Chapter 27: Electromagnetic Induction
Farady’s Law
 Discovery of Farady’s law of induction
Farady’s Law
 Farady’s
law of induction
An emf in volts is induced in a circuit that is equal to the time rate of
change of the total magnetic flux in webers threading (linking) the circuit:
dB
 
dt
The flux through the circuit may be changed in several different ways
1) B may be made more intense.
2) The coil may be enlarged.
3) The coil may be moved into a region of stronger field.
4) The angle between the plane of the coil and B may change.
Farady’s Law
 Farady’s
law of induction (cont’d)

ds
Farady’s Law
 Induced
electric field
Consider work done in moving a test charge around the loop in one
revolution of induced emf.
Work done by emf
:
Work done by electric field:
W  emf  q0 


 F  ds  q0  E  ds q0 E (2r )
emf = 2rE for a circular loop
In general :
 
  emf   E  ds
 
d B
E

d
s



dt
Faraday’s law rewritten
For a circular current
loop
Lenz’s Law
 Direction
of induced emf and Lenz’s law
Extend Farady’s law to solenoids with N turns:
dB
 N
dt
Why the minus sign and
what does it mean?
Number of turns
Lenz’s Law
The sign of the induced emf is such that it tries to produce a current that
would create a magnetic flux to cancel (oppose) the original flux change.
or – the induced emf and induced current are in such a direction as to
oppose the change that produces them!
Lenz’s Law
 Example
1
B due to induced current
B due to induced current
Lenz’s Law
 Example
2
• The bar magnet moves towards loop.
• The flux through loop increases, and an emf induced in the loop produces
current in the direction shown.
• B field due to induced current in the loop (indicated by the dashed lines)
produces a flux opposing the increasing flux through the loop due to the
motion of the magnet.
Motional Electromotive Force
 Origin
of motional electromotive force I
FE
FB
Motional Electromotive Force
 Origin
of motional electromotive force I (cont’d)
Motional Electromotive Force
 Origin
of motional electromotive force II
v: constant
.
B
Bind
Motional Electromotive Force
 Origin
of motional electromotive force II (cont’d)
Motional Electromotive Force
 Origin
of motional electromotive force II (cont’d)

E

E
 
d  (  B)  ds in general

 
   (  B)  ds : motional emf for a closed conducting loop

Motional Electromotive Force
 Origin
of motional electromotive force II (cont’d)
Motional Electromotive Force
 Origin
of motional electromotive force II (cont’d)
Motional Electromotive Force
 Origin
of motional electromotive force III
Motional Electromotive Force
 Origin
of motional electromotive force III (cont’d)
Motional Electromotive Force
 Origin
of motional electromotive force III (cont’d)
Motional Electromotive Force
 A bar
magnet and a loop (again)
In this example, a magnet is being
pushed towards a closed loop.
The number of field lines linking
the loop is evidently increasing.
There is relative motion between
the loop and the field lines and an
observer at any point in the metal
of the loop, or the charges in the
loop, will see an E field
E  υobs  B
Also we have
 E  ds  
the Faraday emf.
Motional Electromotive Force
 Example:
A bar magnet and a loop (cont’d)
For the example just considered, let us see
what happens in a small interval dt. The
relative displacement υloopdt causes a small
area of B field to enter the loop. For a length
dL of the loop the ddΦB passing inside is
d[(dA)B] = dL υloopdt sinθ B. We can see this
as
d (d  B )  ddA  B  (dL  υ loop dt )  B  dL  (υ loop dt  B)
dB
d
 dL  E  d 
dt
Integrating this expression right round the circuit ( i.e. over dL) shows
that this υ×B interpretation recovers Faraday’s law. You will also see
that the sign of E is consistent with Lenz’s Law.
loop
Motional Electromotive Force
 Example:
A generator (alternator)
top
The armature of the generator is rotating
in a uniform B field with angular velocity
ω this can be treated as a simple case of
the E = υ×B field.
On the ends of the loop υ×B is
perpendicular to the conductor so does not
contribute to the emf. On the top υ×B is
parallel to the conductor and has the value
E = υB cos θ = ωRB cos ωt. The bottom
conductor has the same value of E in the
opposite direction but the same sense of
circulation.
 AB 
 E  ds  2 LRB cos t   AB cos t
v
v
B
bottom
the Faraday emf.
Eddy Current

