Chapter 32 Inductance
32.1 Self-Induction and Inductance
Self-Inductance
Φ m = BA = µ0 nIA ∝ I --> Φ m = LI
The unit of the inductance is henry (H). 1H = 1
Wb
T ⋅ m2
=1
A
A
When the current in the circuit is changing, the magnetic flux is also changing.
dΦ d (LI )
dI
dΦ
dI
=
=L
--> The induced emf should be ε = −
= −L
dt
dt
dt
dt
dt
The self inductance of a infinite long solenoid:
Φ m = NAB = NAµ0
ε = −L
N
N2
Φ
N2
I = µ0 A
I --> L = m = µ0 A
l
l
I
l
dI
dt
Considering the inductor having an internal resistance r, the potential difference is:
∆V = ε − Ir = − L
dI
− Ir
dt
Example: Model a long coaxial cable as two thin, concentric,
cylindrical conducting shells of radii a and b and length l. The
conducting shells carry the same current in opposite directions.
Calculate the inductance L of this cable.
Calculate magnetic flux:
µI
B= 0
2πr
Φ = LI
µ0 I
µ I b
drdz = 0 l ln 
2πr
2π
a
0 a
l b
Φ = ∫∫
L=
µ 0l  b 
ln 
2π  a 
1
32.2 RL Circuits
Use Kirchhoff’s rule:
r r
E
∫ ⋅ dl = 0
ε 0 − IR − L
dI
=0
dt
dI
dI
= 0 --> ε 0 − IR = L
dt
dt
L d (ε 0 − IR )
--> dt = −
R (ε 0 − IR )
Differential Eq: ε 0 − IR − L
--> dt = L
--> −
dI
(ε 0 − IR )
R
R
− t
− t
 ε − IR 
R
ε 
 --> ε 0 − IR = ε 0e L --> I = 0 1 − e L 
t = ln 0

L
R 
 ε0 

Time constant: τ =
L
R
Example: Find the total energy dissipated in the resistor R, when the current in the
inductor decreases from its initial value of I0 to 0?
I=
ε0
R
e
R
− t
L
∞
, P = I 2 R --> U = ∫ I 02e
R
−2 t
L
0
Rdt =
1 2
LI 0
2
32.3 Energy in a Magnetic Field
Obtain the magnetic energy from the emf induced by self inductance.
Φ m = LI --> The induced emf is ε = −
dΦ m
dI
= −L
dt
dt
The energy dissipated or the power is P = IV = Iε = − LI
dI
dt
The total energy when the current has reached its final value If is:
t =tf
 dI 
U = ∫  LI dt =
dt 
t =0 
I = If
1
∫ LIdI = 2 LI
2
I =0
Calculate the magnetic energy by obtaining the energy stored in the self inductor of an
infinite solenoid.
2
B = µ0 nI , Φ = nlA(µ0 nI ) = LI
U=
1 2 1
LI = µ0 n 2 ( Al )I 2 = umV
2
2
um =
1
B2
µ0 n 2 I 2 =
2
2µ0
<---->
1
ue = ε 0 E 2 (Do you remember how to get this?)
2
Example: A certain region of space contains a uniform magnetic field of 0.020 T and a
uniform electric field of 2.5 X 106 N/C. Find (a) the total electromagnetic density.
(
)(
1
1
ue = ε 0 E 2 = 8.85 × 10 −12 2.5 × 106
2
2
)
2
= 27.7 J/m3
B2
(0.02) = 159J/m3
=
2 µ0 2 4π × 10 − 7
2
um =
(
)
32.4 Mutual Inductance
Mutual Inductance
The magnetic field of loop1 is: B ~
The flux at 2 is Φ 2 ~
µ0 I1
2R
The flux at 1 is Φ1 ~ µ0
πR = µ0
2
µ0 I1
2R
πR 2
2
loop 2
I1 = I1M 12
loop 1
πR 2
I 2 = I 2 M 21
2
The concept of inductance: Φ = MI
M 12 = M 21 --> The mutual inductance is determined when the geometrical
configuration between the two loops is given.
B in 1 due to 2: B = µ0
N2
I2
l
( )
 N

Flux in 1: Φ = N1 πr12  µ0 2 I 2  = I 2 M 21
l


B in 2 due to 1: B = µ 0
N1
I1
l
( )
 N 
Flux in 2: Φ = N 2 πr12  µ0 1 I1  = I1M 12
l 

