Download Lecture12: Inductance, RL Circuits

Physics 121 - Electricity and Magnetism
Lecture 12 - Inductance, RL Circuits
Y&F Chapter 30, Sect 1 - 4
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Inductors and Inductance
Self-Inductance
RL Circuits – Current Growth
RL Circuits – Current Decay
Energy Stored in a Magnetic Field
Energy Density of a Magnetic Field
Mutual Inductance
Summary
Copyright R. Janow – Fall 2015
Induction: basics
•
Magnetic Flux:

 
dB  B  dA  B  n̂dA

B
•
Faraday’s Law: A changing magnetic
flux through a coil of wire induces an
EMF in the wire, proportional to the
number of turns, N.
•
Lenz’s Law: The current driven by an induced EMF creates
an induced magnetic field that opposes the flux change.
Eind  N
n̂
dB
dt
Bind & iind oppose changes in B
•
Induction and energy transfer: The forces on the loop
oppose the motion of the loop, and the power required to
sustain motion provides electrical power to the loop.
 
P  F  v  Fv
P  i
  Blv
•
Transformer principle: changing current i1 in primary
induces EMF and current i2 in secondary coil.
•
Generalized Faraday Law:
A
changing magnetic flux creates
non-conservative electric field.
 
   E  ds   N dB
dt
Copyright R. Janow – Fall 2015
Changing magnetic flux directly induces electric fields
A thin solenoid, cross section A, n turns/unit length
• zero B field outside solenoid
B   0In
• inside solenoid:
Flux through a
conducting loop:
conducting loop, resistance R
  BA   0nIA
Current I varies with time  flux varies  EMF is
induced in wire loop:
d
dI
   0nA
dt
dt
E
I'  ind
induced in the loop is:
R
Eind  
Current
Bind
If dI/dt is positive, B is growing, then Bind opposes change and I’ is counter-clockwise
What makes the induced current I’ flow, outside solenoid?
•
•
•
•
•
•
B = 0 there so it’s not the Lorentz force
An induced electric field Eind along the loop causes current to flow
It is caused directly by d/dt within the loop path
E-field is there even without the conductor (no current flowing)
Electric field lines here are loops that don’t terminate on charge.
E-field is a non-conservative (non-electrostatic) field as the line
integral around a closed path is not zero
 Eind 

dB
E

d
s


loop ind
dt
Generalized Faradays’ Law
(hold loopCopyright
path R.
constant)
Janow – Fall 2015
Example: Find the electric field induced by
changing magnetic flux

Eind 

 B  ds  0ienc

dB
Eind  ds  
loop
dt

Find the magnitude E of the induced electric field at
points within and outside the magnetic field.
Assume: dB/dt = constant within the circular shaded
area. E must be tangential: Gauss’ law says any normal
component of E would require charge enclosed. |E| is
constant on the circular integration path due to symmetry.


