Ben Gurion University of the Negev
www.bgu.ac.il/atomchip, www.bgu.ac.il/nanocenter
Physics 2 for Electrical Engineering
Lecturers: Daniel Rohrlich , Ron Folman
Teaching Assistants: Ben Yellin, Yoav Etzioni
Grader: Gady Afek
Week 9. Inductance – Self-inductance • RL circuits • Energy in a
magnetic field • mutual inductance • LC circuits • RLC circuits
Sources: Halliday, Resnick and Krane, 5th Edition, Chap. 36.
Self-inductance
We have already seen several circuit elements:
battery
capacitor
switch
resistor
and now we are going to get to know the
inductor
Self-inductance
Inductors come in various sizes and colors:
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Self-inductance
Let’s see what an inductor does in a circuit.
Self-inductance
Let’s see what an inductor does in a circuit. When current
starts to flow through an inductor, it induces a magnetic flux
that changes that same current. Hence the term “self”
inductance.
The inductance L of an inductor is defined according to the
“emf” E that it produces for a given change in current:
dI
E  L
,
dt
which is analogous to the definition of capacitance:
q
V 
C
Like C, L is defined to be positive.
.
Self-inductance
What is the self-inductance of a long solenoid?
Self-inductance
What is the self-inductance of a long solenoid?
d B
, to the solenoid, by
We apply Faraday’s law, E   N
dt
calculating the magnetic flux in the solenoid: ΦB = πr2B,
where N is the total number of coils in the solenoid, r is its
radius and n is the number of coils per unit length. Also, the
formula for B inside a solenoid is B = μ0 nI.
Self-inductance
What is the self-inductance of a long solenoid?
d B
, to the solenoid, by
We apply Faraday’s law, E   N
dt
calculating the magnetic flux in the solenoid: ΦB = πr2B,
where N is the total number of coils in the solenoid, r is its
radius and n is the number of coils per unit length. Also, the
formula for B inside a solenoid is B = μ0 nI. But what is N
doing in the formula for E?
Self-inductance
What is the self-inductance of a long solenoid?
d B
, to the solenoid, by
We apply Faraday’s law, E   N
dt
calculating the magnetic flux in the solenoid: ΦB = πr2B,
where N is the total number of coils in the solenoid, r is its
radius and n is the number of coils per unit length. Also, the
formula for B inside a solenoid is B = μ0 nI. The inductor
“links” with itself N times!
Self-inductance
What is the self-inductance of a long solenoid?
d B
From B = μ0nI, E   N
and ΦB = π r2B we have
dt
d B
2 dB
2 dI
E  N
  r N
 μ 0 n r N
,
dt
dt
dt
from which we infer L = μ0 nπ r2N for a long solenoid.
Self-inductance
dI
Since the formula for inductance is E   L
and the “emf” E
dt
has units of volts, the unit of inductance must be volt/
(ampere/second). This unit is called the henry and denoted H:
H = V·s/A .
Joseph Henry
Self-inductance
Example 1: (a) Calculate the inductance L of a solenoid with
100 coils/cm if the volume inside the solenoid is 10–6 m3.
(b) The current in the solenoid is decreasing by 0.50 A/s. What
is the induced “emf”?
Self-inductance
Example 1: (a) Calculate the inductance L of a solenoid with
100 coils/cm if the volume inside the solenoid is 10–6 m3.
(b) The current in the solenoid is decreasing by 0.50 A/s. What
is the induced “emf”?
dI
2
2 dI
 μ 0 n ( r l )
,
Answer: (a) Note E  μ 0 n r N
dt
dt
2
where l is the length of the solenoid; thus π r2l is the volume,
and we have
L = μ0 n2(π r2l) = (4π × 10–7 T·m/A)(108/m2 )(10–6 m3)
= 4π × 10–5 T·m2/A
and we check via “E + v × B” that T·m2/A = V·s/A.
Self-inductance
Example 1: (a) Calculate the inductance L of a solenoid with
100 coils/cm if the volume inside the solenoid is 10–6 m3.
(b) The current in the solenoid is decreasing by 0.50 A/s. What
is the induced “emf”?
Answer: (b) We substitute L = 4π × 10–5 V·s/A back into the
definition E = –L(dI/dt) to obtain E = 2π × 10–5 V.
Self-inductance
Example 2: We fill the solenoid with (a) a diamagnetic or
(b) a paramagnetic or (c) a ferromagnetic material. What
happens to L?
Self-inductance
Example 2: We fill the solenoid with (a) a diamagnetic or
(b) a paramagnetic or (c) a ferromagnetic material. What
happens to L?
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RL circuits
Now back to the RL circuit:
When the switch or circuit
breaker is closed, we have
(summing potentials around
the circuit):
dI
Ebat  IR  L
0 ,
dt
where Ebat is a constant, the
“emf” of the battery. We
d  Ebat 
R  Ebat 
I
  I 
.
can write this as



