thur. nov. 5, 2009physics 208, lecture 191 from last time… faraday: motional emf moving conductor...

Thur. Nov. 5, 2009 Physics 208, Lecture 19 1

From last time…

Faraday:

€

ε =−d

dtΦB

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

€

qr v ×

r B

Charges move, build up at ends of metal

Equilibrium: electric force cancels magnetic force

€

rF E = −

r F B ⇒ qE = qΔV /l = −qvB so ΔV = −vl B

Tue. Nov. 2, 2009 Physics 208, Lecture 18 3

QuestionTwo 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

lI

€

l

N turns

Wire turns Surface for flux


‘Self’-flux in a solenoid

Flux through one turn

€

Bsolenoid A =μoNA

lI

=Flux through entire solenoid

€

=NBsolenoid A =μoN

2A

lI

€

Bsolenoid =μoN

lI N=# of turns, =length of solenoid

€

l

inductance

€

=LI

Lsolenoid = μoN2A /l


Inductance: a general result

Flux = (Inductance) X (Current)

€

=LI

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

€

Δ=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

l

€

=LI

Lsolenoid = μoN2A /l

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

VbattR

L

IVb

Va

Try to change current You increase R in time:

Current through inductorstarts 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 = −LdI

dt


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 = −LdI

dtIdt = −LIdI

€

UL = LIdI =1

20

I

∫ LI2 Energy stored in inductor

€

dUL = −dW


Magnetic energy density

Energy stored in inductor

€

UL =1

2LI2

Solenoid inductance

€

Lsolenoid = μo

N 2A

l

Energy stored in solenoid

€

Usolenoid =1

2μo

AlμoNI

l

⎛

⎝ ⎜

⎞

⎠ ⎟2

Bsolenoid

€

Usolenoid

Al=

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

Al=

Bsolenoid2

2μo

€

Bsolenoid =μoNI

lB 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 = −LdI

dt

VL + −IR( ) = 0⇒ dI /dt = −IR

L


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( )


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?

+

-

€

dI

dt= −I

R

L⇒

€

dI = −IR

Ldt

Time constant

€

τ =L /R


RL circuits

I(t)

+

-

€

I = Ioe−t /(L / R ) = Ioe

−t /τ

€

τ =L /R

Time constant

thur. nov. 5, 2009physics 208, lecture 191 from last time… faraday: motional emf moving conductor...

Documents

current slide

flux slide

solenoid flux

motional emf slide

time flux

b slide

fixed current

flux li