Physics 121 - Electricity and Magnetism
Lecture 12 - Inductance, RL Circuits
Y&F Chapter 30, Sect 1 - 4
•
•
•
•
•
•
•
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
Copyright R. Janow – Spring 2017
Recap: Faraday’s Law of Induction
•
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 – Spring 2017
Changing magnetic flux directly induces electric fields
conducting loop, resistance R
A thin solenoid, n turns/unit length
• Solenoid cross section A
• Zero B field outside solenoid
• Field inside solenoid:
B  
Flux through the wire loop:
in
  BA  0niA
0
Changing current i  changing flux 
EMF is induced in wire loop:
d
di
   0nA
dt
dt
Eind
induced in the loop is: i' 
R
Eind  
Bind
Current
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 outside solenoid, so it’s not the Lorentz force
An electric field Eind is induced directly by d/dt within the loop
Eind causes current to flow in the loop
Eind is there even without the wire loop (no current flowing)
Electric field lines are loops that don’t terminate on charge.
Eind is a non-conservative (non-electrostatic) field as the line
integral (potential) 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 – Spring 2017
Example: Find the electric field induced by a
region of 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 across the circular shaded area.
• E must be tangential: Gauss’ law says any normal
component of E would require enclosed charge.
• |E| is constant on a 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
E
r dB
2 dt
The magnitude of induced electric field grows
linearly with r, then falls off as 1/r for r>R
Copyright R. Janow – Spring 2017
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 (or Bind) 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/flux in itself.
• Changing flux induces flux opposing the change, along with opposing EMF, current, and induced field.
• The “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
Eind = - L di/dt
L is the inductance
Copyright R. Janow – Spring 2017
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 – Spring 2017
Example: Find the Self-Inductance of a solenoid
Field:
+
-
L
B  μ0in where n 
N
# turns


unit Length
NA ALL N turns contribute
Flux in just
ΦB  BA  μ0i
to self-flux through
one turn:

RL
ONE turn
Apply definition of self-inductance:
 N turns
 Area A
 Length l
 Volume V = Al

•
NΦB
N2 A
2
L
 μ0
 μ0n V
•
i

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
Eind
B
above
2
di
 NA di   μ0N A di
 N.  μ0

 L

 dt 

dt
dt

Note: Inductance per unit length
has same dimensions as 0
L
A
 μ0N2 2


T.m
H

A. – Springm2017
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
(ideal solenoid)
(abstraction):
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

4π  10
L
 VIND  EL  ir  measured terminal voltage
 Direction of ir depends on current
 Direction of E depends on di/dt
L
 Constant current implies
, induced voltage EL = 0
Inductor behaves like a wire with resistance r
Vind
Copyright R. Janow – Spring 2017
Lenz’s Law applied to Back EMF
i
+
If i is increasing:
EL
Di / Dt
-

 0
dt
EL opposes increase in i
Power is being stored in B field of inductor
i
If i is decreasing:
Di / Dt
dΦB
EL
+

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 – Spring 2017
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 – Spring 2017
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
i
Apply:
i
EL   L
di
dt
0
Substitute:
dt
EL  50 m H  10
Am p
means that current is increasing
and to the right
Δi
Δt
  0.5 Volts
se c

 1 Am p
0.1 s ec
 10
Am p
s ec
Negative result means that
induced EMF is opposed to
both di/dt and i.
Devices:
•
•
•
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.
Copyright R. Janow – Spring 2017
Inductors in Circuits—The RL Circuit
Basic series RL circuit:
•
•
•
•
A battery with EMF E drives a current
around the loop
Changing current produces a back EMF or
sustaining EMF EL in the inductor.
Derive circuit equations using Kirchhoff’s loop
rule.
Convert to differential equations and solve
(as for RC circuits).
New for Kirchhoff rule:
When traversing an inductor in the same
direction as the assumed current insert:
di
EL   L
dt
Copyright R. Janow – Spring 2017
Series LR circuit
+
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 previous 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 – Spring 2017
LR circuit: decay phase solution
R
• After growth phase equilibrium, switch from a to b, battery out
• Current i0 = E / R initially still flows CW through R
• Inductance tries to maintain current using stored energy
• Polarity of EL reversed 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 decay solution
• At t = 0+: large current implies large di / dt, so EL is large (now driving current)
• 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
EL  E e
 t / tL
VR  E L iR
t
2t
3t
t
Compare to RC circuit, decay
Q(t )  CEe  t / RC
RC Copyright
 capacitive
time cons tan t
R. Janow – Spring 2017
LR circuit: growth phase solution
Loop Equation is :
E - iR  EL  0
EL (t )   L
Substitute :
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: back EMF opposes battery, current is small, di / dt is large,.
Current starts from zero, grows as a saturating exponential.