dt 
R 
L
R 
RL circuits
The solution is
Ebat
I (t ) 
 const. e  Rt / L ,
R
and from t = 0 we have
Ebat 

const.   I (0) 
.

R 

So if I(0) = 0 (if there is
no current at t = 0, because
the switch was open until
then), then
Ebat
I (t ) 
1  e  Rt / L .
R


RL circuits


Ebat
1  e  Rt / L :
Here is a graph of I (t ) 
R
I(t)
Ebat/R
0.63 Ebat/R
L/R
t
RL circuits
dI Ebat  Rt / L
And here is a graph of

e
:
dt
L
dI/dt
Ebat/L
t
RL circuits
Example 1: A switch controls the current in an RL circuit with
large L. Is sparking more likely when you initially close the
switch or when you open it after t >> L/R?
RL circuits
Example 1: A switch controls the current in an RL circuit with
large L. Is sparking more likely when you initially close the
switch or when you open it after t >> L/R?
Answer: When you close the switch, there is no initial current,
I(0) = 0, so IR = 0 and the potential on the inductor is –Ebat.
The potential over the switch drops immediately to 0 and the
inductor won’t immediately let current through, so sparking
isn’t likely. When you open the switch, there is already a
steady current Ebat/R and the inductor does not allow the
current to drop to 0 immediately, so charge builds up on the
switch and may cause sparking.
RL circuits
Example 2: Consider the setup below; assume that Switch 1
has been closed for a long time while Switch 2 has been open.
At time t = 0, Switch 1 is opened and Switch 2 is closed. What
is the current in the upper loop for t > 0?
Switch 2
Switch 1
RL circuits
Answer: At t = 0, the current has reached its maximum value
dI
 0.
Ebat/R and Kirchhoff’s equation for the circuit is IR  L
dt
Thus I (t )  I (0) e  Rt / L  (Ebat / R) e  Rt / L .
Switch 2
Switch 1
RL circuits
Example 3: We repeat the last example with two RL circuits, A
and B. Which circuit has the larger L, if the resistances and
batteries are the same?
I(t)
A
B
Open Switch 1,
close Switch 2
t
RL circuits
Answer: The current depends on an exponential in t/τ where
the time constant τ of the circuit is τ = L/R. Since B has a
larger τ (slower exponential decay), its L is larger.
I(t)
A
B
Open Switch 1,
close Switch 2
t
Energy in a magnetic field
If we multiply each term in the circuit equation
dI
Ebat  IR  L
0
dt
by I and rearrange the terms, we obtain
dI
IEbat  I R  IL
dt
2
.
What is IEbat? It is the rate at which charge exits the battery
at the + end and enters the battery at the – end, times the
difference in electric potential between the two ends. So it is
the rate at which the battery delivers potential energy to the
circuit. What happens to this potential energy?
Energy in a magnetic field
If we multiply each term in the circuit equation
dI
Ebat  IR  L
0
dt
by I and rearrange the terms, we obtain
dI
IEbat  I R  IL
dt
2
.
What is IEbat? It is the rate at which charge exits the battery
at the + end and enters the battery at the – end, times the
difference in electric potential between the two ends. So it is
the rate at which the battery delivers potential energy to the
circuit. What happens to this potential energy? We recognize
I2R as the rate of energy (heat) dissipation in the resistor.
Energy in a magnetic field
If we multiply each term in the circuit equation
dI
Ebat  IR  L
0
dt
by I and rearrange the terms, we obtain
dI
IEbat  I R  IL
dt
2
.
What is IEbat? It is the rate at which charge exits the battery
at the + end and enters the battery at the – end, times the
difference in electric potential between the two ends. So it is
the rate at which the battery delivers potential energy to the
circuit. What happens to this potential energy? The last term,
IL(dI/dt), must be the rate at which energy is stored in the
inductor.
Energy in a magnetic field
What does it mean that IL(dI/dt) is the rate at which energy is
stored by the inductor? As we saw in Example 2, Kirchhoff’s
dI
 0,
equation for an RL circuit without a battery is IR  L
dt
implying that the inductor is the source for the “emf” across the
resistor.
Energy in a magnetic field
Hence if we integrate IL(dI/dt) with respect to time t, starting
from I(0) = 0, we must obtain the total energy U stored in the
inductor at time t or, equivalently, at current I(t):
I 2L
 dI 
U  LI   dt  LI dI 
.
2
 dt 


