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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)