Chapter 30

Inductance

Inductor and Inductance

Capacitor: store electric energyInductor: store magnetic energy

L =NΦB

I: Inductance

Unit: H (henry)

1H =1Tm2 / A

Measure how effective it is at trapping magnetic energy

Inductance of a solenoid

L =NΦB

I

l

n : turns per unit length, l : length, A: area

What is ΦB?

ΦB = BA = μ 0nIA

N = nl

⇒ L =(nl)(μ 0nIA)

I= μ 0n

2Al

or L = μ 0N2A / l

Self-Induction

L =NΦB

I⇔ NΦB =LI

ξL =−dNΦB

dt=−L

dIdt

Always resists the change in current

The sign of ξL

If ξL is positive, then the induced emf points in the same direction as the current.

If ξL is negative, then the induced emf points in the opposite direction as the current.

RL Circuits (“charging”)

ξ −IR − LdI

dt= 0

⇔ I =ξ

R(1− exp(−Rt / L))

Inductive time constant: τ L =L

R

⇒ I =ξ

R(1− exp(−t / τ L ))

Example

ξL = −LdI

dt= −

Lξ

τ LRexp(−t / τ L )

= −Lξ

(L / R)Rexp(−t / τ L ) = −ξ exp(−t / τ L )

ξL (t = 0s) = −ξ

It complete stops the current at t = 0s.

ξL (t → ∞) = 0

It acts just like a usual piece of conductor at equilibrium.

I =ξR(1−exp(−t / τ L ))

Find ξL at time t=0s and ∞.

“Discharging”

ξ −IR − LdI

dt= 0 ("charging", before)

−IR − LdI

dt= 0 ("discharging", here)

⇔ I = I0 exp(−Rt / L)

Inductive time constant: τ L =L

R⇒ I = I0 exp(−t / τ L )

Energy in an inductorUB =

12

LI 2

In a solenoid, L =μ0n2lA

⇒ UB =12μ0n

2lAI 2

Define magnetic energy density, uB =UB

vol=

UB

lA:

⇒ uB =12μ0n

2 I 2 =1

2μ0

(μ0nI )2

⇒ uB =1

2μ0

B2

where we used B=μ0nI for solenoid

Magnetic and Electric energy density

uB =1

2μ0

B2

uE =12ε0E

2

Mutual Inductance

Coil 2 with respect to 1:

M 21 =Φ21

I1

M 21 =Φ21

I1

ξ2 =−dΦ21

dt=−M21

dI1dt

Coil 1 with respect to 2:Similar to before:

ξ1 =−dΦ12

dt=−M12

dI 2dt

It turns out:M12 =M21 ≡M

ξ1 = −MdI2

dt

ξ2 = −MdI1dt

Example

M =πμ0N1N2R2

2

2R1

Note that this equation is only true for a flat coil, not true for solenoid (which you will derive in the homework)

Reminder:Magnetic and Electric energy

uB =1

2μ0

B2

uE =12ε0E

2

UB =12

LI 2

UE =12

CV2

Electromagnetic Oscillations

&&Q =−ω0

2Q ω0 =1

LC

Electromagnetic Oscillations

UTotal =12

LI 2 +12

CV2 =constant

Energy conservation

UTotal =12

LI 2 +12

CV2 =constant

Going around the loopRecall that VC =

qC,VL =−L

dIdt

⇒ −qC−L

dIdt

=0

but I =dqdt

⇒qC+ L

d2qdt2

=0

⇒d2qdt2

=−q

LC

Simple Harmonic Oscillatord 2q

dt 2=−

qLC

Define "natural angular frequency":

ω =1LC

, then we have:

d 2q

dt 2=−ω 2q

This is the same equation as all other simple harmonic oscillators

Solutiond 2q

dt 2=−ω 2q

⇒ q = qp cos(ωt +φ) where qp , φ: constant

I =dqdt

⇒ I =−ωqpsin(ωt+φ)

qp and φ are determined by initial conditions.

In most cases below, we will assum φ =0.

MasteringPhysics

In HW 30, the question “Oscillations in an LC circuit” Part C, use instead:

€

I = −dq

dt

Reason: They wanted you to look at the magnitude of the current only, and without the minus sign I would have been negative.

EnergyUE =

q2

2C=

qp2

2Ccos2 (ωt+φ)

UB =12

LI 2 =12

Lω 2qp2 sin2 (ωt+φ)

but ω =1LC

⇒ UB =qp

2

2Csin2 (ωt+φ)

UTotal =UB +UE =qp

2

2C: constant

Energy conservation

UTotal =UB +UE =qp

2

2C: constant

Damped OscillationsEnergy is dissipated by the resistor

Going around the loop:

Recall that VC =qC,VL =−L

dIdt

⇒ −qC−L

dIdt

−IR=0

but I =dqdt

⇒ Ld2qdt2

+ Rdqdt

+1C

q=0

Solution (damped)

Ld 2q

dt 2+ R

dqdt

+1C

q=0

⇒ q=qpe−Rt/2L cos(ω 't+φ)

where ω '= ω 2 −(R / 2L)2

(Natural) Frequency, Period, etc…I(t) =I psin(ωt+φ)

d

dtsinωt=ω cosωt

sin(ωt∫ )dt=−1ω

cosωt€

ω =2πf =2π

T€

ω : Angular frequency (rad /s ≡ s−1)

T : Period s( )

f : Frequency (Hz)