0
0
This expression is analogous to the expression U = q2/2C for
the energy stored in a capacitor.
Energy in a magnetic field
From U = q2/2C = C(ΔV)2/2 we derived the energy density uE
in an electric field of strength E:
U
C (V ) 2  0 A / d ( Ed ) 2 1
uE 


 0E2
Ad
2 Ad
2 Ad
2
.
Energy in a magnetic field
From U = q2/2C = C(ΔV)2/2 we derived the energy density uE
in an electric field of strength E:
U
C (V ) 2  0 A / d ( Ed ) 2 1
uE 


 0E2
Ad
2 Ad
2 Ad
2
.
Likewise we can translate between I and B to derive the energy
density uB in a magnetic field of strength B. For a long enough
solenoid we have B = μ0 nI and L = μ0 nπ r2N, where N = nl, so
2
2
2

n

r
N
I 2L
B

r
l
2 0
U
 B /  0 n 

2
2
2 0
B2

.
and u B 
 r 2l 2 0
U
Energy in a magnetic field
To summarize:
1
uE   0 E 2
2
B2
uB 
2 0
Mutual inductance
If two electrical coils are close together, a changing flux due to
a changing current in one coil may induce a current in the other
coil, and vice versa. This is the mutual induction of two coils
and depends on their mutual inductance M.
B
Coil 2
Coil 1
Source
Mutual inductance
Let Φ12 denote the magnetic flux through one turn of Coil 2
due to Coil 1. We will assume that Φ12 is proportional to I1,
since Ampère’s law implies that B is proportional to I. We
write N2Φ12 = MI1 where N2 is the number of turns in Coil 2.
It can be shown that the inductance M is mutual: N2Φ12 = MI1
and also N1Φ21 = MI2 with the same M.
Mutual inductance
Given that N2Φ12 = MI1 and N1Φ21 = MI2, we calculate the
induced “emf” in each coil as
dI1
dI 2
E2   M
, E1   M
,
dt
dt
and now the formal similarity of mutual inductance and selfinductance
dI
E  L
dt
is apparent.
Mutual inductance
Given that N2Φ12 = MI1 and N1Φ21 = MI2, we calculate the
induced “emf” in each coil as
dI1
dI 2
E2   M
, E1   M
,
dt
dt
and now the formal similarity of mutual inductance and selfinductance
dI
E  L
dt
is apparent.
Can there be mutual conductance with self-conductance? Can
there be self-inductance without mutual inductance?
Mutual inductance
Example 1: Calculate the mutual inductance M of two coils, if
Coil 1 is a long solenoid of radius r, length l and N1 turns, and
Coil 2 (with N2 turns) encloses Coil 1.
Mutual inductance
Example 1: Calculate the mutual inductance M of two coils, if
