Phys-272 Lecture 18
Mutual Inductance
Self-Inductance
R-L Circuits
Mutual Inductance
If we have a constant current i1 in coil 1,
a constant magnetic field is created and
this produces a constant magnetic flux in
coil 2. Since the ΦB2 is constant, there NO
induced current in coil 2.
If current i1 is time varying, then the ΦB2
flux is varying and this induces an emf ε2
in coil 2, the emf is
dΦ B2
ε 2 = −N2
dt
We introduce a ratio, called mutual inductance, of flux in coil 2
divided by the current in coil 1.
N 2Φ B 2
M 21 =
i1
Mutual Inductance
N 2Φ B 2
mutual inductance, M 21 =
, can now be used in Faraday’s eqn.
i1
M 21 i1 = N 2 Φ B 2
di1
dΦ B 2
di1
M 21
= N2
= −ε 2 ; ε 2 = − M 21
dt
dt
dt
We can also the varying current i2 which creates a changing flux ΦB1
in coil 1 and induces an emf ε1. This is given by a similar eqn.
di2
ε1 = − M12
dt
It can be shown (we do not prove here) that,
M 12 = M 21 = M
The units of mutual inductance is T ⋅ m2/A = Weber/A = Henry
(after Joseph Henry, who missed Faraday’s Law)
Mutual Inductance
The induced emf,
has the following features;
di2
ε 1 = −M
dt
• The induced emf opposes the magnetic flux change (lenz’s Law)
• The induced emf increases if the current changes very fast
• The induced emf depends on M, which depends only the geometry
of the two coils and not the current.
• For a few simple cases, we can calculate M, but usually it is just
measured.
Problem 30.1
Two coils have mutual inductance of 3.25×10−4 H. The current
in the first coil increases at a uniform rate of 830 A/s .
A) What is the magnitude of induced emf in the 2nd coil? Is it
constant?
B) suppose that the current is instead in the 2nd coil, what is the
magnitude of the induced emf in the 1st coil?
di1
ε 2 = −M
dt
A
= −(3.25 × 10 H )(830 ) = − 0.27V
s
−4
di2
ε 1 = −M
= − 0.27 V
dt
Tesla Coil Example
Magnetic field due to coil 1 is
B1 = µ0n1i1 = µ0N1i1 / l
Mutual inductance is,
N 2 Φ B 2 N 2 B1 A
M=
=
i1
i1
N 2 µ0 N1i1 A µ0 N1N 2 A
=
=
i1l
l
The induced emf in coil1 from coil2 is
µ0 N1N 2 A di2
di2
ε1 = − M
=−
dt
l
dt
Nicolai Tesla (1856-1943)
Born in Croatia, graduated from University of
Prague. Arrived in New York with 4cents and went
to work for Edison. Tesla invented polyphase
alternating-current system, induction motor,
alternating-current power transmission, Tesla
coil transformer, wireless communication, radio,
and fluorescent lights. He set up a Tesla coil in
Colorado Springs in 1899, below
is a photo of this lab. He lighted
lamps 40Km away. He also claimed
to receive messages from another
planet!! In honor of his contributions
to electromagnetic phenomena, the
Magnetic field intensity was named
in units of “Tesla”
Applications of Mutual Inductance
iron
• Transformers
– Change one AC voltage into another
• Airport Metal Detectors
– Pulsed current pulsed magnetic field
Induces emf in metal
– Ferromagnetic metals “draw in” more B
larger mutual inductance larger emf
– Emf current (how much, how long it
lasts, depends on the resistivity of the
material)
– Decaying current produces decaying
magnetic field
induces current in receiver coils
– Magnitude & duration of signal depends
on the composition and geometry of the
metal object.
N2
V2 =
V1
N1
∼
V1
V2
ε
N1
N2
(primary) (secondary)
Applications of Mutual Inductance
• Pacemakers
– Not easy to change the battery!
– Instead, use an external AC supply.