Eddy current: Examples
Eddy Current

Eddy current : Examples (cont’d)
Eddy Current

Eddy current prevention
The orange represents a magnetic field pointing into the screen and let say it
is increasing at a steady rate like 100 gauss per sec. Then we put a copper ring
In the field as shown below. What does Faradays Law say will happen?
Current will flow in
the ring. What will
happen If there is
no ring present?
Now consider a
hypothetical path
Without any copper
ring.There will be
an induced Emf with
electric field lines as
shown above.
In fact there
will be many
concentric
circles
everywhere
in space.
The red circuits
have equal areas.
Emf is the same
in 1 and 2, less in 3
and 0 in 4. Note no
current flows.
Therefore, no thermal
energy is dissipated
Example:
x
A magnetic field is  to the board (screen) and uniform inside
a radius R. The magnetic field is increasing at a steady rate.
What is the magnitude of the induced field at a distance r
from the center?
d
x
x
x
r
R
x
x
B

dl
x
x
x
x
Field
circulates
around B
field
Notice that there is no wire or
loop of wire. To find E use
Faraday’s Law.
E is parallel
to dl
 Edl  E 2r  
m
dt
 m  BA  Br 2
d m d
2
2 dB
 ( Br )  r
dt
dt
dt
E 2r  r 2
E
dB
2rE  R 2
dt
r dB
2 dt
dB
dt
rR
R 2 dB
E
2r dt
rR
Example with numbers
Suppose dB/dt = - 1300 Gauss per sec and R= 8.5 cm
Find E at r = 5.2 cm
r dB
E
2 dt
rR E
(0.052m)
0.13T  0.0034 V /m  3.4mv /m
2
Find E at 12.5 cm

R dB
E
rR
2r dt
2
(0.085m) 2
E
0.13T  0.0038 V /m  3.8mV /m
2(0.125m)
Self Inductance
 Self induction
When a current flows in a circuit, it creates a magnetic flux which links its
own circuit. This is called self-induction. (‘Induction’ was the old word for
the flux linkage ΦB).
The strength of B is everywhere proportional to the I in the circuit so we can
write
 B  LI
L is called the self-inductance of the circuit
L depends on shape and size of the circuit. It may also be thought as being
equal to the flux linkage ΦB when I = 1 amp.
The unit of inductance is the henry
Wb
T m2
1H 1
1
A
A
Self Inductance
 Calculation
of self inductance : A solenoid
Accurate calculations of L are generally difficult. Often the answer
depends even on the thickness of the wire, since B becomes strong close
to a wire.
In the important case of the solenoid,
the first approximation result for L is
quite easy to obtain: earlier we had
N
B  0 I

Then,
Hence
N2A
 B  NAB  0
I

B
N2A
L
 0
 0 n 2 A
I

n : the number of turns
per unit length
So L is proportional to n2 and the volume of the solenoid
Self Inductance
 Calculation
of self inductance: A solenoid (cont’d)
N2A
L  0
 0 n 2 A

n : the number of turns
per unit length
Example: the L of a solenoid of length 10 cm, area 5 cm2, with a total
of 100 turns is
L = 6.28×10−5 H
0.5 mm diameter wire would achieve 100 turns in a single layer.
Going to 10 layers would increase L by a factor of 100. Adding an iron
or ferrite core would also increase L by about a factor of 100.
The expression for L shows that μ0 has units H/m, c.f, Tm/A obtained earlier
Self Inductance
 Calculation
of self inductance: A toroidal solenoid
The magnetic flux inside the solenoid:
 B  BA 
 0 NIA
2r
Then the self-inductance of the solenoid:
N B
N2A
L
 0
 0 n 2 A(2 r )
I
2r
If N = 200 turns, A = 5.0 cm2 , and r = 0.10 m:
[4 10 7 Wb/(A  m)](200) 2 (5.0 10 4 m 2 )
L
2 (0.10 m)
 40 10 6 H  40 H.
Then when the current increases uniformly from 0.0 to 6.0 A in 3.0 s,
the self-induced emf E will be:
dI
 L
 (40 10 6 H)(2.0 106 A/s)  80 V.
dt
d B
(   
)
dt
Self Inductance
 Stored energy in magnetic field
Why is L an interesting and very important quantity?
This stems from its relationship to the total energy stored in the B field of the
circuit
which we shall prove below.
1 2
Um 
2
LI
When I is first established, we have a finite
d B
dI
 L  
dt
dt
(self-induced emf)
The source of I does work against the self-induced emf in order to raise I to
its final value.
dU m
dI
  I  LI
dt
dt
Um

0
power = work done per unit time
1 2
dU m  L  IdI  LI  U m
0
2
I
Self Inductance
 Stored energy in magnetic field: Example
Returning to our expression for the energy stored in an inductance we
can use it for the case of a solenoid. Using formulae we have already
obtained for the solenoid
B  0 nI
and
L
B
  0 n 2 A
I
2
 B 
1 2 1
B2
2
Hence: U m  LI   0 n A
 
