Review&for&Exam&III€¦ · Review&for&Exam&III “Leftovers” RCcircuits. Right?Hand?Rules F qv...

17
Basic Laws Basic Definitions Important Derivations Examples + B dl I = μ 0 U B = ! ! μ dF Idl B ! ! ! = × R mv qB = Review for Exam III “Leftovers” RC circuits

Transcript of Review&for&Exam&III€¦ · Review&for&Exam&III “Leftovers” RCcircuits. Right?Hand?Rules F qv...

Page 1: Review&for&Exam&III€¦ · Review&for&Exam&III “Leftovers” RCcircuits. Right?Hand?Rules F qv B! ! ! = × F IL B! ! ! = × r F! ! τ= × B! ! τ= µ× 2 0 ˆ 4 r I d s r dB ×

Basic  Laws

Basic  Definitions

Important  Derivations

Examples

+B dl I• =∫ µ0

U B= − •! !µ

dF Idl B! ! != ×

R mvqB

=

Review  for  Exam  III“Leftovers” RC  circuits

Page 2: Review&for&Exam&III€¦ · Review&for&Exam&III “Leftovers” RCcircuits. Right?Hand?Rules F qv B! ! ! = × F IL B! ! ! = × r F! ! τ= × B! ! τ= µ× 2 0 ˆ 4 r I d s r dB ×

Right-­Hand-­RulesBvqF!!!

×= BLIF!!!

×=

Fr!!

×=τ B!!

×= µτ

20 ˆ4 r

rsdIBd ×=

!!

πµ

Page 3: Review&for&Exam&III€¦ · Review&for&Exam&III “Leftovers” RCcircuits. Right?Hand?Rules F qv B! ! ! = × F IL B! ! ! = × r F! ! τ= × B! ! τ= µ× 2 0 ˆ 4 r I d s r dB ×

Basic  Laws• Biot-­Savart Law:

Direct  calculation  of  B  from  currents.dB I dl r

r

!! !

=×µ

π0

34

x

Rrθ

θ

P

Idx

RHR

We  did  the  integral  but  it  is  messy  and  takes  time  …  so

Page 4: Review&for&Exam&III€¦ · Review&for&Exam&III “Leftovers” RCcircuits. Right?Hand?Rules F qv B! ! ! = × F IL B! ! ! = × r F! ! τ= × B! ! τ= µ× 2 0 ˆ 4 r I d s r dB ×

Basic  Laws• Biot-­Savart Law:

direct  calculation  of  B  from  currents.

• Ampere's  Law:calculation  of  B  in  cases  of  high  symmetry

dB I dl rr

!! !

=×µ

π0

34

B dl I• =∫ µ0

´

If  B  “constant”  along  paththen  integral  is  just  the  path

B IR

=µπ0

2

Page 5: Review&for&Exam&III€¦ · Review&for&Exam&III “Leftovers” RCcircuits. Right?Hand?Rules F qv B! ! ! = × F IL B! ! ! = × r F! ! τ= × B! ! τ= µ× 2 0 ˆ 4 r I d s r dB ×

Basic  Laws• Faraday's  Law:

determines  the  induced  E  field  (or  emf)  from  a  changing  magnetic  flux.

ε = −ddtBΦ

! !E dl d

dtB•∫ = −

Φ

dS

B BΦB B dS≡ •∫! !

Define  Flux

Basic  question:    Does  it  change  in  time  (ie,  increase  or  decrease?)You  will  have  to  figure  that  out

Page 6: Review&for&Exam&III€¦ · Review&for&Exam&III “Leftovers” RCcircuits. Right?Hand?Rules F qv B! ! ! = × F IL B! ! ! = × r F! ! τ= × B! ! τ= µ× 2 0 ˆ 4 r I d s r dB ×

Lenz's  LawThe  induced  current  will  appear  in  such  a  direction  that  itopposes  the  change  in  flux that  produced  it.

vB

S Nv

BN S

Page 7: Review&for&Exam&III€¦ · Review&for&Exam&III “Leftovers” RCcircuits. Right?Hand?Rules F qv B! ! ! = × F IL B! ! ! = × r F! ! τ= × B! ! τ= µ× 2 0 ˆ 4 r I d s r dB ×

Change Area of loop Change magnetic fieldthrough loop

Change orientation of loop relative to B

EMF  Observed  when  dΦ/dt is  non  zeroDirection  from  Lenz’s  Law

Examples

Page 8: Review&for&Exam&III€¦ · Review&for&Exam&III “Leftovers” RCcircuits. Right?Hand?Rules F qv B! ! ! = × F IL B! ! ! = × r F! ! τ= × B! ! τ= µ× 2 0 ˆ 4 r I d s r dB ×

BvqEqF!!!!

×+=

Definitions  &  Derivations

• Lorentz  Force:

Force  in  an  electric  field Force  in  a  magnetic  field

RHR

E x    x    x    x    x    x    x    x    xx    x    x    x    x    x    x    x    xx    x    x    x    x    x    x    x    xx    x    x    x    x    x    x    x    x

Examples  with  Magnetic  Field  only  and  a  Moving charge:

Page 9: Review&for&Exam&III€¦ · Review&for&Exam&III “Leftovers” RCcircuits. Right?Hand?Rules F qv B! ! ! = × F IL B! ! ! = × r F! ! τ= × B! ! τ= µ× 2 0 ˆ 4 r I d s r dB ×

Magnetic  Dipole  Moment

Bx

.FFθ

θ

µ

Direction:  ⊥ to  plane  of  the  loop  in  the  direction  the  thumb  of  right  hand  points  if  fingers  curl  in  direction  of  current.

