from last time… motional emf - physics.wisc.edu · 1 thur. nov. 5, 2009 physics 208, lecture 19 1...
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Thur. Nov. 5, 2009 Physics 208, Lecture 19 1
From last time…
Faraday:
€
ε = −ddtΦB
Motional EMF Moving conductor generates
electric potential difference to cancel motional EMF
Changing flux generates EMF
Motional EMF
Charges in metal feel magnetic force
Tue. Nov. 2, 2009 Physics 208, Lecture 18 2
€
q v × B
Charges move, build up at ends of metal
Equilibrium: electric force cancels magnetic force
€
F E = −
F B ⇒ qE = qΔV / = −qvB so ΔV = −vB
Tue. Nov. 2, 2009 Physics 208, Lecture 18 3
Question Two identical bars are moving through a vertical magnetic field.
Bar (a) is moving vertically and bar (b) is moving horizontally. Which of following statements is true?
A. motional emf exists for (a), but not (b) B. motional emf exists for (b), but not (a) C. motional emf exists for both (a) and (b) D. motional emf exists for neither (a) nor (b)
B(t)
Coil in magnetic field Uniform B-field
increasing in time Flux in z-direction
increasing in time
Thur. Nov. 5, 2009 Physics 208, Lecture 19 4
Lenz’ law: Induced EMF would produce current to oppose change in flux
Induced current
B(t)
+ + +
- - -
But no equilibrium current Charges cannot flow out end
Build up at ends, makes Coulomb electric field
Cancels Faraday electric field
End result No current flowing Electric potential difference
from one end to other Opposes Faraday EMF ΔV = - EMF
Thur. Nov. 5, 2009 Physics 208, Lecture 19 5
Increasing with time
Compare motional EMF
Thur. Nov. 5, 2009 Physics 208, Lecture 19 6
Coil can generate it’s own flux Uniform field inside solenoid Change current -> change flux
€
Binside =µoN
I
€
N turns
Wire turns Surface for flux
2
Thur. Nov. 5, 2009 Physics 208, Lecture 19 7
‘Self’-flux in a solenoid
Flux through one turn
€
Bsolenoid A =µoNA
I
Φ=Flux through entire solenoid
€
= NBsolenoid A =µoN
2A
I
€
Bsolenoid =µoN
I N=# of turns, =length of solenoid
€
inductance
€
Φ = LILsolenoid = µoN
2A /
Thur. Nov. 5, 2009 Physics 208, Lecture 19 8
Inductance: a general result
Flux = (Inductance) X (Current)
€
Φ = LI
Change in Flux = (Inductance) X (Change in Current)
€
ΔΦ = LΔI
Thur. Nov. 5, 2009 Physics 208, Lecture 19 9
Question
The current through a solenoid is doubled. The inductance of the solenoid
A. Doubles
B. Halves
C. Stays the same
Inductance is a geometrical property, like capacitance
Thur. Nov. 5, 2009 Physics 208, Lecture 19 10
Question A solenoid is stretched to twice its length
while keeping the same current and same cross-sectional area. The inductance
A. Increases
B. Decreases
C. Stays the same
€
Bsolenoid =µoNI
€
Φ = LILsolenoid = µoN
2A /
Field, hence flux, have decreased for same current
Fixed current through ideal inductor
For fixed resistor value Current through inductor I = Vbatt/R Flux through inductor = LI Constant current -> Flux through
inductor doesn’t change No induced EMF Voltage across inductor = 0
Thur. Nov. 5, 2009 Physics 208, Lecture 19 11
Vbatt R
L
I
Ideal inductor: coil has zero resistance
Vbatt R
L
I Vb
Va
Try to change current You increase R in time:
Current through inductor starts to decrease
Flux LI through inductor starts to decrease
Faraday electric fields in inductor wires
Induces current to oppose flux decrease
Drive charges to ends of inductor Charges produce
Coulomb electric field Electric potential diff
Thur. Nov. 5, 2009 Physics 208, Lecture 19 12
time
R(t)
€
ΔVL =Vb −Va = −L dIdt
3
Thur. Nov. 5, 2009 Physics 208, Lecture 19 13
Question The potential at a is higher than at b. Which
of the following could be true?
A) I is from a to b, steady
B) I is from a to b, increasing
C) I is from a to b, decreasing
D) I is from b to a, increasing
E) I is from b to a, decreasing
e,.g. current from a to b: current increases, flux to right increases. sign of induced emf such that it would induce current to produce flux to left to oppose change in flux. Electric potential difference opposite to induced EMF, so Va>VB
Thur. Nov. 5, 2009 Physics 208, Lecture 19 14
Energy stored in ideal inductor Constant current (uniform charge motion)
No work required to move charge through inductor
Increasing current: Work required
to move charge across induced EMF
Energy is stored in inductor:
Total stored energy
€
ΔVLq = ΔVLIdt
€
dW = ΔVLIdt = −L dIdtIdt = −LIdI
€
UL = LIdI =120
I
∫ LI2 Energy stored in inductor
€
dUL = −dW
Thur. Nov. 5, 2009 Physics 208, Lecture 19 15
Magnetic energy density
Energy stored in inductor
€
UL =12LI2
Solenoid inductance
€
Lsolenoid = µoN 2A
Energy stored in solenoid
€
Usolenoid =12µo
A µoNI
2
Bsolenoid
€
Usolenoid
A=Bsolenoid2
2µo
Energy density
Thur. Nov. 5, 2009 Physics 208, Lecture 19 16
Question A solenoid is stretched to twice its length
while keeping the same current and same cross-sectional area. The stored energy
A. Increases
B. Decreases
C. Stays the same
€
Usolenoid
A=Bsolenoid2
2µo
€
Bsolenoid =µoNI
B decreases by 2
Energy density decr by 4 Volume increases by 2
Thur. Nov. 5, 2009 Physics 208, Lecture 19 17
Inductor circuit
Induced EMF extremely high Breaks down air gap at switch Air gap acts as resistor
Question Here is a snapshot of an inductor circuit at
a particular time. What is the behavior of the current?
Thur. Nov. 5, 2009 Physics 208, Lecture 19 18
I
Vb
Va
A. Increasing
B. Decreasing
C. Nonzero Constant
D. Must be zero
E. Need more info
€
VL =Vb −Va = −L dIdt
VL + −IR( ) = 0⇒ dI /dt = −I RL
4
Thur. Nov. 5, 2009 Physics 208, Lecture 19 19
Perfect inductors in circuits
Constant current flowing All Voltage drops = 0
I
I? Voltage needed to drive
current thru resistor -IR + ΔVL = 0
€
−IR − LdI /dt = 0+
-
€
dI /dt = −I R /L( )Thur. Nov. 5, 2009 Physics 208, Lecture 19 20
RL circuits
Current decreases in time Slow for large inductance
(inductor fights hard, tries to keep constant current) Slow for small resistance
(little inductor voltage needed to drive current)
I?
+
-
€
dIdt
= −I RL⇒
€
dI = −I RLdt
Time constant
€
τ = L /R
Thur. Nov. 5, 2009 Physics 208, Lecture 19 21
RL circuits
I(t)
+
-
€
I = Ioe−t /(L /R ) = Ioe
−t /τ
€
τ = L /R
Time constant
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