Electromagnetic Induction!
March 13, 2014
Chapter 29
1
Faraday’s Law of Induction
!  The magnitude of the potential difference, ΔVind,
induced in a conducting loop is equal to the time rate of
change of the magnetic flux through the loop.
ΔVind = −
dΦ B
dt
!  We can change the magnetic flux in several ways:
dB
A,θ constant: ΔVind = −Acosθ
dt
dA
B,θ constant: ΔVind = −Bcosθ
dt
A, B constant: ΔVind = ω ABsinθ
!  Where A is the area of the loop, B is the magnetic field strength,
and θ is the angle between the surface normal and the B field vector
March 13, 2014
Chapter 29
2
Lenz’s Law - Four Cases
!  (a) An increasing magnetic field pointing to the right
induces a current that creates a magnetic field to the left
!  (b) An increasing magnetic field pointing to the left induces
a current that creates a magnetic field to the right
!  (c) A decreasing magnetic field pointing to the right induces
a current that creates a magnetic field to the right
!  (d) A decreasing magnetic field pointing to the left induces a
current that creates a magnetic field to the left
March 13, 2014
Chapter 29
3
Self Induction
!  Consider a circuit with current flowing through
an inductor that increases with time
!  The self-induced potential difference will oppose
the increase in current
!  Consider a circuit with current flowing through
an inductor that is decreasing with time
!  A self-induced potential difference will oppose
the decrease in current
!  We have assumed that these inductors are ideal
inductors; that is, they have no resistance
!  Induced potential differences manifest
themselves across the connections of the inductor
March 13, 2014
Chapter 29
5
Mutual Induction
!  Consider two adjacent coils with their central axes aligned
March 13, 2014
Chapter 29
6
Mutual Induction
!  The mutual inductance of coil 2 due to coil 1 is
N 2 Φ1→2
M1→2 =
i1
!  Multiplying both sides by i1 gives
i1 M1→2 = N 2 Φ1→2
!  If i1 changes with time we can write
di1
dΦ1→2
M1→2
= N2
dt
dt
!  Comparing with Faraday’s Law gives us
di1
ΔVind,2 = −M1→2
dt
!  A changing current in coil 1 produces a potential difference
in coil 2
March 13, 2014
Chapter 29
7
Mutual Induction
!  Now reverse the roles of the two coils
March 13, 2014
Chapter 29
8
Mutual Induction
!  Using the same analysis we find
di2
ΔVind,1 = −M 2→1
dt
!  If we switched the indices 1 and 2 and repeated the entire
analysis of the coils’ effects on each other, we could show
that
M1→2 = M 2→1 = M ⇒
di2
di1
ΔVind,1 = −M
ΔVind,2 = −M
dt
dt
!  Here M is the mutual inductance between the two coils
!  One major application of mutual inductance is in
transformers
March 13, 2014
Chapter 29
9
RL Circuits
!  We know that if we place a source of external voltage, Vemf,
into a single loop circuit containing a resistor R and a
capacitor C, the charge q on the capacitor builds up over
time as
q = CVemf (1− e −t/τ RC ) τ RC = RC
!  The same time constant governs the decrease of the initial
charge q0 in the circuit if the emf is suddenly removed and
the circuit is short-circuited
q = q0e −t/τ RC
!  If a source of emf is placed in a single-loop circuit
containing a resistor with resistance R and an inductor with
inductance L, called an RL circuit, a similar phenomenon
occurs
March 13, 2014
Chapter 29
10
RL Circuits
!  Consider a circuit in which a source of
emf is connected to a resistor and an
inductor in series
!  If the circuit included only the resistor,
the current would increase quickly to the
value given by Ohm’s Law when the
switch was closed
!  However, in the circuit with both the
resistor and the inductor, the increasing
current flowing through the inductor
creates a self-induced potential difference
that opposes the increase in current
!  After a long time, the current becomes
steady at the value Vemf/R
March 13, 2014
Chapter 29
11
RL Circuits
!  Use Kirchhoff’s loop rule to analyze this circuit assuming
that the current i at any given time is flowing through the
circuit in a counterclockwise direction
!  The emf source represents a gain in
potential, +Vemf, and the resistor
represents a drop in potential, -iR
!  The self-inductance of the inductor
represents a drop in potential because it is opposing the
increase in current
!  The drop in potential due to the inductor is proportional to
the time rate change of the current and is given by
ΔVind = −L
March 13, 2014
di
dt
Chapter 29
12
RL Circuits
!  Thus we can write the sum of the potential drops around the
circuit loop as
Vemf
di
− iR − L = 0
dt
!  Rewrite this equation as
di
L + iR = Vemf
dt
(differential equation for
current)
!  The solution is
Vemf
Vemf
L
−t/τ RL
−t/( L/R )
i(t) =
1− e
=
1− e
τ RL =
(
)
R
R
R
(
March 13, 2014
)
Chapter 29
13
RL Circuits
!  Now consider the case in which an emf source had been
connected to the circuit and is suddenly removed
!  We can use our previous equation with Vemf = 0 to describe
the time dependence of this circuit
di
di
VL +VR = L + iR = 0 ⇒ L + iR = 0
dt
dt
March 13, 2014
Chapter 29
14
