```Ben Gurion University of the Negev
www.bgu.ac.il/atomchip
Physics 2B for Materials and Structural Engineering
Lecturer: Daniel Rohrlich
Teaching Assistants: Oren Rosenblatt, Shay Inbar
Week 9. Inductance – Self-inductance • RL circuits • Energy in a
magnetic field • mutual inductance • LC circuits • RLC circuits
Source: 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
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:
E  L
dI
,
dt
which is analogous to the definition of capacitance:
V 
q
C
Like C, L is defined to be positive.
.
Self-inductance
What is the self-inductance of a long solenoid?
We apply Faraday’s law, E   N
d B
, to the solenoid, by
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
Self-inductance
What is the self-inductance of a long solenoid?
From B = μ0nI, E   N
E  N
d B
dt
d B
dt
2
  r N
and ΦB = π r2B we have
dB
dt
2
 μ 0 n r N
dI
dt
from which we infer L = μ0 nπ r2N for a long solenoid.
,
Self-inductance
Since the formula for inductance is E   L
dI
dt
and the “emf” E
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”?
2
Answer: (a) Note E   μ 0 n  r N
dI
dt
2
2
  μ 0 n ( r l )
dI
dt
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.
RL circuits
Now back to the RL circuit:
When the switch or circuit
breaker is closed, we have
(summing potentials around
the circuit):
Ebat  IR  L
dI
0 ,
dt
where Ebat is a constant, the
“emf” of the battery. We
E bat 
E bat 
d 
R 
can write this as
I 
   I 
.
dt 
R 
L 
R 
RL circuits
The solution is
I (t ) 
Ebat
R
 const.  e
 Rt / L
and from t = 0 we have
E bat 

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
I (t ) 
Ebat
R
1  e
 Rt / L

.
,
RL circuits
Here is a graph of I ( t ) 
Ebat
R
1  e
 Rt / L
:
I(t)
Ebat/R
0.63 Ebat/R
L/R
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?
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
Ebat/R and Kirchhoff’s equation for the circuit is IR  L
Thus I ( t )  I ( 0 ) e  Rt / L  ( Ebat / R ) e  Rt / L .
Switch 2
Switch 1
dI
dt
 0.
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
Energy in a magnetic field
If we multiply each term in the circuit equation
Ebat  IR  L
dI
0
dt
by I and rearrange the terms, we obtain
2
I E bat  I R  IL
dI
.
dt
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
equation for an RL circuit without a battery is IR  L
dI
 0,
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):
U 

0
 dI 
LI 
 dt 
 dt 
 LI
2
dI 
I L
.
2
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:
uE 
U

C (V )
2

 0 A / d ( Ed ) 2

1
2
0E
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
U 
I L
2
and u B 
U
2
 r l
 B /  0 n 

B
2
20
.
2
2
 0 n r N
2
2

2
B  r l
20
Energy in a magnetic field
To summarize:
uE 
1
uB 
B
2
0E
2
20
2
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
the circuit is
q
L
C
dI
 0,
dt
and since I = dq/dt, it is
q
C
2
L
d q
dt
2
2
 0.
current oscillates!
Solving
d q
dt
2
 
q
LC
, we will find that the
LC circuits
2
The solution of
d q
dt
2

q
LC
is q(t) = qmax sin(ωt + δ), where
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

2

1
LC
, so   1 /
LC and T = 2π/ω = 2π LC .
(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
2
The solution of
d q
dt
2

q
LC
is q(t) = qmax sin(ωt + δ), where
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

2

1
LC
, so   1 /
LC and T = 2π/ω = 2π LC .
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.
The energy of a harmonic oscillator is the sum of its kinetic
and potential energies:
2
1
 dx 
2 2
U  m
  m x
2  dt 
2
1
.
The energy in the LC circuit has the same form:
2
1
 dq 
2
U  L
q
 
2  dt 
2C
1
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
now
q
C
2
L
d q
dt
2
R
dq
 0;
dt
its solution is an exponential
i t  
function q ( t )  q max e
in which ω is complex.
RLC circuits
Substituting this solution into the equation for the circuit, we
find
1
 L
2
 i  R  0;
C
and its solutions are
 i
R
2L

 R 


LC  2 L 
1
2
.
For R = 0 we recover the LC
result.
RLC circuits
Substituting this solution into the equation for the circuit, we
find
1
 L
2
 i  R  0;
C
and its solutions are
 i
R

2L
 R 


LC  2 L 
1
2
.
For 0  R  2 L / C , we get
damped oscillations at a
reduced frequency:
q(t) =
e–Rt/2L
qmax sin(ω't + δ), where ω' =

2
2
 (R / 2L) .
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/ 3 LC
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
C
L
dI 1
dt
 L
d
dt
( I1  I 2 ) 
B
I1
I2
A
q2
C
L
dI 2
dt
.
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
C
L
dI 1
dt
 L
d
dt
( I1  I 2 ) 
q2
L
C
dI 2
.
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 ) ,
C
dt
dt
  1/ 3 LC
q1  q 2
d
L
( I1  I 2 )  0 ,
C
dt
  1/
LC
,
.
```