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Chapter 36 Inductance
Capacitance C
++++++-
q  CV
dU  dqV  qdq C
U
q
0
0
U   dU   qdq / C
di
dB
d
iB



dt
dt
dt
di Inductance
 L  L
dt unit of L is henry (H)
di
dU   dq   idt  L idt
dt
u
U   dU
1 2
1
1
2
 q / C  C (V )  qV
2
2
2
Electric energy
0
1 2
  Lidi  Li
0
2
i
Magnetic energy
Calculating the capacitance
Procedure:
1. Suppose that the capacitor is charged, with ±q on
the two plates respectively.
2. Find the electric field E in the region between the
plates.
3. Evaluate the potential difference between the
positive and negative plates, by using the

formula:
 
V  V  V   E  ds

4.The expected capacitance is then:
C  q V
Calculating the Inductance
Procedure:
1. Suppose i
2. Find the magnetic field B, FB
3. Evaluate the EMF by using the formula:
d B
di
    L
dt
dt
d  Ldi
  Li
L

i
Calculating the Inductance
B   0 ni ΦB  μ 0 niA
NΦB  μ0 NniA
N  nl
NΦB 0 NnAi
2
L

 0 n lA
i
i
L is independent of i and depends
only on the geometry of the device.
Calculating the Inductance
ΦB  BdA


L
i
i

b
a
 0i
ldr  l b
2r
 0 ln
i
2 a
Calculating the capacitance
q

C
V
20l

b
b
a 20 rl dr ln a
q
q
a
a
b
b
Inductance of a Toroid
L?
 0iN
B  0 ni 
2r
b  iN
 0iNh b dr
  b
0
ΦB   B  dA  a Bhdr  a 2r hdr  2 a r
NΦB  0 N h b

ln
L
2
a
i
2
 0iNh b

ln
2
a
Inductors with Magnetic Materials
 0 N Nh h b b
LL m
ln ln
22 a a


B   m B0  L   m L0
2
0
   m0
2
L

i
C
0 A
C
d
C'   e
0 A
d

A
d
Ferromagnetic cores (κm >>1, κm =103 - 104)
provide the means to obtain large inductances.
RC Circuits
Combine Resistor and Capacitor in Series
a
Switch at position (a)
(b)

b
R
C
LR Circuits
q(t )  Ce(1t/ RCet / RC )
q(tt)/ RC
t / RC
Vc (t )  e   (1  e
)
C
dq   t /RCt / RC
i (t ) 
ee
dt R R
RC Circuits
Combine Resistor and Capacitor in Series
at→∞, i=ε/R. Switch at position (a)
di
di
  iR  L


iR

L

0
R
b

dt
dt
C  di  dt  di   R dt

  iR L
L
t=0, i=0
i
R
LR Circuits
i
t
di
R
ΔVR=-iR
 dt t =L/R
inductive
0 i  time constant
0
L
tR
R
t


L
t )
i

(
1

e
)
i 
R
R  t
R
ln
i
L

R
di
i  L
dt
RC Circuits
Combine Resistor and Capacitor in Series
at→∞, i=ε/R.
Switch at position (b)
R

t

d
i
L
R
b
iR  L  0 i  e

dt
R
C
t=0, i=0
LR Circuits
t =L/R
i

R
e


t  0, i 
R
ΔVR=iR
i
di
i  L
dt
t  , i  0
t
t
22
1
ddqqq dq22 q
2




0

L


q
2
2
2
LC
C
dt
dt
CL
dt
q  q0 cos(t   )
L-C circuit
i ?
L
C
dq
i
 q0 sin( t   )
Energy
conservation
dt
q
i
q
di
L 0
C
dt
dq
i
dt
t
1 q02
1 2 2 2
2
UE 
cos (t   ) U B  Lq0  sin (t   )
2
2C
Electric energy
Magnetic energy
Damped and Forced oscillations
d 2 x b dx
2



x0
2
dt
m dt
x(t )  xm e bt / 2 m cos(t  )
If there are resistances
in circuit, the U is no longer
constant.
   m cos  t
Resonance 
1
2
b  R , m  L,  
L
LC
C
 Rt / 2 L
R
q(t )  q0e
cos(t  )
  2  ( R/ 2 L) 2
di q
dq
 L   iR  0 i  
dt C
dt
d 2 q R dq 1