i(t )  iinf 1  e
tL  L/R
 t / tL

iinf 
iinf
E
R
 inductive 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 – Spring 2017
Example: For growth phase find back EMF EL as a function of time
Use growth phase solution i(t) 
di
E
(1  e  t / tL ) EL   L
R
dt
At t = 0: current = 0



L  0.1H
E  5V
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
R  1Ω
+
E
i(t  0)  (1  e0 )  0
R
Derivative :
S
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 – Spring 2017
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 (ordinary or broken) 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 – Spring 2017
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 (ordinary or broken) 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 – Spring 2017
Extra
Summarizing RL circuits growth phase
• When t / tL is large: i   Inductor acts like a wire. i   (1  e  Rt / L )
R
i = 0.
Inductor acts like
an open circuit.
•
When t / tL is small:
•
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 – Spring 2017
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 – Spring 2017
Energy stored in inductors
Recall: Capacitors store energy in their electric fields
UE  electrostatic potentialenergy
UE 
2
1Q
2 C
1
2
 CV 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
1 2
UB  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 decreases as current decreases. It powers
the persistent EMF during the inductor’s decay phase
R. Janow – Spring 2017
• The B field stores energy with or without aCopyright
conductor
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
1
U

L = 53
mH
E = 12 V
R  0.35 Ω
 31 J
b) How long (tf) does it take to store half of this energy?
At t f :
1
UB  U
2
1 2 1 1
Lif    L  i2 
2
2 2

if 
i
2
i
 t /t
if    i (1  e f L )

2
 t /t
2
1
e f L 

 1  1/ 2
2
2
take natural log of both sides
t  - tL ln(1  1 / 2 )  1.23 tL
f
tL  L / R  53 x 10-3 / 0.35
 0.15 sec.
Copyright
R. Janow – Spring 2017
Mutual Inductance
• Example: a pair of co-axial, flux-linked coils
• di1/dt in the first coil induces current i2 in the
second coil, in addition to self-induced effects.
• M21 depends on geometry only, as did L, C, and R
• Changing current in primary (i1) creates varying
flux through coil 2  induced E field 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 linked flux is the same (smaller coil) if coils reverse roles, so…..
di
proof not obvious
E1   M12 2
M12  M21  M
Copyright R. Janow – Spring 2017
dt
Calculating the mutual inductance M
Outer coil 1 is a flat loop of N1 turns (not
a solenoid)
B1  Field inside loop 1 near center

 0i1N1
PRIMARY
large coil 1
N1 turns
radius R1
Assume uniform B
Use arc formula with f=2
SECONDARY
small coil 2
N2 turns
radius R2
2R1
Flux through one turn of flat inner coil 2
depends on B1 and smaller area A2
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  M R. Janow – Spring 2017
i1
i2
Air core Transformer Example: Concentric coils
OUTER COIL – IDEAL SOLENOID
• Coil diameter D = 3.2 cm
• nouter = 220 turns/cm = 22,000 turns/m
• Current iouter 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.
nout 
Nout
Lout
FIND INDUCED EMF IN INNER COIL DURING THIS PERIOD
• Coil diameter d = 2.1 cm = .021 m, Ainner =  d2/4, short length
• Ninner = 130 turns = total number of turns in inner coil
Din  flux change through one turn of inner coil during Dt
DBout Ain
 0 nout Diout
Diout
Din
Ein  - Nin
 - Nin
 - Nin Ain
-M
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 – Spring 2017