Chapter 32

Inductance

Self-inductance• Some terminology first:

• Use emf and current when they are caused by batteries or other sources

• Use induced emf and induced current when they are caused by changing magnetic fields

• It is important to distinguish between the two situations

Self-inductance• When the switch is closed, the current

does not immediately reach its maximum value

• Faraday’s law can be used to describe the effect

• As the current increases with time, the magnetic flux through the circuit loop due to this current also increases with time

• This increasing flux creates an induced emf in the circuit

Self-inductance• The direction of the induced emf is

such that it would cause an inducedcurrent in the loop, which would establish a magnetic field opposing the change in the original magnetic field

• The direction of the induced emf is opposite the direction of the emf of the battery

• This results in a gradual increase in the current to its final equilibrium value

• This effect of self-inductance occurs when the changing flux through the circuit and the resultant induced emf arise from the circuit itself

Self-inductance• The self-induced emf εL is always proportional to the

time rate of change of the current. (The emf is proportional to the flux change, which is proportional to the field change, which is proportional to the current change)

• L: inductance of a coil (depends on geometric factors)

• The negative sign indicates that a changingcurrent induces an emf in opposition to thatchange

• The SI unit of self-inductance: Henry

• 1 H = 1 (V · s) / AJoseph Henry1797 – 1878

L

d Iε L

dt

Inductance of a Coil

• For a closely spaced coil of N turns carrying current I:

• The inductance is a measure of the opposition to a change in current

dt

dN B

L

dt

dIL Bd

L

NdI BL

NI

I

NL B

dtdI

L L

/

Chapter 32Problem 4

An emf of 24.0 mV is induced in a 500-turn coil when the current is changing at a rate of 10.0 A/s. What is the magnetic flux through each turn of the coil at an instant when the current is 4.00 A?

Inductance of a Solenoid• Assume a uniformly wound solenoid having N turns and

length ℓ (ℓ is much greater than the radius of the solenoid)

• The flux through each turn of area A is

• This shows that L depends on thegeometry of the object

BAB nIA0 IAl

N0

I

NL B

I

IAlN

N

0

l

AN 20

l

ANL

20

Inductor in a Circuit• Inductance can be interpreted as a measure of

opposition to the rate of change in the current (while resistance is a measure of opposition to the current)

• As a circuit is completed, the current begins to increase, but the inductor produces a back emf

• Thus the inductor in a circuit opposes changes in current in that circuit and attempts to keep the current the same way it was before the change

• As a result, inductor causes the circuit to be “sluggish” as it reacts to changes in the voltage: the current doesn’t change from 0 to its maximum instantaneously

RL Circuit• A circuit element that has a large self-inductance is

called an inductor

• The circuit symbol is

• We assume the self-inductance of the rest of the circuit is negligible compared to the inductor (However, in reality, even without a coil, a circuit will have some self-inductance

• When switch is closed (at time t = 0),the current begins to increase, and atthe same time, a back emf isinduced in the inductor that opposesthe original increasing current

RL Circuit• Applying Kirchhoff’s loop rule to the circuit in the

clockwise direction gives 0 IRL

0 IRdt

dIL

dt

dI

L

IR

IR

dI

L

dt

IR

IRd

L

Rdt

)( )ln()ln( ConstIR

L

Rt

)( IRConste L

Rt

0)0( tI

Conste 0

1

Const

)(1

IR

L

Rt

eR

I 1

RL Circuit• The inductor affects the current exponentially

• The current does not instantly increase to its final equilibrium value

• If there is no inductor, the exponential term goes to zero and the current would instantaneously reach its maximum value as expected

• When the current reaches its maximum, the rate of change and the back emf are zero

L

Rt

eR

I 1

RL Circuit• The expression for the current can also be

expressed in terms of the time constant , of the circuit:

• The time constant, , for an RL circuit is thetime required for the current in the circuitto reach 63.2% of its final value

L

Rt

eR

I 1

LR

t

eR

I 1

RL Circuit• The current initially increases very rapidly and then

gradually approaches the equilibrium value

• The equilibrium value of the current is /R and is reached as t approaches infinity

Chapter 32Problem 13

Consider the circuit shown in the figure. Take ε = 6.00 V, L = 8.00 mH, and R = 4.00 Ω. (a) What is the inductive time constant of the circuit? (b) Calculate the current in the circuit 250 μs after the switch is closed. (c) What is the value of the final steady-state current? (d) How long does it take the current to reach 80.0% of its maximum value?