Coil 1 is a long solenoid of radius r, length l and N1 turns, and
Coil 2 (with N2 turns) encloses Coil 1.
Answer: If Coil 2 encloses Coil 1, then all the magnetic flux in
Coil 1 links all the turns of Coil 2, and so Φ12 is just the flux
πr2B of the solenoid. Thus N2Φ12 = N2πr2B = N2 (πr2μ0 N1/l) I1,
so for this case M = πr2μ0 N1N2/l.
Mutual inductance
Example 2: The battery inside this electric toothbrush, in its
handle, is charged by mutual induction. The base has a coil
that just fits into the handle. Given
that Nbase = 800, Nhandle = 1500,
that the cross-sectional area of the
coils is 1.0 × 10–4 m2, and that
l = 0.02 m, what is the mutual
inductance of the two solenoids?
Mutual inductance
Example 2: The battery inside this electric toothbrush, in its
handle, is charged by mutual induction. The base has a coil
that just fits into the handle. Given
that Nbase = 800, Nhandle = 1500,
Coil 1 (base)
that the cross-sectional area of the
Nbase
coils is 1.0 × 10–4 m2, and that
l = 0.02 m, what is the mutual
Coil 2 (handle)
Nhandle
inductance of the two solenoids?
l
Answer: We already calculated
M = πr2μ0 N1N2/l for a solenoid;
we now have πr2 = 1.0 × 10–4 m2,
N1N2 = 1.2 × 106 and l = 0.02 m,
so M = 7.5 mH.
LC circuits
At right is an LC circuit:
Let’s assume that the
capacitor is charged when
the switch is closed at t = 0.
For t > 0 the equation for
q
dI
L
 0,
the circuit is
C
dt
and since I = dq/dt, it is
q
d 2q
d 2q
q
L
 0. Solving

, we will find that the
2
2
C
LC
dt
dt
current oscillates!
LC circuits
The solution of
d 2q
q

is q(t) = qmax sin(ωt + δ), where
LC
dt 2
qmax , ω and δ are constants: qmax is the amplitude of the charge
oscillations, ω is their angular frequency and δ is their phase.
Substituting this solution into the differential equation, we find
1
 
, so   1 / LC and T = 2π/ω = 2π LC .
LC
2
(What about units? The units of L are the henry, H = V·s/A;
the units of C are the farad, F = C/V. So LC has units s2 and
1 / LC has units s–1 = Hz.)
LC circuits
The solution of
d 2q
q

is q(t) = qmax sin(ωt + δ), where
LC
dt 2
qmax , ω and δ are constants: qmax is the amplitude of the charge
oscillations, ω is their angular frequency and δ is their phase.
Substituting this solution into the differential equation, we find
1
 