3
( ) N lN
M 12 = M 21 = µ0 πr12
1
2
32.5 Oscillations in an LC Circuit
Kirchhoff’s Rule:
dI Q
d 2Q Q
− =0
L 2 + =0
dt C
dt
C
Compare with: F = ma = −kx
−L
d 2x
m 2 + kx = 0 solution: x(t ) = A cos(ωt + φ )
dt
L
d 2Q Q
+ = 0 solution: Q (t ) = Qmax cos(ωt + φ ) , ω =
dt 2 C
I = −ωQmax sin (ωt + φ )
U = UC + U L =
=
1
LC
Q2 L 2
L
1 2
2
+ I =
Qmax cos 2 (ωt + φ ) + ω 2Qmax
sin 2 (ωt + φ )
2C 2
2C
2
2
Qmax
2C
4
Apply the Kirchhoff’s loop rule:
−L
I
dI Q
− =0
dt C
Loop direction
d 2Q Q
--> L 2 + = 0 (2nd Differential Equation, compare with
dt
C
d 2x
harmonic oscillation: F = ma = m 2 = − kx with the answer of x = A cos(ωt + δ ) ,
dt
where ω =
k
)
m
Remember the pattern of this differential equation:
d 2Q Q
L 2 + = 0 --> the solutions are periodical functions and the useable functions
dt
C
are sin ( x ) ( cos( x ) ), exp(ix ) ( exp( x ) ), …
Guess that the answer is Q = A cos(Bt + C ) . (Here A and C can be determined by
initial conditions.)
--> − LB 2 A cos(Bt + C ) +
1
A cos(Bt + C ) = 0 --> B =
C
--> Q = A cos(ωt − δ ) & I =
1
≡ω
LC
dQ
= −ωA sin (ωt − δ ) = − I peak sin (ωt )
dt
Capacitor --> Electric Field
--> Potential Energy
Inductor --> Moving of
Charges --> Kinetic Energy
5
The average energy stored in the capacitor (inductor) is
Q2 1 2
( LI ).
2C 2
The instantaneous energy transferring in the circuit is:
2
Q 2 1 2 A2 cos 2 (ωt − δ ) 1
A2 Q peak 1 2
2 2
2
+ LI =
+ Lω A sin (ωt − δ ) =
=
= LI peak
2C 2
2C
2
2C
2C
2
What are physical pictures of Q peak and I peak ?
Example: A 2-µF capacitor is charged to 20 V and the capacitor is then connected
across a 6-µH inductor. (a) What is the frequency of oscillation? (b) What is the peak
value of the current?
2
2
1
CV 2 Q peak LI peak
, (b)
=
=
2
2C
2
LC
(a) ω =
Simple AM Radio receiver:
32.6 The RLC Circuit
Kirchhoff’s Rule:
− IR − L
dI Q
− =0
dt C
Power Consideration:
P = LI
dI Q
+ I = −I 2R
dt C
d 2Q
dQ Q
L 2 +R
+ =0
dt
dt C
Compare with damped oscillation:
m
d 2x
dx
+ b + kx = 0
2
dt
dt
x(t ) = Ae
b 2 < 4mk
Q (t ) = Qmax e
−
R
t
2L
x(t ) = Ae nt , mn 2 + bn + k = 0
−
b
t
2m
 4mk − b 2
cos

2m

 4L / C − R2
cos

2L

n=
− b ± b 2 − 4mk
2m

t



t


6
RLC Circuit (Damped Oscillation)
Diff Eq: − L
dI Q
− − IR = 0
dt C
I
Loop direction
d 2Q
dQ Q
--> L 2 + R
+ = 0 (Damped Oscillations: F = ma = − kx − bv )
dt
dt C
1
d 2Q Q
( L 2 + = 0 ).
dt
C
LC
for solving the differential equation
The natural frequency (no resistaor) is: ω0 =
We guess a solution of Q = Ae Bt
L
d 2Q
dQ Q
+R
+ = 0.
2
dt
dt C
1  Bt
 2
 LB + RB +  Ae = 0 --> B =
C

− R ± R2 − 4
2L
L
2
C =− R ±  R  − 1
 
2L
 2 L  LC
2
1
 R 
over-damped:   −
>0
 2 L  LC
7
2
1
 R 
under-damped:   −
<0
 2 L  LC
2
under-damped solution: Q = Ae
−
2  R 
R
t ± i ω 0 − 2 L  t


2L
e
The energy distributed in the circuit elements is:
d 2Q
dQ Q
dI
Q dQ
L 2 +R
+ = 0 --> LI
+ RI 2 +
=0
dt
dt C
dt
C dt
d  1 2 Q2  2
 LI +
+I R =0
-->
dt  2
2C 
8