q
 0m cos  t
2
dt
L dt LC
Energy Storage in a Magnetic Field
K a
di
  iR  L  0
dt
di
2
i  i R  Li
dt
ΔVR=-iR
 i  L
i= (dq/dt)= (dq)/dt, the power by the emf device.
i2R, the power consuming in the resistor.
Li(di/dt), the rate at which energy is stored in the space
of the inductor, it can be put out, when switch to b
dU
di
 Li
dt
dt
dU  Lidi
0
 dU  
U
0
i
Lidi
1 2
U  Li
2
di
dt
2
q
1q
UE 
2 C q
energy is stored in the electric field
1 2
U B  Li
2
L   0 n 2lA
inductance
UB
i
energy is stored in the magnetic field
i
magnetic field
1
uE   0 E 2
2
UB
UB 
2
B
U B 0 n 2
uB 

i 
2 0
V
2
2
 0 n 2lA
2
i2
B  0 ni
Analogy to Simple Harmonic Motion
x  xm sin( t  )
2
d x
2
 x 0
2
dt
k
 
m
2
dx
v
 xm  cos(t  )
dt
Us 
1 2
kx
2
K
1 2
mv
2
q  q0 sin( t  )
2
d q
1
2
2
 q  0  
2
LC
dt
q  x Lm
i v 1 C k
dq
i
 q0 cos(t  )
dt
2
1q
UE 
2C
UB 
1 2
Li
2
Electric Field

E
q
40 r


F  qE
2
rˆ
Magnetic Field
  0 qv  rˆ
B
2
4r

  0ids  rˆ
dB 
4r 2

 
F  qv  B

 
dF  ids  B
Electric Field


P  qd
1
p
E
40 x 3
 
t  P E

Magnetic Field

  iA


0 
B

3
2z
  
t  B
Electric Field
 

E  E0  E '
Magnetic Field
 

B  B0   0 M
 1 
E
E0 ( e  1)


B   m B0 ( m  1,  m  1)
   e 0
   m 0
e


 Es  dl  0
  q
 E  ds 
0
 
 B  dl  0i Ampere Law
 
 B  ds  0
Gauss Law
Electric Field
Magnetic Field
Induction

 
d
B 
 Ei  dl   i   dt   t  dS
d  (v  B)  ds
   (v  B)  ds
Electric Field
Magnetic Field
N B
di
  L
L
dt
i
q
q  C V C 
V
q(t )  C (1  e
t / RC
2
1q
UE 
2C
1
2
uE   0 E
2
)
i

R
(1  e
R
 t
L
1 2
U B  Li
2
1 B2
uB 
2 0
)
Electric Field
Magnetic Field
1
 
LC
2
2
2
0
q  q0 sin( t  )
dq
i
 q0 cos(t  )
dt
q
1q
2
UE 

sin (t   )
2C
2C
1 2 1 2 2 2
U B  Li  Lq  cos (t   )
2
2
0
qx
iv
Lm 1 C k
Example
UE  ?
2
q
1
q
1
) 2 2 2 2
uE   0 E 2   0 (
8  0l r
2
20lr
2
q 2 dr
q2
dU B  uB dV 2 2 2 (2rl )(dr )
40l r
8  0l r
2
b q
q2
b
dr

ln
UB  
a 4 l r
40l a
0
20l
q
q
 b

C
q
b
V
a 20 rl dr ln
a
a
b
q2
UE 
2C
q2
b

ln
40l a
Example
UB  ?
2
1 B2

i
1  0i 2
uB 

(
)  02 2
b
2 0
8 r
2 0 2r
a
 0i 2
 0i 2l dr
dU B  uB dV  2 2 (2rl )(dr ) 
8 r
4 r
2
2
b  i l dr

i
l b
0
0
UB  

ln
a 4
r
4
a
ΦB  BdA
L
2

2
Li

i
l b
0
i
i
UB 

ln
b  i
2
4
a
0


a
2r
i
ldr
 0l b

ln
2 a
Exercises
P839-841 9, 10, 23, 41
Problems
P842
3, 5