electromagnetic induction - michigan state university · mutual induction ! using the same analysis...
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March 13, 2014 Chapter 29 1
Electromagnetic Induction!
March 13, 2014 Chapter 29 2
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.
! We can change the magnetic flux in several ways:
! 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
ΔVind = −
dΦB
dt
A,θ constant: ΔVind = −Acosθ dB
dt
B,θ constant: ΔVind = −Bcosθ dA
dt
A,B constant: ΔVind =ωABsinθ
March 13, 2014 Chapter 29 3
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 5
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
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
! Multiplying both sides by i1 gives
! If i1 changes with time we can write
! Comparing with Faraday’s Law gives us
! A changing current in coil 1 produces a potential difference in coil 2
March 13, 2014 Chapter 29 7
M1→2 =
N2Φ1→2
i1
i1M1→2 = N2Φ1→2
M1→2
di1
dt= N2
dΦ1→2
dt
ΔVind,2 = −M1→2
di1
dt
Mutual Induction ! Now reverse the roles of the two coils
March 13, 2014 Chapter 29 8
Mutual Induction ! Using the same analysis we find
! If we switched the indices 1 and 2 and repeated the entire analysis of the coils’ effects on each other, we could show that
! 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
ΔVind,1 = −M2→1
di2
dt
M1→2 = M2→1 = M ⇒
ΔVind,1 = −M di2
dt ΔVind,2 = −M di1
dt
March 13, 2014 Chapter 29 10
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
! 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
! 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
q =CVemf 1− e−t/τRC( ) τRC = RC
q = q0e−t/τRC
March 13, 2014 Chapter 29 11
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 12
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 di
dt
March 13, 2014 Chapter 29 13
RL Circuits
! Thus we can write the sum of the potential drops around the circuit loop as
! Rewrite this equation as
(differential equation for current)
! The solution is
Vemf − iR− L di
dt= 0
L di
dt+ iR =Vemf
i(t)= Vemf
R1− e−t/ L/R( )( ) = Vemf
R1− e−t/τRL( ) τRL =
LR
March 13, 2014 Chapter 29 14
! 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
RL Circuits
VL +VR = L di
dt+ iR = 0⇒ L di
dt+ iR = 0
March 13, 2014 Chapter 29 15
RL Circuits ! The solution to this differential equation is
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
i(t)= i0e−t/τRL
March 13, 2014 Chapter 29 18
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
! 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
UE =
12
q2
C
March 13, 2014 Chapter 29 19
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
! Integrating this power over the time it takes to reach a final current yields the energy stored in the magnetic field of the inductor
! The energy stored in the magnetic field of a solenoid is
L didt
+ i 0( ) =Vemf = L didt
P =Vemfi = L didt
⎛⎝⎜
⎞⎠⎟ i
UB = Pdt
0
t
∫ = L ′i d ′i0
i
∫ = 12
Li2
UB =
12
Lsolenoidi2 = 12µ0n2Ai2 = 1
2µ0
N 2
Ai2
March 13, 2014 Chapter 29 20
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
! Since B = μ0ni, we can write
! Although we derived this using an ideal solenoid, it is valid for magnetic fields in general
uB =
UB
volume=
12µ0n2Ai2
A= 1
2µ0n2i2
uB =
12µ0
B2
March 13, 2014 Physics for Scientists & Engineers, Chapter 22 21
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 22
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
! The current as a function of time for this circuit is P t( ) =Vemfi t( )
i t( ) = Vemf
R1− e−t/τRL( ) τRL = L / R
March 13, 2014 Physics for Scientists & Engineers, Chapter 22 23
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
SIMPLIFY
! Substituting and rearranging gives us
! Which evaluates to
W = P t( )dt
0
T
∫
W = Vemf
2
R1− e−t/τRL( )dt
0
T
∫
W = Vemf2
Rt +τRLe−t/τRL⎡⎣ ⎤⎦0
T
W = Vemf2
RT +τRL e−T /τRL −1( )⎡⎣ ⎤⎦
March 13, 2014 Physics for Scientists & Engineers, Chapter 22 24
Work Done by a Battery CALCULATE
! First we calculate the time constant
! Now the work
ROUND
! We round our result to three significant figures
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
τRL = L / R = 2.20 H( ) / 160.0 Ω( ) =1.375 ⋅10−2 s
W =
40.0 V( )2
160.0 Ω0.0165 s+ 1.375 ⋅10−2 s( ) e−0.0165 s/1.375⋅10−2 s −1( )⎡⎣
⎤⎦ = 0.06891 J
W = 0.0689 J
March 13, 2014 Physics for Scientists & Engineers, Chapter 22 25
Work Done by a Battery ! The average current if the current increased linearly with
time is
! The work would then be
! This is less than, but close to, our calculated result ! We are confident in our answer
iave =12
i T( ) = 12
Vemf
R1− e−T /τRL( )
iave =12
40.0 V160.0 Ω
1− e−0.0165 s/1.375⋅10−2 s( ) = 0.0874 A
W = PT = iaveVemfT = 0.0874 A( ) 40.0 V( ) 0.0165 s( ) = 0.0577 J
March 13, 2014 Chapter 29 26
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 27
! Higher storage density!
! Computer hard drives over 250 GB
! Giant magneto-resistance