MUTUAL INDUCTANCE M
Coil 2
N2 turns
Coil 1
N1 turns
i1(t)
The flux passing through each of the N2 turns
of coil 2 due to the current in coil 1 is Φ 2.
Clearly, N2Φ2 is proportional to i1, thus
N2 Φ2 = M i1
where M is called the mutual inductance
The mutual inductance is defined by
N 2Φ 2
N1Φ1
M=
=
i1
i2
The inductance M depends only on geometry.
Units:
Henry (H)
1 H = 1 Wb/A = 1 V·s/A
By Faraday’s Law, there is an emf induced
across the N2 turns of coil 2
d ( N 2Φ 2 )
d ( M i1 )
dΦ2
E2 = N 2 d t =
=
dt
dt
di
E 2 = M d t1
Similarly,
di
E 1 = M d t2
EXAMPLE:
Two coils, one wrapped on top of the other
Cross-sectional area of coils is A
Coil 2
N2 turns
A
Coil 1
N1 turns
l
i1
The magnetic field inside the solenoid, coil 1, is
N
B1 = µ0 1 i1
l
All this field passes through coil 2; thus,
N 2Φ 2
N 2B1A
M=
=
i1
i1
M =
µ 0 N 1N 2 A
l
M depends only on the geometry
SELF INDUCTANCE L
A
N turns
i
i
The magnetic field inside the coil is B and is
proportional to i. The flux through each turn
is Φ = BA and is also proportional to i. Thus,
the total flux linking the turns of a coil is
Φ = Li
NΦ
where the constant of proportionality L is the
self inductance.
L=
NΦ
i
By Faradays Law, the coil must induce an emf
in itself given by
dΦ
d NΦ
d Li
E = N d t = ( d t ) = (d t )
di
E = L dt
The direction can be obtained via Lenz’s Law
EXAMPLE:
L for a solenoid of cross-sectional area A with
N turns in a length l.
N
B = µ0 i
l
NB A
N A µ0N i
NΦ
L=
=
=
l
i
i
i
µ0 N 2A
L=
l
Note, L depends only on geometry
ENERGY STORED IN A MAGNETIC FIELD
When the current i in an inductor is changing
di
the induced emf is
at a rate
dt
di
E = L dt
The rate energy is supplied to the inductor is
di
P = E i = Li
dt
The energy supplied in a time dt is
dW = P dt = L i di
The total work done (the energy stored in the
inductor) to increase the current from 0 to
some final current I is
I
W = ∫ d W = ∫ L i d i = 12 LI2
0
The work W done to build up the current in
the inductor is equal to the energy U stored in
the inductor
U = 12 LI2
Compare this to the energy stored in a
capacitor when the charge is increased from 0
to some final charge Q
U = 12 1 Q2
C
= 12 CV2
ENERGY DENSITY IN A MAGNETIC FIELD
Consider the solenoid inductor L with N turns,
cross-sectional area A, and length l.
µ0 N 2A
L=
l
N
Bl
B = µ0 I ⇒ I =
µ0N
l
Thus,
2
2
2


B
µ
N
A
B
l

 = 1 Al
U = 12 LI2 = 12  0

2 µ0
 l   µ0N
The volume of the inductor is Al, hence, the
energy density is
2
B
u = 12
µ0
This is analogous to the energy density due to
the electric field
u = 12 ε 0E 2
SERIES LR CIRCUIT
c
b
L
V
-
+
R
i
i
di
V – iR – L = 0
dt
dt
di
=
V - iR L
i
t
0
0
di′
dt ′
=
∫ V - i ′R ∫ L
V - i R t
=
– 1 ln 
R  V  L
a
i(t) = V (1 − e − t τ )
R
where τ = L is the inductive time constant.
R
V
I=2.00
R
1.50
1.00
0.50
0.00
0
1
2
3
4
t in units of τ
5
Now, U = 12 Li2 and the rate energy changes
(the power) in the inductor is
di  di
dU
= i ( V − i R ) = iV–i2R
= Li = i L
d t  dt 
dt
dU
iV = i 2 R + dt
Power
from
source
Power
dissipated
in R
Power
supplied
to L
c
b
L
i
R
i( 0 ) = I
a
di
<0
dt
i
di
iR + L = 0
dt
i
t
I
0
d i′
R
=
–
∫ i′ ∫ L d t ′
i( t ) = I e − t τ
I
2.00
1.50
1.00
0.50
0.00
0
1
2
3
4
t in units of τ
5
LC CIRCUIT
L
i
-q +q
C
We take q(0)=Q, and i(0)=0
di q
–L + =0
dt C
Since
i(t) = –
dq
dt
2
q
d q
2 + LC = 0
dt
Try solution of the form
ωt + φ)
q(t) = Q0 cos(ω
where Q0 and φ are determined by the initial
conditions, e.g. q(0)=Q, and i(0)=0
ωt + φ) +
– Q0 ω2 cos(ω
1
ωt + φ) = 0
Q cos(ω
LC 0
 −ω 2 + 1  cos(ω
ωt + φ) = 0

LC 
Thus, we have the resonance condition
1
ω = 2π
πf=
LC
and
ωt + φ)
q(t) = Q0 cos(ω
dq
ωt + φ)
i(t) = –
= Q0 ω sin(ω
dt
The energy stored in the capacitor is
2
2
q
(
t
)
Q
ωt + φ)
= 0 cos2(ω
UE(t) = 12
C
2C
The energy stored in the inductor is
2 2
L
ω
Q0
1
2
ωt + φ)
sin2(ω
UB(t) = 2 Li =
2
Q 02
ωt + φ)
=
sin2(ω
2C
The total energy is
U = UE(t) + UB(t)
Q 02
ωt + φ) + sin2(ω
ωt + φ)}
=
cos2(ω
{
2C
Q 02
=
2C
RESONANCE FREQUENCY
1
1
f = 2π LC
Spring:
1 k
f=
2ππ m
Pendulum:
1 g
f=
2ππ l
Electric Oscillations
2
Q
1
W=
2C
Q
W= 1 LI 2
2
I
v
2
Q
1
W=
2C
Q
I
W= 1 LI 2
2
v