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


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


Increasing with time

Compare motional EMF


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


‘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 /


Inductance: a general result

  Flux = (Inductance) X (Current)

€

Φ = LI

  Change in Flux = (Inductance) X (Change in Current)

€

ΔΦ = LΔI


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


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


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


time

R(t)

€

ΔVL =Vb −Va = −L dIdt

3


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


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


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


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


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?


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


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


RL circuits

I(t)

+

-

€

I = Ioe−t /(L /R ) = Ioe

−t /τ

€

τ = L /R

  Time constant

From last time… Motional EMF - physics.wisc.edu · 1 Thur. Nov. 5, 2009 Physics 208, Lecture 19 1...

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