibrium current when it is connected to a 1.50V battery and (c)the time required for it to reach 99% of this equilibrium current. (a)Using the definition of resistance, R ≡ ρ l A
= ρ πDN πa^2
where a is the
radius of the wire. Assuming the coils are close packed 2aN = l where l is the length of the solenoid. Solving, R = ρ π/^ DN π / (^) ( (^) 2Nl)
2 =^
4 ρDN^3 l^2
= 4(1.7x
(0.100)^2
(b)After a while the current will build up. It can grow until all the voltage is dropped across the resistor. According to Ohm's Rule, ε = Io R ⇒ Io = ε R
= 1.50V
= 22.1mA.
(c)Starting with the equation for a charging LR circuit I = Io 1 − e
− RL t
and solving for t,
I Io
= 1 − e
− RL t ⇒ e
− RL t = 1 − I Io
⇒ − R
L
t = ln 1 − I Io
⇒ t = − L R
ln 1− I Io
Putting in the numbers, t = − 9.87x
− 4
ln 1( − 0.99) = 66.8μs
I
t
I o
I
I o
t
L
ε (^) Rint
7. Energy Storage in Inductors and B-fields
The energy stored in the inductor can be found by starting with the loop theorem, V (^) L + VR + V (^) C = 0 ⇒ L dI dt
+ IR + Q
C
Multiplying through by the current, LI dI dt
+ I^2 R + I Q
C
= 0. The middle term is the power consumed by the resistor. It's a good guess to assume the other terms are the power flow into the capacitor and the inductor. Since we know the answer for the capacitor let's start there, P (^) C = I Q C
⇒ dUC dt
= dQ dt
Q
C
⇒ 0 dU (^) C
UC ∫ =^
C
0 QdQ
Q ∫ ⇒^ UC^ =^
1 2
Q^2
C
Since this is the right answer, we should play the same game with the inductance term,
P (^) L = LI dI dt
⇒ dUL dt
= LI dI dt
⇒ 0 dU (^) L
UL ∫ =^0 LIdI
I ∫ ⇒^ UL^ =^
1 2 LI
U (^) L = 12 LI^2 Energy Storage in an Inductor
Example 7: For the inductor circuit of example 6 (L=987μH, R=68.0 Ω , ε =1.50V) find the energy stored when the current reaches equilibrium. From example 2 the equilibrium current is 22.1mA. Using the energy stored in the inductor, U (^) L = 12 LI^2 = 12 (987x10−^6 )(0.0221)^2 = 0.241μJ.
If we associate this energy with the magnetic field created by the inductor instead of associating it with the
inductor itself, we can find the energy density in magnetic fields, um ≡ U vol
1 2 LI
2 1 4 πD
(^2) l. Using the
inductance of a solenoid found in example 1,
L = μoN
(^2) πD 2 4 l
⇒ um =
1 2
μoN^2 πD^2 4 l
I
2
1 4 πD
(^2) l =
1 2 μoN
2 I 2
l^2
= (^2) μ^1 o^ μo^ NI l
2 .
The stuff in parentheses is the magnetic field due to the solenoid so,
um = B
2 2 μo
Energy Density in a Magnetic Field
Recall the energy density in the electric field is given by the similar expression, ue = 12 εoE^2.
Example 8: In some region of space the electric field and the magnetic field give equal contributions to the energy density. Find the ratio of E to B. Given, um = ue ⇒ B
2 2 μo^ =^
1 2 εoE
2 ⇒ E^2
B^2 =^
μoεo
⇒ E
B
μoεo
Putting in the numbers, 1 μoεo
(4πx10−^7 )(8.85x10−^12 )
= 3.00x10^8 m / s.
The speed of light! Hmmm.
L
C
R
