chapter 30 – self inductance, inductors & dc circuits...
TRANSCRIPT
Chapter 30 – Self Inductance, Inductors & DC Circuits Revisited
Self-Inductance and Inductors • Self inductance determines the magnetic flux in a circuit due to
the circuit’s own current.
• Self-induced emf – A circuit’s self-inductance will give rise to an emf if the current is changing with time.
• Inductor - a circuit device designed to have a particularly high self-inductance relative to the rest of the circuit. It is typically a solenoid.
Every circuit has some self inductance!
�B = LI
E = �d�B
dt= �L
dI
dt)
I~E · ~dl = �L
dI
dt
Example: Estimate the self inductance of a 2-cm long solenoid have 200 windings per cm and a radius of 0.3 cm.
� = µ0nIN⇡R2
L = �/I = µ0nN⇡R2
L = (4⇡ ⇥ 10�7)(2⇥ 104)(4⇥ 102)⇡(0.003)2 = 0.3 mH
CT 33.29
What is the current through the resistor a very long time after the switch is closed?
L = 10H V = 10V
R = 20Ω
A) 0 A B) 0.5 A C) 1 A D) 10 A E) other
Clicker Question
CT 33.28
What is the current through the resistor immediately after the switch is closed ?
L = 10H V = 10V
R = 20Ω
A) 0 B) 0.5 A C) 1 A D) 10 A E) other
Clicker Question
Clicker question What is the voltage at the top and the bo1om of the inductor immediately a8er switch is closed?
1. Vtop = 10 V, Vbo,om= 0
2. Vtop = 0, Vbo,om = 0
3. Vtop = 10 V, Vbo,om= 10 V
4. WTF (huh?)?! This seems wrong. Seems to depend on how I look at it…
L = 10H V = 10V
R = 20Ω
• Faraday’s Law of induction for the circuit:
• Self-induced inductance causes line integral to be non-zero!
• E-field no longer conservative!
Thinking of circuits solely in terms of voltages (and using Kirchoff’s Loop Law) is problematic since the electric field is no longer conservative (it has curl)
I~E · ~dl = �L
dI
dt L = 10H V = 10V
R = 20Ω
X
i
�Vi 6= 0!
Where is the electric field?
Note: It is not correct to think of a voltage difference existing across the inductor (as many books do)
E downward in battery
• E-field in battery • E-field in resistor (if
there is a current) • That’s it! No E-field
anywhere else!!!
I~E · ~dl = �L
dI
dt
Integral of E-field across resistor + integral of E-field across battery = induced emf.
IR� Eb = �LdI
dt
Solution to the Circuit:
timescale to reach steady state:
� = L/R
I(t) =EbR
[1� exp(�Rt/L)]
IR� Eb = �LdI
dt
Follow-up question
Immediately after the switch is closed, what is the reading of the voltmeter V2?
The switch has bin in position a for a long time. What happens when switch is switched to position B?
I(t) =�Vbat
Rexp(�Rt/L)
IR = �LdI
dt
I~E · ~dl = �L
dI
dt
Why do you get a spark as you unplug your hair dryer while it is running?
Clicker Ques,on
When switch is opened, what happens to the 20-‐Ohm bulb?
a) bulb instantly goes out (zero luminosity)
b) luminosity of bulb momentarily stays the same and then fades
c) Luminosity of bulb instantly increases and then fades
Suppose the two resistors are light-bulbs. The switch has been closed for a long time.
Energetics How much energy is dissipated by the resistor after the switch is in position b?
I(t) =�Vbat
Rexp(�Rt/L)
Where does the energy (converted to Ohmic dissipation) come from?
P = I2(t)R = I20R exp[�2Rt/L]
E =
Z 1
0P (t) dt = I20R
Z 1
0exp[�2Rt/L] dt
E =L
2I20
Energy of Magnetic Field • The idea is that magnetic fields are a form of energy.
• Inside the solenoid, there is an energy density uB due to the B field.
• Always true (general result is based on a much more formal derivation)
1
2
LI2= uB ⇥ volume
uB =B2
2µ0
L = µ0nN⇡R2 = µ0nnV
1
2LI2 =
1
2µ0(nI)
2V =1
2
B2
µ0
(V = ⇡R2d)
V
Solar Flare
The LC Circuit
Suppose switch closed at t = 0 with capacitor initially charged to Q. What is q(t)?
Ld2q
dt2+
q
C= 0
� =r
1LC
I = �dq
dtLooks like harmonic oscillator!
m
d
2x
dt
2+ kx = 0 (mass on spring)
��E · �dl = � q
C= �L
dI
dt
q(t) = Q0 cos(!t)
Energy in an LC circuit oscillates between electric and magnetic
• similar to mechanical harmonic oscillator (kinetic and potential)
LRC Circuit:
With resistance in the circuit, LC oscillations damp out.
Similar to the damped harmonic oscillator:
underdamped case:
I
I⇥E · d� = VC + IR = �L
dI
dt
Ld2q
dt2+ R
dq
dt+
1C
q = 0
q(t) = q0 exp(�R/(2L)t) cos
r1
LC� R2
4L2t
!
+q
-q
I
I =dq
dt
md2x
dt2+ b
dx
dt+ kx = 0
Back emf in motor circuit Circuits involving motors contain a rotating coil in a B field
The rotating coil will generate an emf due to induction.
This emf causes the current to be smaller than what it would be if it weren’t rotating (Lenz’s law).
For this reason we call it a back emf.