Energy Stored in a Magnetic Field

• In a circuit with an inductor, the battery must supply more energy than in a circuit without an inductor

• I is the rate at which energy is being supplied by the battery

• Part of the energy supplied by the battery appears as internal energy in the resistor

• I2R is the rate at which the energy is being delivered to the resistor

0 IRdt

dIL RI

dt

dIILI 2

Energy Stored in a Magnetic Field

• The remaining energy is stored in the magnetic field of the inductor

• Therefore, LI (dI/dt) must be the rate at which the energy is being stored in the magnetic field dU/dt

0 IRdt

dIL RI

dt

dIILI 2

dt

dIIL

dt

dU L 2

2LIU L

Energy Storage Summary

• A resistor, inductor and capacitor all store energy through different mechanisms

• Charged capacitor stores energy as electric potential energy

• Inductor when it carries a current, stores energy as magnetic potential energy

• Resistor energy delivered is transformed into internal energy

Mutual Inductance• The magnetic flux through the area enclosed by a

circuit often varies with time because of time-varying currents in nearby circuits

• This process is known as mutual induction because it depends on the interaction of two circuits

• The current in coil 1 sets up amagnetic field

• Some of the magnetic field lines passthrough coil 2

• Coil 1 has a current I1 and N1 turns;Coil 2 has N2 turns

Mutual Inductance• The mutual inductance of coil 2 with respect to coil 1 is

• Mutual inductance depends on the geometry of both circuits and on their mutual orientation

• If current I1 varies with time, theemf induced by coil 1 in coil 2 is

1

12212 I

NM

dt

dN 1222

2

11212 N

IM

2

1122 N

IM

dt

dN

dt

dIM 112

dt

dIM 1122

2

21121 I

NM

LC Circuit• A capacitor is connected to an inductor in an LC

circuit

• Assume the capacitor is initially charged and then the switch is closed

• Assume no resistance and no energy losses to radiation

• The current in the circuit and thecharge on the capacitor oscillatebetween maximum positive andnegative values

LC Circuit• With zero resistance, no energy is transformed into

internal energy

• Ideally, the oscillations in the circuit persist indefinitely (assuming no resistance and no radiation)

• The capacitor is fully charged and the energy in the circuit is stored in the electric field of the capacitor

Q2max / 2C

• No energy is stored in the inductor

• The current in the circuit is zero

LC Circuit• The switch is then closed

• The current is equal to the rate at which the charge changes on the capacitor

• As the capacitor discharges, the energy stored in the electric field decreases

• Since there is now a current, someenergy is stored in the magneticfield of the inductor

• Energy is transferred from theelectric field to the magnetic field

LC Circuit• Eventually, the capacitor becomes fully discharged

and it stores no energy

• All of the energy is stored in the magnetic field of the inductor and the current reaches its maximum value

• The current now decreases in magnitude, recharging the capacitor with its plates having opposite their initial polarity

• The capacitor becomes fullycharged and the cycle repeats

• The energy continues to oscillatebetween the inductor and the capacitor

LC Circuit• The total energy stored in the LC circuit remains

constant in time

• Solution:

22

22 LI

C

QU 0

22

22

LI

dt

d

C

Q

dt

d

dt

dU

0dt

dILI

dt

dQ

C

Q0

2

2

dt

QdL

C

Q

LC

Q

dt

Qd

2

2

LC

tQtQ cos)( max

LC Circuit• The angular frequency, ω, of the circuit depends on

the inductance and the capacitance

• It is the natural frequency of oscillation of the circuit

• The current can be expressed as a function of time:

LC

tQtQ cos)( max tQtQ cos)( max

1ωLC

tQdt

d

dt

dQtI cos)( max tQ sinmax

tItI sin)( max

LC Circuit• Q and I are 90° out of phase with each other, so when

Q is a maximum, I is zero, etc.

tQtQ cos)( max

tItI sin)( max

Energy in LC Circuits• The total energy can be expressed as a function of

time

• The energy continually oscillatesbetween the energy stored in theelectric and magnetic fields

• When the total energy is stored inone field, the energy stored in theother field is zero

22

22 LI

C

QU t

LIt

C

Q 22

max22

max sin2

cos2

Energy in LC Circuits

• In actual circuits, there is always some resistance

• Therefore, there is some energy transformed to internal energy

• Radiation is also inevitable in this type of circuit

• The total energy in the circuit continuously decreases as a result of these processes

Chapter 32Problem 38

An LC circuit consists of a 20.0-mH inductor and a 0.500-μF capacitor. If the maximum instantaneous current is 0.100 A, what is the greatest potential difference across the capacitor?

RLC Circuit• The total energy is not constant, since there is a

transformation to internal energy in the resistor at the rate of dU/dt = – I2R

• Radiation losses are still ignored

• The circuit’s operation can be expressed as:

RIdt

dILI

dt

dQ

C

Q 2

IRdt

dIL

C

Q

02

2

Rdt

dQ

dt

QdL

C

Q

RLC Circuit

• Solution:

• Analogous to a damped harmonic oscillator

• When R = 0, the circuit reduces to an LC circuit (no damping in an oscillator)

02

2

Rdt

dQ

dt

QdL

C

Q

teQtQ dL

Rt

cos)( 2max

2

2

1

L

R

LCd

Answers to Even Numbered Problems

Chapter 32:

Problem 2

1.36 µH