, so   1 / LC and T = 2π/ω = 2π LC .
LC
2
Note: While the charge is q(t) = qmax sin(ωt + δ), the current is
I = dq/dt = ωqmax cos(ωt + δ) = Imax cos(ωt + δ). This means
that the charge and current are π/2 or T/4 out of phase.
LC circuits
The formulas for q and I in an LC circuit are analogous to the
formulas for the position x and momentum p of a body in a
harmonic oscillator. Thus, we can intuitively understand LC
circuits by analogy with harmonic oscillators.
I=0
qmax
v=0
E
t=0
–qmax
xmax
LC circuits
The formulas for q and I in an LC circuit are analogous to the
formulas for the position x and momentum p of a body in a
harmonic oscillator. Thus, we can intuitively understand LC
circuits by analogy with harmonic oscillators.
I = Imax
v = vmax
q=0
t = T/4
x=0
B
LC circuits
The formulas for q and I in an LC circuit are analogous to the
formulas for the position x and momentum p of a body in a
harmonic oscillator. Thus, we can intuitively understand LC
circuits by analogy with harmonic oscillators.
I=0
–qmax
E
v=0
t = T/2
qmax
xmax
LC circuits
The formulas for q and I in an LC circuit are analogous to the
formulas for the position x and momentum p of a body in a
harmonic oscillator. Thus, we can intuitively understand LC
circuits by analogy with harmonic oscillators.
I = Imax
v = vmax
q=0
t = 3T/4
x=0
B
LC circuits
The formulas for q and I in an LC circuit are analogous to the
formulas for the position x and momentum p of a body in a
harmonic oscillator. Thus, we can intuitively understand LC
circuits by analogy with harmonic oscillators.
I=0
qmax
v=0
E
t=T
–qmax
xmax
LC circuits
The formulas for q and I in an LC circuit are analogous to the
formulas for the position x and momentum p of a body in a
harmonic oscillator. Thus, we can intuitively understand LC
circuits by analogy with harmonic oscillators.
The energy of a harmonic oscillator is the sum of its kinetic
and potential energies:
2
1  dx 
1
U  m   m 2 x 2 .
2  dt 
2
The energy in the LC circuit has the same form:
2
1  dq 
1 2
U  L  
q
2  dt 
2C
In both systems, the total energy is constant.
.
RLC circuits
Just as harmonic motion may be forced or damped, so can an
electrical circuit. Adding a resistor to an LC circuit turns it into
an RLC circuit, and the resistor damps the oscillations.
The equation for the circuit is
q
d 2q
dq
now
L
R
 0;
2
C
dt
dt
its solution is an exponential
it 
q
(
t
)

q
e
function
max
in which ω is complex.
RLC circuits
Substituting this solution into the equation for the circuit, we
1
find  L 2  iR  0;
C
this is a quadratic equation
and its solutions are
R
 i

2L
2
1  R 

 .
LC  2 L 
For R = 0 we recover the LC
result.
RLC circuits
Substituting this solution into the equation for the circuit, we
1
find  L 2  iR  0;
C
this is a quadratic equation
and its solutions are
R
 i

2L
2
1  R 

 .
LC  2 L 
For 0  R  2 L / C , we get
damped oscillations at a
reduced frequency:
q(t) =
e–Rt/2L
2
2


(
R
/
2
L
)
.
qmax sin(ω't + δ), where ω' =
RLC circuits
Substituting this solution into the equation for the circuit, we
1
find  L 2  iR  0;
C
this is a quadratic equation
and its solutions are
R
 i

2L
2
1  R 

 .
LC  2 L 
For 0  R  2 L / C , we get
damped oscillations at a
reduced frequency:
q(t) =
e–Rt/2L
2
2


(
R
/
2
L
)
.
qmax sin(ω't + δ), where ω' =
Halliday, Resnick and Krane, 5th Edition, Chap. 36, Prob. 13:
Three identical inductors (with inductance L) and two identical
capacitors (with capacitance C) are connected as shown. Show
the circuit has two different angular frequencies,   1/ 3LC
and   1/ LC .
I1
I2
Halliday, Resnick and Krane, 5th Edition, Chap. 36, Prob. 13:
Answer: The three paths from “A” to “B” must have the same
potential difference. We therefore obtain
q1
dI1 q2
dI 2
d
L

L
  L ( I1  I 2 ) .
C
dt
C
dt
dt
B
I1
I2
A
Halliday, Resnick and Krane, 5th Edition, Chap. 36, Prob. 13:
Answer: The three paths from “A” to “B” must have the same
potential difference. We therefore obtain
q1
dI1 q2
dI 2
d
L

L
  L ( I1  I 2 ) .
C
dt
C
dt
dt
We can now derive equations for q1 – q2 and for q1 + q2 with
different angular frequencies:
q1  q 2
d
d
 L ( I1  I 2 )  2 L ( I1  I 2 ) ,   1/ 3LC ,
C
dt
dt
q1  q2
d
 L ( I1  I 2 )  0 ,
  1/ LC .
C
dt