electromagnetic induction - michigan state university · mutual induction ! using the same analysis...

March 13, 2014 Chapter 29 1

Electromagnetic Induction!


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


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


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


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


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


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


ΔVind,1 = −M2→1

di2

dt

M1→2 = M2→1 = M ⇒

ΔVind,1 = −M di2

dt ΔVind,2 = −M di1

dt


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


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


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


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


!  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


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


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


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


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


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


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( )⎡⎣ ⎤⎦


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


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


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


!  Higher storage density!

!  Computer hard drives over 250 GB

!  Giant magneto-resistance

electromagnetic induction - michigan state university · mutual induction ! using the same analysis...

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