• Torque  on  loop  is  then:

• Note:  if  loop  consists  of  N  turns,  µ =  NAI

Magnitude: µ = AI

τ = AIB sinθ Þ! ! !τ µ= × B

Remember  this:    The  torque  always  wants  to  line  µ up  with  B!

Definitions  &  Derivations

RHR

I

Page 10: Review&for&Exam&III€¦ · Review&for&Exam&III “Leftovers” RCcircuits. Right?Hand?Rules F qv B! ! ! = × F IL B! ! ! = × r F! ! τ= × B! ! τ= µ× 2 0 ˆ 4 r I d s r dB ×

Potential  Energy  of  Dipole

Bx

.FFθ

θ

µ

• Work  is  required  to  change  the  orientation  of  a  magnetic  dipole  in  the  presence  of  a  magnetic  field.

• Define  potential  energy  U  (with  zero  at  position  of  max  torque) corresponding  to  this  work.

∫≡ θτdU Þ BU!!

•−= µ

Remember  this:    The  torque  always  wants  to  line  µ up  with  B …which minimizes the potential energy!

Page 11: Review&for&Exam&III€¦ · Review&for&Exam&III “Leftovers” RCcircuits. Right?Hand?Rules F qv B! ! ! = × F IL B! ! ! = × r F! ! τ= × B! ! τ= µ× 2 0 ˆ 4 r I d s r dB ×

Examples  with  RC  circuits

• Charging  a  capacitor:

• Discharging  a  capacitor:R

I I

a

b + +-­ -­

RI I

a

bε = +Rdq

dtqC

Rdqdt

qC

+ = 0

Loop  rule  !

Page 12: Review&for&Exam&III€¦ · Review&for&Exam&III “Leftovers” RCcircuits. Right?Hand?Rules F qv B! ! ! = × F IL B! ! ! = × r F! ! τ= × B! ! τ= µ× 2 0 ˆ 4 r I d s r dB ×

Charging                            DischargingRC

t

2RC

0

0 1 2 3 40

0.5

1

t/RC

Q f( )x

x

-­ε/R

I

t

q

0

0 1 2 3 4

0.5

11

0.0183156

f( )x

40 x

q = C εe -t/RC

I dqdt R

e t RC= = − −ε /

RC

t

q

2RC

00 1 2 3 4

0

0.5

1

t/RC

Q f( )x

x

I

0 t

ε/R

0 1 2 3 4

0.5

11

0.0183156

f( )x

40 x

( )q C e t RC= − −ε 1 /

I dqdt R

e t RC= = −ε /

Page 13: Review&for&Exam&III€¦ · Review&for&Exam&III “Leftovers” RCcircuits. Right?Hand?Rules F qv B! ! ! = × F IL B! ! ! = × r F! ! τ= × B! ! τ= µ× 2 0 ˆ 4 r I d s r dB ×

Examples

• Orbit  of  charged  particle  in  uniform  magnetic  field: x    x    x    x    x    x

x    x    x    x    x    xx    x    x    x    x    x

v

F

B

q

x    x    x    x    x    xx    x    x    x    x    xx    x    x    x    x    x

FvR

R mvqB

=

Page 14: Review&for&Exam&III€¦ · Review&for&Exam&III “Leftovers” RCcircuits. Right?Hand?Rules F qv B! ! ! = × F IL B! ! ! = × r F! ! τ= × B! ! τ= µ× 2 0 ˆ 4 r I d s r dB ×

Examples

• B  for  straight  wire:× R

(Ampere)

• B  for  ¥ current  sheet:

• B  for  ¥ solenoid: xxx xx•• • ••

ii

xxxx

x

xx

xxxx

i

B IR

=µπ0

2

(Biot-­Savart)

x

Rrθ

θ

P

Idx

B = µ0 ni

2

B = µ0 ni

Page 15: Review&for&Exam&III€¦ · Review&for&Exam&III “Leftovers” RCcircuits. Right?Hand?Rules F qv B! ! ! = × F IL B! ! ! = × r F! ! τ= × B! ! τ= µ× 2 0 ˆ 4 r I d s r dB ×

Examples:  B  from  long  straight  wire

B = µ0 I2 π r

a 2

• Inside  the  wire:  (r  <  a)

• Outside  the  wire:  (r>a)

B = µ0 I2πr

Page 16: Review&for&Exam&III€¦ · Review&for&Exam&III “Leftovers” RCcircuits. Right?Hand?Rules F qv B! ! ! = × F IL B! ! ! = × r F! ! τ= × B! ! τ= µ× 2 0 ˆ 4 r I d s r dB ×

Examples

• Force  on  parallel  current-­carrying  conductors  of  length  L:

• Currents  in  oppositedirections  Þ⇒ repulsive  force.

• Currents  in  same  direction  Þ attractive  force.

πd2LIIµBLIF ba0

abb =×=!!! L d

×F

Ib

Ia×

F

• Force  on  current  carrying  wire  in  a  magnetic  field:

dF Idl B! ! != × RHR

Page 17: Review&for&Exam&III€¦ · Review&for&Exam&III “Leftovers” RCcircuits. Right?Hand?Rules F qv B! ! ! = × F IL B! ! ! = × r F! ! τ= × B! ! τ= µ× 2 0 ˆ 4 r I d s r dB ×

Examples• Current  induced  by  pulling  coil  through  magnetic  field:

RLBv

dtd

RRI =

Φ==1εIs  the  flux  changing?

If  so,  is  it  increasing?  Decreasing?How  can  you  express  this  quantitatively?

The  current  through  this  bar  results  in  a  force  on  the  bar  

RILBvRvBLFvP 2=⎟

⎠⎞

⎜⎝⎛==

BLIF!!!

×=