 Eind   E  ds   Eds  E ds  E(2r )
For r > R: B  BA  B(R 2 )
dB
E(2r )  dB / dt  R 2
dt
For r < R: B  BA  B(r 2 )
E(2r )  dB / dt  r 2
dB
dt
R 2 dB
E
2r dt
r dB
E
2 dt
The magnitude of induced electric field grows
linearly with r, then falls off as 1/r for r>R
Copyright R. Janow – Fall 2015
RELATED FARADAYS LAW EFFECTS IN COILS:
Mutual-induction between coils: di1/dt in “transformer primary” induces EMF
and current i2 in “linked” secondary coil (transformer principle).
N
Self-induction in a single Coil: di/dt produces “back
EMF” due to Lenz & Faraday Laws: ind opposes d/dt due
to current change. Eind opposes di/dt.
• ANY magnetic flux change is resisted, analogous to inertia
• Changing current in a single coil generates changing magnetic field.
• Changing Flux induces flux opposing the change, along with opposing EMF,
current, and induced field..
• This back EMF limits the rate of current (and therefore flux) change in the circuit
• For increasing current,
back EMF limits the
rate of increase
• For decreasing current,
back EMF sustains the
current
Inductance measures opposition to the rate of change of current
Copyright R. Janow – Fall 2015
Definition of Self-inductance
Recall capacitance: depends only on geometry
It measures charge stored per volt
Self-inductance depends only on coil geometry
It measures flux created per ampere
number of turns
L
self-inductance
SI unit of
inductance:
N B
i
C
L
Q
V
linked flux
unit current
Joseph Henry
1797 – 1878
flux through one turn depends
on current & all N turns
cancels current dependence in flux above
2
1 Henry  1 H.  1 T.m / Ampere  1 Weber / Ampere
 1 Volt.sec / Ampere (.sec)
Why choose
this definition?
Cross-multiply……. Take time derivative
Li  N B
di
EL   L
dt
di
dB
L
N
 - EL
dt
dt
• L contains all the geometry
• EL is the “back EMF”
Another form of Faraday’s
Law!
Copyright
R. Janow – Fall 2015
Example: Find the Self-Inductance of a solenoid
Field:
+
-
L
Flux in just
one turn:
RL
B  μ0in where n 
ΦB
NA
 BA  μ0i

Apply definition of self-inductance:
 N turns
 Area A
 Length l
 Volume V = Al

•
NΦB
N2 A
2
L
 μ0
 μ0n V
•
i

N
# turns


unit Length
ALL N turns
contribute to selfflux through ONE
turn
Depends on geometry
only, like capacitance.
Proportional to N2 !
Check: Same formula for L if you start with Faraday’s Law for B:
Eind  N
dΦB
dt
for solenoid use
B
above
2
di
 NA di   μ0N A di
Eind  N.  μ0

 L

 dt 

dt
dt

Note: Inductance per unit length
has same dimensions as 0
L
A
 μ0N2 2


T.m
H

A. – Fall 2015
m
Copyright R. Janow
[ 0 ] 
Example: calculate self-inductance L for an ideal solenoid
N  1000 turns, radius r  0.5 m, length   0.2 m
l
L 

2
μ 0N A


L  49.4  10
Ideal inductor
(abstraction):
4π  10
3
7
6
2 2
 10  π  (5  10 )
0.2
Henrys 49.4 m illi - Henrys
 Internal resistance r  0 (recall ideal battery)
 B  0 outside
 B  μ0in inside (ideal solenoid)
Non-ideal inductors have
internal resistance:
r
L
 VIND  EL  ir  measured voltage
 Direction of ir depends on current
 Direction of E depends on di/dt
L
 If current i is constant, then induced E  0
L
Vind
Inductor behaves like a wire with resistance r
Copyright R. Janow – Fall 2015
Induced EMF in an Inductor
12 – 1: Which statement describes the current through the inductor
below, if the induced EMF is as shown?
A.
B.
C.
D.
E.
Rightward and constant.
Leftward and constant.
Rightward and increasing.
Leftward and decreasing.
Leftward and increasing.
L 
di
EL   L
dt
Copyright R. Janow – Fall 2015
Lenz’s Law applied to Back EMF
i
+
If i is increasing:
EL
Di / Dt
-

EL opposes increase in i
Power is being stored in B field of inductor
If i is decreasing:
-
EL
+
 0
dt
i
Di / Dt
dΦB

dΦB
 0
dt
EL opposes decrease in i
Power is being tapped from B field of inductor
What if CURRENT i is constant?
EL = 0, inductance acts like a simple wire
Copyright R. Janow – Fall 2015
Example: Current I increases uniformly from 0 to 1 A. in
0.1 seconds. Find the induced voltage (back
EMF) across a 50 mH (milli-Henry) inductance.
+
EL
i defines positive direction
-
di
dt
i
and tow ard the right
i
Apply:
EL   L
di
Substitute:
dt
EL  50 m H  10
 0 m eansthat current i is increasing
Am p
se c
Δi
Δt
  0.5 Volts