– Alternating current
alternating B
alternating ФB inside “wearer”
induces AC current to power
pacemaker
~
Self Inductance
We previously considered induction between 2 coils. Now we consider
the situation where a single isolated coil induces emf on itself. This is
Called “back emf” and if the current changes, there is a self induced emf
that opposes the change in current. We form the same ratio, now called
Self-Inductance, L,
NΦ B
L=
i
and we have the back emf,
di
ε = −L
dt
Behavior of isolated coil in circuits
Resistor with current I
has potential drop, V=iR
from a to b
Coil with
a) constant current i has NO
Voltage drop
b) di/dt>0, potential decreases
from a to b, V=Ldi/dt
c) di/dt<0, potential increases
from a to b, V=-L|di/dt|
Remember, emf in coil opposes current change.
Self inductance of long solenoid
• Long Solenoid:
l
N turns total, radius r, Length l
r
N
r << l ⇒ B = µ 0
I
l
N turns
For a single turn, A = π r 2 ⇒ φ = BA = µ 0
N
Iπ r 2
l
The flux through a turn is given by:
Φ B = µ0
N
Iπr 2
l
Inductance of solenoid can then be calculated as:
2
NΦ B
N2 2
N
L≡
= µ0
πr = µ 0   lπr 2
I
l
 l 
Clicker:
Two simple pieces of wire A and B are shaped
into almost complete loops. The loose ends of
each loop are connected to identical batteries.
Assume the loops have the same total
resistance, and that they do not interfere with
each other.
2) Which loop has the greatest flux through it (assume the loops
have the same current in them)?
a) A
b) B
c) same
Recall the B-field of current loop (at the center): B = µ0I/(2R)
r r
2
The area of the loop is A = π R Flux is Φ B ≡ ∫ B • dS
⇒ The flux through the loop increases with R.
⇒ The flux through loop A is bigger than the flux through loop B.
3) Which loop has the greatest self inductance?
a) A
b) B
c) same
Self inductance is defined as
so if I is the same, LA > LB.
L ≡
Φ
I
B
Clicker problem
• Consider the two inductors shown:
l
– Inductor 1 has length l, N total turns
and has inductance L1.
– Inductor 2 has length 2l, 2N total turns
and has inductance L2.
– What is the relation between L1 and
L2?
(a) L2 < L1
(b) L2 = L1
2l
r
r
N turns
(c) L2 > L1
r
2N turns
Clicker problem
• Consider the two inductors shown:
l
– Inductor 1 has length l, N total turns
r
and has inductance L1.
N turns
– Inductor 2 has length 2l, 2N total turns
and has inductance L2.
– What is the relation between L1 and
L2? (a) L2 < L1
(c) L2 > L1
(b) L2 = L1
2l
r
r
2N turns
• To determine the self-inductance L, we need to determine the flux ΦB
which passes through the coils when a current I flows: L ≡ NΦB / I.
• To calculate the flux, we first need to calculate the magnetic field B
produced by the current: B = µ0(N/l)I
• i.e., the B field is proportional to the number of turns per unit
length.
• Therefore, B1 = B2. But does that mean L1 = L2?
How to combine inductors
• To calculate L, we need to calculate the
flux.
• Since B1=B2, the flux through any
given turn is the same in each
inductor
l
2l
r
r
N turns
r
2N turns
• There are twice as many turns in inductor 2; therefore the net flux
through inductor 2 is twice the flux through inductor 1! Therefore,
L2 = 2L1.
Inductors in series add (like resistors): Leff = L1 + L2
And inductors in parallel
add like resistors in parallel:
1
1 1
= +
Leff L1 L2
Self Inductance of toroidal solenoid
The magnetic field in a toroid was
µ0 Ni
B=
2πr
and the net mag.flux is
µ0 Ni
Φ B = BA =
A
2πr
Hence the self inductance is,
2
N µ0
NΦ B Nµ0 Ni
=
A=
A
L=
i
i 2πr
2πr
Example 30.3 N=200, A=5cm2, r=0.1m
2002 4π × 10−7
−4
L=
2π (0.1)
5 × 10
= 40µH
Energy of an Inductor
• How much energy is stored
in an inductor when a
current is flowing through it?
