Chapter 33 & 34 Review

Chapter 33: The Magnetic Field

Cyclotron Motion

Newton’s Law mra =

rF

Both a and F directed toward center of circle

ra =

v2

rcyc

rF = qvB

rF =q(

rv×

rB)

rcyc

Equate

W=

v

rcyc=qB

m= 2p fcyc

rcyc =

v

W=

v

2p fcyc

What is the force on those N electrons?

Force on a single electron

rF = - e

rv¥

rB( )

Force on N electrons

rF = - eN

rv¥

rB( )

Now use eN

rv = I l

Lets make l a vector - eNrv = I

rl

points parallel to the wire in the direction of the current when I is positive

Force on wire segment:

rFwire = I

rl ¥

rB

rFwire = I

rl ¥

rB

Comments:Force is perpendicular to both B and lForce is proportional to I, B, and length of line segment

Superposition: To find the total force on a wire you must break it into segments and sum up the contributions from each segment

rFtotal = I

rli ¥

rB(

rri )

segments- iÂ

rB(

rr) =

μ0

4πqrv×r̂r2

rE(

rr) =

q4πε0

r̂r2

Electric Field Magnetic Field

q

r̂

rv

What are the magnitudes and directions of the electric and magnetic fields at this point? Assume q > 0

r

Comparisons: both go like r-2, are proportional to q, have 4π in the denominator, have funny Greek letters

Differences: E along r, B perpendicular to r and v

Magnetic Field due to a current

Magnetic Field due to a single charge

rB(

rr) =

μ0

4πqrv×r̂r2

If many charges use superposition

rB(

rr) =

μ0

4πqi

rvi ×r̂i

ri2

charges−i∑

#2 q2 , v2

#1 q1, v1

#3 q3 , v3

r3

r2

r1

Where I want to know what B is

r̂3

r̂2

r̂1

For moving charges in a wire, first sum over charges in each segment, then sum over segments

rB(

rr) =

μ0

4πqi

rvi ×r̂i

ri2

chargesineachsegment−i

∑⎛

⎝

⎜⎜

⎞

⎠

⎟⎟sgments−j

∑ =μ0

4πI

rlj ×r̂j

rj2

sgments−j∑

qirv i ×r̂i

ri2

chargesineachsegment−i

∑ =I

rlj ×r̂j

rj2

Summing over segments - integrating along curve

rB(

rr) ==

μ0

4πI

rlj ×r̂j

rj2

sgments−j∑ →

μ0

4πIdrl ×r̂r2—∫

drl

r̂

I r

Integral expression looks simple but…..you have to keep track of two position vectors

Biot Savart law

which is where you want to know B

which is the location of the line segment that is contributing to B. This is what you integrate over.

rr

r′r

rr

r′r

Magnetic field due to an infinitely long wire

x

z

I

drl = d ¢z k

r̂

Current I flows along z axis

I want to find B at the point

rB(

rr) ==

μ0

4πIdrl ×r̂r2∫

rr = xi + 0 j + 0k

I will sum over segments at points r′r = 0i + 0 j + ′z k

r̂ =

rr -

r¢r

rr -

r¢r

=xi - ¢z k

x2 + ¢z 2

r =

rr -

r¢r = x2 + ¢z 2

rB =

μ0 I2πr

r

compare with E-field for a line charge

rE =

λ2πε0r

rB(

rr) =

μ0q1

4πr2

rv×r̂

Gauss’ Law:

rB⋅d

rA—∫ =0

rB(

rr)⋅d

rs=—∫ μ0 I through

Biot-Savart Law implies Gauss’ Law and Amperes Law

But also, Gauss’ law and Ampere’s Law imply the Biot -Savart law

Ampere’s Law

rB(

rr) =

μ0q1

4πr2

rv×r̂

Electric field due to a single chargeMagnetic field due to a single loop of current

QuickTime™ and aGraphics decompressor

are needed to see this picture.