 1 Am p
0.1 sec
 10
Am p
sec
Negative result
means that induced
EMF is opposed to
both di/dt and i.
Copyright R. Janow – Fall 2015
Inductors in Circuits—The RL Circuit
• Inductors, sometimes called “coils”, are common circuit components.
• Insulated wire is wrapped around a core.
• They are mainly used in AC filters and tuned (resonant) circuits.
Analysis of series RL circuits:
E
•
A battery with EMF
drives a current around the loop
•
•
Changing current produces a back EMF or sustaining EMF EL in the inductor.
Derive circuit equations: apply Kirchoff’s loop rule, convert to differential
equations (as for RC circuits) and solve.
New rule: when traversing an inductor in the same
direction as the assumed current, insert:
di
EL   L 2015
Copyright R. Janow – Fall
dt
Series LR circuits
+
i
a
R
E
-
b
L
i
EL
• Find time dependent behavior
• Inductance & resistance + EMF
• Use Loop Rule
di
EL   L
dt
NEW TERM
FOR
KIRCHHOFF
LOOP RULE
Given E, R, L: Find i, EL, UL for inductor as functions of time
Growth phase, switch
to “a”. Loop equation:
di
E  iR  L
 0
dt
Decay phase, switch
to “b”, exclude E,
Loop equation:
di
 iR  L
 0
dt
• i through R is clockwise and growing: EL opposes E
• As t  infinity, current is large & stable, di/dt  0
Back EMF EL 0, i  E / R,
L acts like an ordinary wire
• At t = 0, rapidly growing current but i = 0, EL= E
L acts like a broken wire
Energy is stored in L & dissipated in R
• Energy stored in L will be dissipated in R
• EL acts like a battery maintaining current
At t = 0 current i = E / R, unchanged, CW
• Current begins to collapse
Current  0 as t  infinity – energy is depleted
Copyright R. Janow – Fall 2015
LR circuit: decay phase solution
R
• After growth phase equilibrium, switch from a to b, battery out
• Current i0 = E / R initially still flowing CW through R
• Inductance tries to maintain current using stored energy
• Polarity of EL reverses versus growth. Eventually EL 0
Loop Equation is :
- iR  EL  0
EL (t )   L
Substitute :
di
dt
b
EL
L
+
i
Circuit Equation:
di
R
  i
dt
L
di/dt <0
during decay,
opposite to
current
• First order differential equation with simple exponential solution
• At t = 0: large current implies large di / dt, so EL is large
• As t  infinity: current stabilizes, di / dt and current i both  0
Current decays exponentially:
i(t )  i0e  t / tL
tL  L/R
i0 
E
i0
e 1  0.37
i
R
 inductive time constant
EMF EL and VR also decay exponentially:
di
E  t / tL

e

dt
RtL
t
2t
3t
t
 t / tL
Compare to RC circuit, decay
VR  E L iR
RC Copyright
 capacitive
time cons tan t
R. Janow – Fall 2015
EL  E e
Q(t )  CEe  t / RC
LR circuit: growth phase solution
Loop Equation is :
Substitute :
E - iR  EL  0
EL (t )   L
Circuit Equation:
L di
E
i 

R dt
R
di
dt
• First order differential equation again - saturating exponential solutions
• As t  infinity, di / dt approaches zero, current stabilizes at iinf = E / R
• At t = 0: current is small, di / dt is large, back EMF opposes battery.
Current starts from zero, grows as a saturating exponential.

i(t )  iinf 1  e
 t / tL
tL  L/R  inductive

iinf 
iinf
E
R
time constant
• i = 0 at t = 0 in above equation  di/dt = E/L
fastest rate of change, largest back EMF
Back EMF EL decays exponentially
di
E

e  t / tL  E L   E e  t / tL
dt R tL
Voltage drop across resistor VR= -iR
i
1  e 1  0.63
t
2t
3t
t
Compare to RC circuit, charging