• Start with loop
rule: ε = IR + L
I
R
b
dI
dt
L
ε
• Multiply this equation by I:
εI = I 2 R + LI
I
a
dI
dt
• From this equation, we can identify PL, the rate at which
energy is being stored in the inductor:
dU
dI
PL =
•
= LI
dt
dt
We can integrate this equation to find an expression for U, the
energy stored in the inductor when the current = I:
U
I
U = ∫ dU = ∫ LI dI
0
0
⇒
1 2
U = LI
2
Magnetic field Energy in a toroid
Consider a toroid magnet, the B field
πr (ex.28.11). The energy
is , B = µ0NI/2π
is,
1 2 1  µ0 N 2 A  2
I
U = LI =
2
2  2πr 
Substituting the B field into the
Eqn., we have,
2 2 2
2 2
2


U
1  µ0 N I 
1  µ0 N I  B
=
=
=
A2πr 2  (2πr )2  2µ0  (2πr )2  2µ0
U
U
B2
=
=
= Energy density
A2πr volume 2µ0
Energy in Electric Fields and Magnetic Fields
In chapter 24.3, we discussed energy in a parallel plate with area A
and separation d, The electric field energy in the capacitor was
1
2 1
2 1 ε0 A
2 1
2
U = CV = C ( Ed ) = 
( Ed ) = ε 0 ( Ad )E
2
2
2 d 
2
U energy ε 0 2
=
= E
Ad volume 2
Now we find the magnetic field energy in the toroid magnet is
U
energy
1 2
=
=
B
A2πr volume 2µ0
r2 r2
The B , E fields are proportional to the energy density
Announcements
R-L Circuits
(many iclicker examples)
Hints for Homework X
Some small inductors
Inductor in a
power supply
Charging Mat uses induction
Inductors in
Circuits
General rule: inductors resist change in current
• Attached to a current source
– Initially, the inductor behaves like an open switch.
– After a long time, the inductor behaves like an ideal wire.
• Disconnected from a current source
– Initially, the inductor behaves like a current source.
Multi-part clicker
• At t=0 the switch is thrown from
position b to position a in the
circuit shown:
2A
– What is the value of the current I
∞
I
a
I
R
b
L
ε
a long time after the
R
switch is thrown?
(a) I∞ = 0
2B
–
(b) I∞ = ε/2R
(c) I∞ = 2εε/R
What is the value of the current I0 immediately after the switch is thrown?
(a) I0 = 0
(b) I0 = ε/2R
(c) I0 = 2εε/R
• After a long time the switch is opened.
2C
–What is the value of the current I0 just after the switch
is opened?
(a) I0 = 0
(b) I0 = ε/2R
(c) I0 = 2εε/R
Multi-part clicker
I
a
I
R
b
• At t=0 the switch is thrown from position b to
position a in the circuit shown:
– What is the value of the current I∞ a long time
2A after the switch is thrown?
(a) I∞ = 0
(b) I∞ = ε/2R
L
ε
R
(c) I∞ = 2εε/R
• A long time after the switch is thrown, the current approaches
an asymptotic value: as t → ∞, dI/dt → 0.
• As dI/dt → 0, the voltage across the inductor → 0. Therefore,
I∞ = ε/2R.
Multi-part clicker
• At t=0 the switch is thrown from
position b to position a in the
circuit shown:
I
a
I
R
b
ε
L
2B – What is the value of the current I0 immediately
after the switch is thrown?
(a) I0 = 0
(b) I0 = ε/2R
(c) I0 = 2εε/R
• Just after the switch is thrown, the rate of change of current is as large as it can be (we had been
assuming it was ∞!)
• The inductor limits dI/dt to be initially equal to ε/L. The voltage across the inductor = ε; the current,
then, must be 0!
• Another way: the moment the switch is thrown, the current tries to generate a huge B-field. There
is a huge change in flux through coil—an emf is generated to oppose this. Initially, then, no current
flows through no voltage drop across the resistors.
Multi-part clicker
• After a long time the switch is
opened.
2C
I
a
I
R
b
ε
L
–What is the value of the current I0
just after the switch is opened?
(a) I0 = 0
(b) I0 = ε/2R
(c) I0 = 2εε/R
• Just after the switch is thrown, the inductor induces an emf to
keep current flowing: emf = L dI/dt (can be much larger than ε)
• However, now there’s no place for the current to go charges build up on switch contacts high voltage across
switch gap
•If the electric field exceeds the “dielectric strength”
(~30 kV/cm in air) breakdown SPARK!