A

2
2
20
 0 n 
2
U
B
Energy per unit u  m 
m
volume in the field
A 20
Self Inductance
 Inductor
A circuit device that is designed to have a particular inductance is called
an inductor or a choke. The usual symbol is:
dI
Vab  Va  Vb  L
dt
If dI / dt  0,
Vab  0 potential drops from a to b
If dI / dt  0,
Vab  0 potential increases from a to b
a
I
variable
source
of emf
L
b
Mutual Inductance
 Transformer and mutual inductance
The classic examples of mutual inductance are transformers for power
conversion and for making high voltages as in gasoline engine ignition.
A current I1 is flowing in the
primary coil 1 of N1 turns
and this creates flux B which
then links coil 2 of N2 turns.
The mutual inductance M2 1
is defined such that the
induction Φ2 is given by
2  L2 I 2  M 21I1
Also
1  L1I1  M12 I 2
M2 1—Mutual
Inductance of the coils
Generally, M 1 2 = M 2 1
Mutual Inductance
 Changing current and induced emf
Consider two fixed coils with a varying current I1 in coil 1 producing
magnetic field B1. The induced emf in coil 2 due
to B is proportional
  1
to the magnetic flux through coil 2:
 2   B1  dA2  N 22
2 is the flux through a single loop in coil 2 and N2 is the number of loops
in coil 2. But we know that B1 is proportional to I1 which means that 2 is
proportional to I1. The mutual inductance M is defined to be the constant of
proportionality between 2 and I1 and depends on the geometry of the situation.
 2 N 22
M

I1
I1
d 2
d 2 dI1
dI1
2  

 M
;
dt
dI1 dt
dt
d 2
M
dI1
The induced emf is proportional to M and to the rate of change of the
current .
Mutual Inductance
 Example
Now consider a tightly wound concentric
solenoids. Assume that the inner solenoid
carries current I1 and the magnetic flux
on the outer solenoid B2is created due
to this current. Now the flux produced
by the inner solenoid is:
B1  0 n1I1 where n1  N1 / 
The flux through the outer solenoid due to this magnetic field is:
 B2  N 2 B1 A1  N2 B1 ( r12 )  0 n2 n1( r12 ) I1
M 21 
 B2
I1
 0 n2 n1( r12 ) ; in general M 21  M 12  M .
Mutual Inductance
 Example of inductor: Car ignition coil
Two ignition coils, N1=16,000 turns, N2=400 turns wound over each other.
l=10 cm, r=3 cm. A current through the primary coil I1=3 A is broken in
10-4 sec. What is the induced emf ?
 B2
dI1
 2   M 12
; M 21 
 0 n2 n1( r12 )
dt
I1
dI1
 3 10 4 A s -1
dt
 2  6,000 V
Spark jumps across gap in a spark plug and ignites a gasoline-air mixture
The R-L Circuit
 Current growth in an R-L circuit
Consider the circuit shown. At t < 0 the
switch is open and I = 0.
The resistance R can include the resistance of
the inductor coil.
The switch closes at t = 0 and I begins to increase, Without the inductor the full
current would be established in nanoseconds. Not so with the inductor.
Kirchhoff’s Loop Rule:
Multiply by I:
dI
 0  IR  L
0
dt
dI
 0 I  I R  LI
dt
2
Power balance
The R-L Circuit
 Current growth in an R-L circuit (cont’d)
dI
 0 I  I R  LI
dt
Power supplied
by the battery
2
Power dissipated as heat in the resistor
Um
If energy in inductor is
dU m
dI
 LI
dt
dt
then:
or
dU m  L I dI
Integrate from t = 0 (I = 0) to t =  (I = If)
U mf
U mf 

0
If
dU m 
1 2
L
I
dI

LI f

2
0
So, the energy stored in an inductor
carrying current I is :
Um 
1 2
LI
2
Rate at which energy
is stored up in the
inductor.
The R-L Circuit
 Current growth in an R-L circuit (cont’d)
Kirchhoff’s Loop Rule:
 0  IR  L
At t = 0+, I = 0
0
 dI 

 
 dt 0 L
Current in an
LR circuit as
function of
time
dI
0
dt
I then increases until
finally dI/dt = 0
If 
0
R
Compare with:
The R-L Circuit
 Current growth in an R-L circuit (cont’d)