Q(t )  CE 1  e  t / RC
RC  capacitive time cons tan t
Copyright R. Janow – Fall 2015
Example: For growth phase find back EMF EL as a function of time
Use growth phase solution i(t) 
E
(1  e  t / tL )
R
At t = 0: current = 0
S
+
di E (-)(-)  t / tL

e
dt R tL



where
t
L
tL 
R
L acts like a broken wire at t = 0
L

L
R

0.1H
1Ω
 0 .1 s e c
t
At t  0 : EL   E
Back EMF EL equals the battery potential
causing current i to be 0 at t = 0
iR drop across R = 0
EL
i
-
Back EMF is ~ to rate of change of current
di
EL  L
  Ee  t / tL ;
dt
L  0.1H
E  5V
E
i(t  0)  (1  e0 )  0
R
Derivative :
R  1Ω
EL
- 0.37 V0
-E
After a very long (infinite) time:
 Current stabilizes, back EMF=0
E
i 
 5A
R
 L acts like an ordinary wire at t = infinity
Copyright R. Janow – Fall 2015
Current through the battery - 1
12 – 2: The three loops below have identical inductors, resistors,
and batteries. Rank them in terms of current through the battery
just after the switch is closed, greatest first.
A.I, II, III.
B.II, I, III.
C.III, I, II.
D.III, II, I.
E.II, III, I.
I.

II.
III.
Hint: what kind of wire does L act like?
i(t)  iinf 1  e  t / tL

iinf 
E
R eq
tL  L/R eq
 inductive time constant
Copyright R. Janow – Fall 2015
Current through the battery - 2
12 – 3: The three loops below have identical inductors, resistors,
and batteries. Rank them in terms of current through the battery a
long time after the switch is closed, greatest first.
A.
B.
C.
D.
E.
I, II, III.
II, I, III.
III, I, II.
III, II, I.
II, III, I.
I.

II.
III.
Hint: what kind of wire does L act like?
i(t)  iinf 1  e  t / tL

iinf 
E
R eq
tL  L/R eq
 inductive time constant
Copyright R. Janow – Fall 2015
Extra
Summarizing RL circuits growth phase

• When t is large: i  
Inductor acts like a wire.
i  (1  e  Rt / L )
•
R
When t is small (zero), i = 0.
Inductor acts like
an open circuit.
•
The current starts from zero and
increases up to a maximum of i   / R
with a time constant given by
L
Inductive time constant
tL 
R
Compare: tC  RC Capacitive time constant
•
The voltage across the resistor is
R
VR  iR  (1  eRt / L )
•
The voltage across the inductor is
VL    VR    (1  eRt / L )   eRt / L
Copyright R. Janow – Fall 2015
Summarizing RL circuits decay phase
Extra
The switch is thrown from a to b
•
Kirchoff’s Loop Rule for growth
was:
  iR  L di  0
dt
•
Now it is:
•
di
0
dt
The current decays exponentially:
iR  L
i
R
Voltage across resistor also decays:
VR  iR   eRt / L
•
VR (V)
•
 e Rt / L
Voltage across inductor:
VL  L
di
 d eRt / L   eRt / L
 L
dt
R dt
Copyright R. Janow – Fall 2015
Energy stored in inductors
Recall: Capacitors store energy in their electric fields
UE  electrostatic potentialenergy
2
1Q
1
UE 
 CV2
2 C
2
uE 
electrostatic energy density
derived
UE
1
2
using p-p
uE 
  E
Volume 2 0
capacitor
Inductors also store energy, but in their magnetic fields
Magnetic Potential Energy – consider power into or from inductor
dUB
di
1
Power 
 ELi  Li
 UB   dUB  L  idi  Li2
2
dt
dt
UB  magnetic potentialenergy
UB 
1 2
Li
2
uB  magnetic energydensity
B2
uB 

Volume 2
UB
0
derived
using
solenoid
• UB grows as current increases, absorbing energy
• When current is stable, UB and uB are constant
• UB diminishes when current decreases. It powers
the persistent EMF during the decay phase for the inductor
Copyright R. Janow – Fall 2015
Sample problem: energy storage in magnetic field
of an inductor during growth phase
a) At equilibrium (infinite time) how much energy is stored in the coil?
i

