Inductance
Important fact: Magnetic Flux ΦB is proportional to the
current making the ΦB
All our equations for B-fields show that B α I
v
v v
v µ I dl × rˆ
dB =
∫ B ⋅ dl = µ I
0
4π
r
0 thru
2
Biot-Savart
Ampere
Inductance
Flux Φ B ∝ B ∝ I
Flux Φ B ∝ I
Assumes leaving everything else
the same.
If we double the current I, we will
double the magnetic flux through
any surface.
i
1
Self-inductance
Self-Inductance (L) of a coil of wire
Φ B ≡ LI
ΦB
L≡
I
This equation defines self-inductance.
Note that since ΦB α I, L must be
independent of the current I.
L has units [L] = [Tesla meter2]/[Amperes]
New unit for inductance = [Henry].
Inductors (coil of wire)
An inductor is just a coil of wire.
The Magnetic Flux created by
the coil, through the coil itself:
v v
Φ B = ∫ B ⋅ dA
This is quite hard to calculate
for a single loop. Earlier we
calculated B at the center, but
it varies over the area.
2
Inductors
Consider a simpler case of a solenoid
v
| Binside |= µ0nI = µ0
N
I
L
Recall that the B-field inside a
solenoid is uniform!
N = number of loops
L = length of solenoid
* Be careful with symbol L !
Inductors
v
| Binside |= µ0nI = µ0
v
N
I
L
v
Φ B = N ∫ B ⋅ dA = N BA = N ( µ0nI ) A
L=
Self-Inductance
ΦB
= µ0 NnA = µ0 An 2 L
I
Length
3
Clicker Question
Inductor 1 consists of a single loop of wire. Inductor 2 is
identical to 1 except it has two loops on top of each other.
How do the self-inductances of the two loops compare?
A) L2 = 2 L1
I
B) L2 > 2L1
Coil 1
I
C) L2 < 2L1
Coil 2
Answer: L2 > 2L1 , in fact L2 = 4 L1. The self
inductance L increases by 4. L =Φ/i. If we keep i
fixed, but double the number N of turns, Φ
increases by 4. Total flux. N doubles, but Φ1 = BA
also doubles because when we double the number
of turns, then B is doubled.
Clicker Question
Two long solenoids, each of inductance L, are connected
together to form a single very long solenoid of inductance
Ltotal. What is Ltotal?
A) 2L
B) 4L
C) 8L
D) none of these/don't know
+
L
=
L
Ltotal = ?
Answer: 2L. The inductance of a solenoid is L = µon2A l,
where n is the number of turns per length n = N/l and l is the
length. In this case, we did not change n, but l (length)
doubled, so L doubles.
4
Self-induced EMF
What does this inductance tell us?
ΦB
I
Φ B = LI
L=
dΦ B
dI
=L
dt
dt
L is independent of time.
Depends only on geometry of inductor
(like capacitance).
Self-induced EMF
dΦ B
dI
=L
dt
dt
−ε = L
ε = −L
Recall Faraday’s Law
ε =−
dΦ B
dt
dI
dt
dI
dt
Changing the current in an inductor creates
an EMF which opposes the change in the
current.
Sometimes called “back EMF”
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Self-induced EMF
ε = −L
dI
dt
It is difficult (requires big external Voltage) to change quickly
the current in an inductor.
The current in an inductor cannot change instantly.
If it did (or tried to), there would be an infinite back EMF. This
infinite back EMF would be fighting the change!
Magnetic Field energy
Recall that for a capacitor C, there is stored potential
energy in the electric field.
1
U = CV 2
2
The energy is stored in the electric field and the density
is:
v 2
U
1
uE =
Volume
= ε0 | E |
2
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Magnetic Field energy
The power supplied to an inductor is
P = Vi = Li
di
dt
The energy dU supplied to an inductor in an interval dt is:
dU = Pdt = Lidi
The total energy U supplied while the current increases
from 0 to a final value is
I
U = ∫ Lidi =
0
1 2
LI
2
Magnetic energy density
For an inductor L, with current I, there is stored energy in the
magnetic field.