L dI
I  0
R dt
R
dI
 0 
I  R 



R
dt
L
Integrating between (I = 0, t = 0) and (I = I, t = t)
 I  0 / R 
R
ln 
 t
L
  0 / R 
The R-L Circuit
 Current growth in an R-L circuit (cont’d)
Now we raise e to the power of each side
R

t
 0 / R 
L
e
I




/
R
0


I
0

R
R

0 Lt
 e
R
R
 t
0 
1  e L 
I 

R 

The R-L Circuit
 Discharging an R-L circuit
Add switch S2 to be able to remove
the battery. And add R1 to protect
the battery so that it is protected
when both switches are closed.
First S1 has been closed for a long
enough time so that the current is
steady at its final value I0.
At t=0, close S2 and open S1 to effectively remove the battery. Now the
circuit abcd carries the current I0.
Kirchhoff’s loop rule:
 IR  L
dI
0
dt
I  I 0 e  Rt / L
The R-L Circuit
 Discharging an R-L circuit (cont’d)
Now let’s calculate the total heat produced in resistance R when the current
decreases from I0 to 0.
Rate of heat production: P 
dW
 I 2R
dt

Energy dissipated as heat in the resistor:
W   dW   I 2 Rdt
0
The current as a function of time:
The total energy:
W  I e
I  I 0 e  Rt / L
2  2 Rt / L
0
1 2
Rdt  LI 0
2
The total heat produced equals the energy originally stored in the inductor
The L-C Circuit
 Complex number and plane
Complex number : z = x + iy
real part Re(z)=x, imaginary part Im(z)=y
The L-C Circuit
 Simple harmonic oscillation
The L-C Circuit
 Simple harmonic oscillation (cont’d)
The L-C Circuit
 Simple harmonic oscillation (cont’d)
The L-C Circuit
 An L-C circuit and electrical oscillation
Consider a circuit with an inductor and a
capacitor as shown in Fig. Initially the
capacitor C carries charge Q0
S
dI
L
dt
At t=0 the switch closes and charge flows
through inductor producing self-induced emf.
The current I is by definition:
Kirchhoff’s loop rule:
dQ
I
dt
dI Q
L  0
dt C
d 2Q Q
L 2  0
dt
C
Acceleration equation for
a mass on a spring
d2 x k
c.f. 2  x  0
dt
m
The L-C Circuit
 An L-C circuit and electrical oscillation (cont’d)
d 2Q
1

Q  Q
2
d t
LC
d 2x
k
2
c.f.


x



x
2
dt
m
The solution of this equation is simple harmonic motion.
Q  A cos(t   )
c.f. x  A cos(t   )
Now let’s figure out what A and  are. For that choose initial condition
as: I(0)=0 and Q(0)=Q0. Then A=Q0 and 0.
Q(t )  Q0 cos(t ) , I (t )  Q0 sin( t )  Q0 cos(t   / 2)
The charge and current are 90o out of phase with the same angular frequency 
Iis at maximum when Q=0, and Q is at maximum when I=0.
The L-C Circuit
 An L-C circuit and electrical oscillation (cont’d)
Q(t )  Q0 cos(t ) , I (t )  Q0 sin( t )  Q0 cos(t   / 2)
The charge and current are 90o out of phase with the same angular frequency 
Iis at maximum when Q=0, and Q is at maximum when I=0.
-I(t)
The L-C Circuit
 An L-C circuit and electrical oscillation (cont’d)
The electric energy in the capacitor:
1
1 Q 2 1 Q02
U e  QVc 

cos 2 (t )
2
2 C 2 C
The electric energy oscillates between its maximum Q02 and 0.
The magnetic energy in the inductor:
1 2 1
1 Q02
1
2 2
2
2
U m  LI  L Q0 sin (t ) 
sin (t )  
2
2
2 C
LC
The magnetic energy oscillates between its maximum Q02 /(2C) and 0.
Utot=Ue+Um constant
Ue(t) U (t)
m
The L-R-C Circuit
 Another differential equation
The L-R-C Circuit
 Another differential equation (cont’d)
The L-R-C Circuit
 Another differential equation (cont’d)
The L-R-C Circuit
 An L-R-C circuit and electrical damped oscillation
At t=0 the switch is closed and a capacitor
with initial charge Q0 is connected in series
across an inductor.
Initial condition:
Q0  Q0 ; I (0)  0
A loop around the circuit in the direction of the
current flow yields:
Q
dI
 L  IR  0
C
dt
Since the current is flowing out of the capacitor,
d 2Q R dQ
1


Q0
2
dt
L dt
LC
I 
dQ
dt
The L-R-C Circuit
 An L-R-C circuit and electrical damped oscillation (cont’d)
If R2< 4LC, the solution is:
Q(t )  Q0 e  Rt /(2 L ) cos  ' t where  '   2  [ R /( 2 L)]2 and   1 / LC
Note that if R=0,no damping occurs.