RL Circuits, Quantitative
a
I
I
R
• At t=0, the switch is closed and
the current I starts to flow.
b
L
ε
• Loop rule:
dI
ε − IR − L
=0
dt
Note that this equation is identical in form to that for the RC circuit with the
following substitutions:
RC:
Q
dQ
ε− −R
=0
C
dt
Therefore, τ RC = RC
⇒ RC→
→RL:
⇒
τ RL
L
=
R
R→ L
1
→R
C
Q→I
RL Circuits
• To find the current I as a function of
time t, we need to choose an exponential
solution which satisfies the boundary
a
I
I
R
condition:
b
ε
τRL
ε
dI
(t = ∞) = 0 ⇒ I (t = ∞) =
R
dt
ε
• We therefore write: I = 1 − e − Rt / L
R
(
=L
R
L
)
• The voltage drop across the inductor is given by:
dI
VL = L
= εe − Rt / L
dt
RL Circuit (ε on)
L/R
ε/R
ε/
Current
I =
ε
(
1− e
R
− Rt / L
Sketch curves !
2L/R
)
I
Max = ε/R
63% Max at t=L/R
0
Voltage on L
t
ε
dI
VL = L
= ε e − Rt / L
dt
VL
Max = ε/R
37% Max at t=L/R
0
t
RL Circuits
• After the switch has been
in position a for a long
time, redefined to be t=0, it
is moved to position b.
• Loop rule:
a
I
I
R
b
dI
IR + L
=0
dt
• The appropriate initial condition is: I ( t = 0 ) =
• The solution then
must have the form:
L
ε
I =
ε
R
e − Rt / L
dI
VL = L
= −ε e − Rt / L
dt
ε
R
RL Circuit (ε off)
ε/R
ε/
Current
I =
ε
R
e − Rt
L/R
Sketch curves !
2L/R
/L
Max = ε/R
I
37% Max at t=L/R
0
Voltage on L
dI
VL = L
= − ε e − Rt / L
dt
t
0
VL
Max = -ε
37% Max at t=L/R
-εε
t
ε on
L/R
ε/R
ε/
I
0
ε off
2L/R
I=
ε/R
ε/
ε
(
1− e
R
− Rt / L
)
t
ε
VL
0
L/R
2L/R
I =
I
0
ε
R
e − Rt / L
t
0
dI
VL = L = εe−Rt / L
dt
t
VL
dI
VL = L = −εe−Rt / L
dt
-εε
t
Clicker
• At t=0, the switch is thrown
from position b to position a
as shown:
»Let tI be the time for
circuit I to reach 1/2 of its
asymptotic current.
– Let tII be the time for circuit II to reach 1/2 of
its asymptotic current.
– What is the relation between tI and tII?
I
a
R
b
I
ε
(b) tII = tI
(c) tII > tI
L
R
a
I
I
L
b
ε
II
L
(a) tII < tI
I
R
Clicker
•
At t=0, the switch is thrown from position b to position a as
shown:
– Let tI be the time for circuit I to reach 1/2 of its
asymptotic current.
– Let tII be the time for circuit II to reach 1/2 of its
asymptotic current.
– What is the relation between tI and tII?
I
a
R
b
I
ε
(a) tII < tI
(c) tII > tI
•
We must determine the time constants of the two
circuits by writing down the loop equations.
dI
I: ε − I R − L − I R = 0
dt
dI
dI
ε −L
− IR − L
=0
II:
dt
dt
τI =
τ II
L
R
a
(b) tII = tI
I
I
I
L
b
ε
II
R
L
L
2R
2L
=
R
This confirms that
inductors in series add!
Another inductor example
At t = 10 hrs the switch is opened,
abruptly disconnecting the battery from the
circuit. What will happen to all the energy
stored in the solenoid?
Energy stored in the inductor: U = 1/2 L I2
When the switch is opened, this energy is dissipated in the resistor.
An inductor doesn’t like change!!!
When the switch is opened, the inductor will try to
maintain the current that was flowing through it before the
switch is opened. Since the battery is disconnected from the
circuit, the energy which is necessary to keep current
flowing through the resistor is provided by the inductor.