1
U=
For a coil this is
2
1 µ0 N 2 A 2
I
U=
2 2πr
The energy density is
But we know that
LI 2
U
U
1
N 2I 2
=
= µ0
Volume 2πrA 2 ( 2πr )2
µ NI
B= 0
2πr
B2
→ uB =
2 µ0
uB =
7
Clicker Question
The same current i is flowing through solenoid 1 and solenoid
2. Solenoid 2 is twice as long and has twice as many turns
as solenoid 1, and has twice the diameter. (Hint) for a
solenoid B = µo n i )
What is the ratio of the magnetic energy contained in
solenoid 2 to that in solenoid 1, that is, what is UU
I
A) 2
B) 4
1
C) 8
D) 16
2
I
E) None of these.
2
1
Answer: There is 8 times as much magnetic field energy in the large
solenoid as in the small solenoid.
The B-field is the same in both solenoids (same n = turns/length so same B
=µo n I) so both solenoids contain the same energy per volume u = U/vol =
B2/(2µo). The larger solenoid has 8 times the volume (2X the length, 4X
the cross-sectional area) so it has 8 the energy.
Magnetic Field energy
It takes work to get current flowing through an inductor.
You must work against the back EMF which opposes any
change in the current.
That work = potential energy stored = U = ½ LI2
And is thus stored in the inductor’s magnetic field.
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Inductors as circuit elements
What do these inductors do in circuits?
Just recall that the EMF or Voltage across an inductor is:
ε = −L
dI
dt
So, when we add them to circuits, we can apply the usual
Kirchhoff’s Voltage Law and include the inductors.
Inductors in a circuit
Battery=V
Consider a circuit with a battery, resistor and inductor (RL circuit)
switch
a
b
R
L
Switch is in position (a) for a long time.
In steady state (after a long time), the current will no longer be
changing and thus the inductor looks like a regular wire!
9
Battery=V
RL circuit
switch
a
R
b
L
If after a long time the inductor acts like a wire:
I=
∆V = V − IR = 0
V
R
Battery=V
RL circuit
switch
a
b
R
L
At t=0, move the switch to (b).
Normally one might expect there to immediately be zero current.
However, the inductor has stored energy!
10
Battery=V
LR circuit
∆V = − IR − L
di
⎛R⎞
= −⎜ ⎟i
dt
⎝L⎠
switch
a
R
b
L
dI
=0
dt
We need to solve this differential
equation for i(t).
RL circuit
dI
⎛R⎞
= −⎜ ⎟ I
dt
⎝L⎠
R⎞
⎟t
⎝L⎠
⎛
I ( t ) = I 0e
−⎜
⎛
I (t ) = I 0e
−t / ⎜
⎝
L⎞
⎟
R⎠
where
I0 =
V
R
Current exponentially decays with
Time Constant =τ= L/R (units of seconds).
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Clicker Question
Rank in order, from largest to smallest, the
time constants
in the three circuits.
A.
B.
C.
D.
E.
τ1 > τ2 > τ3
τ2 > τ1 > τ3
τ2 > τ3 > τ1
τ3 > τ1 > τ2
τ3 > τ2 > τ1
L⎞
⎟
⎝R⎠
⎛
i (t ) = i0 e
−t / ⎜
Clicker Question The switch in the circuit below is closed at t=0.
R = 20Ω
V = 10V
L = 10H
What is the initial rate of change of
current di/dt in the inductor,
immediately after the switch is
closed ?
(Hint) what is the initial voltage
across the inductor?)
A) 0 A/s
B) 0.5A/s
C)1A/s
D) 10A/s
E) None of these.
Answer: 1A/s. By the Loop Law, V(battery voltage) = iR +
Ldi/dt. Immediately after the switch is closed , the current is
zero (because the current thru the inductor cannot change
instantly), so iR is zero, so V = L di/dt. So di/dt = V/L =
10V/10H = 1 A/s.
12
Clicker Question
An LR circuit is shown below. Initially the switch is open. At
time t=0, the switch is closed. What is the current thru the
inductor L immediately after the switch is closed (time = 0+)?
R1 = 10Ω
V = 10V
L = 10H
R2 = 10Ω
A) Zero
B) 1 A
C) 0.5A
D) None of these.
Clicker Question
An LR circuit is shown below. Initially the switch is open. At
time t=0, the switch is closed.
R1 = 10Ω
V = 10V
L = 10H
R2 = 10Ω
After a long time, what is the current from the battery?
A) 0A
B) 0.5A
C) 1.0A
D) 2.0A
E) None of these.
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