U



E
R
1
2
(Coil acts like a w ire) 
Li
12
0.35
 34.3 A
2
3
2
 x 53 x 10
x (34.3)

2
U

L = 53
mH
E = 12 V
1
R  0.35 Ω
 31 J
b) How long (t1/2) does it take to store half of this energy?
At t1/2 : UB  21 U
i1/2 
e

1
2
2
Li1/2

1 1

2 2
 L  i2

i1/ 2 
i
2
i
 t /t
 i (1  e 1/2 L )
2
 t1 / 2 / tL

2
1

2
2
 1  1/ 2
take natural log of both sides
t1 / 2  - tL ln(1  1 /
2 )  1.23 t
tL  L / R  53 x 10-3 / 0.35
 0.15 sec.
Copyright
R. Janow – Fall 2015
Mutual Inductance
• Example: a pair of co-axial coils
• di/dt in the first coil induces current in the
second coil, in addition to self-induced effects.
• M21 depends on geometry only, as did L and C
• Changing current in primary (i1) creates varying
flux through coil 2  induced EMF in coil 2
Definition:
number of turns
in coil 2
mutual
inductance
N2  21
M21 
i1
flux through one turn
of coil 2 due to all N1
turns of coil 1
current in coil 1
cross-multiply
M21 i1  N2 21
time derivative
M21
di1
d 21
 N2
dt
dt
di1
E2   M21
dt
• M21 contains all the
geometry
• E2 is EMF induced
in 2 by 1
Another form of Faraday’s Law!
The smaller coil radius determines how much flux is linked, so…..
di
proof not obvious
E1   M12 2
M12  M21  M
Copyright R. Janow – Fall 2015
dt
Calculating the mutual inductance M
Outer coil 1 is a short loop of N1 turns
(not a long Solenoid)
B1  Field inside loop 1 near center

 0i1N1
large coil 1
N1 turns
radius R1
(primary)
small coil 2
N2 turns
radius R2
(secondary)
Assume uniform B
Use arc formula with f=2
2R1
Flux through inner flat coil 2 (one turn) depends
on smaller area A2 & B1
μ iN
Φ21  B1A2  0 1 1 R 22 (for each loop in coil 2 )
2R1
Current i1 in Loop 1 is changing:
E2
 induced voltage in loop 2
 M21  M 
Summarizing
results for mutual
inductance:
μ0N1N2 R 22 di1
dΦ21
di
 N2

 M21 1
dt
2R1
dt
dt
μ0N1N2
R 22
2R 1
E2  - M21
M21 
smaller radius (R2)
determines the linkage
di1
dt
M21 
N2Φ21
i1
N2Φ21 N1Φ12

 MCopyright
12  MR. Janow – Fall 2015
i1
i2
Air core Transformer Example: Concentric coils
OUTER COIL – IDEAL SOLENOID
• Coil diameter D = 3.2 cm
• nout = 220 turns/cm = 22,000 turns/m
• Current iout falls smoothly from
1.5 A to 0 in 25 ms
• Field uniform across outer coil is:
Bout   0 iout (t) nout
Bout
as function of time
 0  4  x 10-7 nout 
Nout
L out
FIND INDUCED EMF IN INNER COIL DURING THIS PERIOD
• Coil diameter d = 2.1 cm = .021 m, Ain =  d2/4, short length
• Nin = 130 turns = total number of turns in inner coil
Din  flux change through one turn of inner coil during Dt
DBout Ain
 n Di
Di
Din
Ein  - Nin
 - Nin
 - Nin A in 0 out out  - M out
Dt
Dt
Dt
Dt
 1.5
 Ein  - 4   10-7  2.2x10 4  130   (.021/ 2)2 
 75 mV
2.5 x10  3
Direction: Induced B is parallel to Bouter which is decreasing
Would the transformer work if we reverse the role of the coils?
Copyright R. Janow – Fall 2015