RL Circuits
!  The solution to this differential equation is
i(t) = i0e −t/τ RL
where the initial conditions when the emf was connected
can be used to determine the initial current, i0 = Vemf /R
!  The current drops
with time exponentially
with a time constant
τRL = L / R
!  After a long time
the current in the
circuit is zero
March 13, 2014
Chapter 29
15
Energy of a Magnetic Field
!  We can think of an inductor as a device that can store
energy in a magnetic field in the manner similar to the way
we think of a capacitor as a device that can store energy in
an electric field
!  The energy stored in the electric field of a capacitor is
1 q2
UE =
2C
!  Consider the situation in which an inductor is connected to
a source of emf
!  The current begins to flow through the inductor producing a
self-induced potential difference opposing the increase in
current
March 13, 2014
Chapter 29
18
Energy of a Magnetic Field
!  The instantaneous power provided by the emf source is
the product of the current and voltage in the circuit
assuming that the inductor has no resistance
di
di
+ i ( 0 ) = Vemf = L
dt
dt
⎛ di ⎞
P = Vemf i = ⎜ L ⎟ i
⎝ dt ⎠
L
!  Integrating this power over the time it takes to reach a final
current yields the energy stored in the magnetic field of the
inductor
1 2
U B = ∫ P dt = ∫ Li ′ di ′ = Li
0
0
2
t
i
!  The energy stored in the magnetic field of a solenoid is
2
1
1
1
N
U B = Lsolenoidi 2 = µ0n 2 Ai 2 = µ0
Ai 2
2
2
2

March 13, 2014
Chapter 29
19
Energy density of a Magnetic Field
!  The magnetic field occupies the volume enclosed by the
solenoid
!  Thus, the energy density of the magnetic field of the
solenoid is
1
2
2
µ
n
Ai
0
UB
1
2
uB =
=
= µ0n 2i 2
volume
A
2
!  Since B = μ0ni, we can write
1 2
uB =
B
2 µ0
!  Although we derived this using an ideal solenoid, it is valid
for magnetic fields in general
March 13, 2014
Chapter 29
20
Work Done by a Battery
!  A series circuit contains a battery that
supplies Vemf = 40.0 V, an inductor with
L = 2.20 H, a resistor with R = 160.0 Ω,
and a switch, connected as shown
PROBLEM
!  The switch is closed at time t = 0
!  How much work is done between t = 0 and t = 0.0165 s?
SOLUTION
THINK
!  When the switch is closed, current begins to flow and power
is provided by the battery
!  Power is the voltage times the current at any given time
!  Work is the integral of the power over time
March 13, 2014
Physics for Scientists & Engineers, Chapter 22
21
Work Done by a Battery
SKETCH
!  The sketch shows the
current as a function
of time
RESEARCH
!  The power in the
circuit at any time t
after the switch is closed is
P (t ) = Vemf i (t )
!  The current as a function of time for this circuit is
Vemf
−t/τ RL
i (t ) =
1−
e
τ RL = L / R
(
)
R
March 13, 2014
Physics for Scientists & Engineers, Chapter 22
22
Work Done by a Battery
!  The work done by the battery is the integral of the power
over the time T in which the circuit has been in operation
T
W = ∫ P (t ) dt
0
SIMPLIFY
!  Substituting and rearranging gives us
W=∫
T
0
2
Vemf
1− e −t/τ RL ) dt
(
R
!  Which evaluates to
2
Vemf
W=
R
2
Vemf
W=
R
March 13, 2014
⎡⎣t + τ RL e
−t/τ RL
T
⎤⎦0
⎡⎣T + τ RL ( e −T /τ RL −1) ⎤⎦
Physics for Scientists & Engineers, Chapter 22
23
Work Done by a Battery
CALCULATE
!  First we calculate the time constant
τRL = L / R = ( 2.20 H ) / (160.0 Ω ) = 1.375⋅10−2 s
!  Now the work
2
40.0 V ) ⎡
(
W=
0.0165 s +
160.0 Ω ⎣
(1.375⋅10 s)(e
−2
−0.0165 s/1.375⋅10−2 s
)
−1 ⎤ = 0.06891 J
⎦
ROUND
!  We round our result to three significant figures
W = 0.0689 J
DOUBLE-CHECK
!  To double-check, let’s assume that the current in the circuit
is constant in time and equal to half of the final current
March 13, 2014
Physics for Scientists & Engineers, Chapter 22
24
Work Done by a Battery
!  The average current if the current increased linearly with
time is
1
1 Vemf
iave = i (T ) =
1− e −T /τ RL )
(
2
2 R
1 40.0 V
−0.0165 s/1.375⋅10−2 s
iave =
1− e
= 0.0874 A
2 160.0 Ω
(
)
!  The work would then be
W = PT = iaveVemf T = ( 0.0874 A )( 40.0 V )( 0.0165 s ) = 0.0577 J
!  This is less than, but close to, our calculated result
!  We are confident in our answer
March 13, 2014
Physics for Scientists & Engineers, Chapter 22
25
Modern Applications
!  Computer hard drives, videotapes, audio tapes, magnetic
strips on credit cards
!  Storage of the information is accomplished by using an
electromagnet in the “write-head”
!  A current that varies in time is sent to the electromagnet
and creates a magnetic field that magnetizes the
ferromagnetic coding of the storage medium as it passes by
!  Retrieval of the information
reverses the process of storage
!  As the storage medium passes by
the “read head”, the magnetization
causes a change of the magnetic
field inside the coil
!  This change of the magnetic field
induces a current in the read head which is then processed
March 13, 2014
Chapter 29
26
!  Higher
storage
density!
!  Computer
hard
drives over
250 GB
!  Giant
magnetoresistance
March 13, 2014
Chapter 29
27