Guassian surfaces

rB⋅d

rA—∫ =0

rE⋅d

rA—∫ =

Qin

ε0

QuickTime™ and aGraphics decompressor

are needed to see this picture.

drs

rE(

rr)

drs

drs

rB(

rr)

0 =

rE(

rr)⋅d

rs—∫

rB(

rr)⋅d

rs—∫ =μ0 I

rB(

rr)⋅d

rs=—∫ μ0 I through

rB(

rr)⋅d

rs=—∫ Bl

μ0Ithrough = μ 0NI

B=μ0NI

l=μ0 (N / l)I

# turns per unit length

rB(

rr) =

μ0q1

4πr2

rv×r̂

Gauss’ Law:

rB⋅d

rA—∫ =0

rB(

rr)⋅d

rs=—∫ μ0 I through

Biot-Savart Law implies Gauss’ Law and Amperes Law

But also, Gauss’ law and Ampere’s Law imply the Biot -Savart law

Ampere’s Law

rB(

rr) =

μ0q1

4πr2

rv×r̂

Chapter 34: Faraday’s Law of Induction

loop

rE +

rv×

rB( )—∫ ⋅d

rS=-

ddt

Φ =−ddt

rB⋅d

rA

Area∫

Faraday’s Law for Moving Loops

Magnetic Flux

Φ =

rB ⋅d

rA

S∫

Some surface

Remember for a closed surface Φ =0

rB d

rA

Magnetic flux measures how much magnetic field passes through a given surface

Open surface

rB⋅d

rA—∫ =0

Closed surface

Suppose the rectangle is oriented do that are parallel

Φ =

rB ⋅d

rA

S∫ =

rB A =

rB ab

Rectangular surface in a constant magnetic field. Flux depends on orientation of surface relative to direction of B

rB and d

rA

Lenz’s LawIn a loop through which there is a change in magnetic flux, and EMF is induced that tends to resist the change in flux

What is the direction of the magnetic field made by the current I?

A. Into the pageB. Out of the page

Reasons Flux Through a Loop Can Change

A. Location of loop can change

B. Shape of loop can change

C. Orientation of loop can change

D. Magnetic field can change

d

dtΦ =

ddt

rB⋅d

rA

Area∫

loop

rE +

rv×

rB( )—∫ ⋅d

rS=-

ddt

Φ =−ddt

rB⋅d

rA

Area∫

Faraday’s Law for Moving Loops

Faraday’s Law for Stationary Loops

loop

rE⋅d

rS—∫ ==−

∂rB∂t

⋅drA

Area∫

QuickTime™ and a decompressor

are needed to see this picture.

QuickTime™ and a decompressor

are needed to see this picture.

QuickTime™ and a decompressor

are needed to see this picture.

QuickTime™ and a decompressor

are needed to see this picture.

QuickTime™ and a decompressor

are needed to see this picture.

QuickTime™ and a decompressor

are needed to see this picture.

QuickTime™ and a decompressor

are needed to see this picture.

QuickTime™ and a decompressor

are needed to see this picture.

QuickTime™ and a decompressor

are needed to see this picture.

L

VB

R

VRVL

I I Now I have cleaned things up making use of IB=-IR, IR=IL=I.

Now use device laws:VR = RIVL = L dI/dt

KVL: VL + VR - VB = 0

LdI

dt+ RI - VB = 0

This is a differential equation that determines I(t). Need an initial condition I(0)=0

LdI (t)

dt+ RI (t)- VB = 0, I (0) = 0

Solution:

I(t) =VBR

1- e- t /t( )

t = (L / R)

Let’s verify

This is called the “L over R” time.

Current starts at zero

Approaches a value VB/R

What is the voltage across the resistor and the inductor?

I(t) =VBR

1- e- t /t( )

VR = RI(t) = VB 1- e- t /t( )

VL = LdI

dt= VBe

- t /t

L

VB

R

VRVL

I I Initially I is small and VR is small.All of VB falls across the inductor, VL=VB.Inductor acts like an open circuit.

Time asymptotically I stops changing and VL is small.All of VB falls across the resistor, VR=VB. I=VB/RInductor acts like an short circuit.

Let’s take a special case of no current initially flowing through the inductor

d 2V (t)

dt 2+V (t)

LC= 0

V (0) = VC (0)

IL (0) = 0 = - CdV

dt t= 0

Solution

V (t) = VC (0)cos(wt)

w= 1 / LC

A:

B: V (t) = VC (0)sin(wt)

Initial charge on capacitor

Current through Inductor and Energy Stored

Energy

UC =1

2CV 2

UL =1

2LIL

2

t

+

IV2

V1

Foolproof sign convention for two terminal devices

1. Label current going in one terminal (your choice).

2. Define voltage to be potential at that terminal wrt the other terminal

V= V2 -V1

3. Then no minus signs

V = RI

V = L

dI

dt

I = C

dV

dt

P = VI

Power to deviceKVL Loop

Contribution to voltage sum = +V