Physics for Scientists and Engineers II , Summer Semester 20091

Lecture 11: June 15th 2009

Physics for Scientists and Engineers II

Physics for Scientists and Engineers II , Summer Semester 20092

Force on a Charged Particle in a Magnetic Field

) and both lar toperpendicu is F ofdirection the(still,

sin of anglean at are, :3 Case

0 :2 Case

) and both lar toperpendicu is F ofdirection (the :1 Case

:field magnetic a through movesit as velocity a has charge a Assume

Bv

vBqFBv

FBv

BvvBqFBv

vq

B

B

B

Θ=⇒Θ

=⇒⊥

=⇒⊥

Θ vB

Θ= sinvBqFB

q

BvqF B ×=

BF

Physics for Scientists and Engineers II , Summer Semester 20093

Force on a Positive Charge

Θ vB

q+

BvqF B ×+=

BF

Physics for Scientists and Engineers II , Summer Semester 20094

Force on a Negative Charge

Θ vB

q−

BvqF B ×−=BF

Physics for Scientists and Engineers II , Summer Semester 20095

Units of Magnetic Field

G101T :(G) gauss"" isunit common Another

)(11s

mCN1 are B of Units

4=

==⇒

=⇒=

teslaTAmN

vqFBvBqF B

B

Physics for Scientists and Engineers II , Summer Semester 20096

Motion of a Charged Particle in a Uniform Magnetic Field

Convention: Magnetic field vector points into the page

Magnetic field vector points out of the page

+q v

BF

Physics for Scientists and Engineers II , Summer Semester 20097

Motion of a Charged Particle in a Uniform Magnetic Field

+q v

BF

v

BF+q

r

v BF+q

Physics for Scientists and Engineers II , Summer Semester 20098

Motion of a Charged Particle in a Uniform Magnetic Field

qBm

vrT

mqB

rvqBmvr

rvmqvB

maFF B

πωππ

ω

222

)frequency"cyclotron ("

2

===

==⇒

=⇒

=⇒

==∑

Physics for Scientists and Engineers II , Summer Semester 20099

Uniform Magnetic Field Plus Electric Field

+q v

BF

E

EF

The charge will pass through this crossed magnetic and electric fieldIn a straight line under the following condition: B

Ev

BvqEqF

=⇒

=×+= 0

Velocity selector

+q

Physics for Scientists and Engineers II , Summer Semester 200910

The Mass Spectrometer Principle

q

Detector array

r qBmvr =

0BEv = E

rBBqm

qBBmEr 0

0

=⇒=

B

0B

Physics for Scientists and Engineers II , Summer Semester 200911

Magnetic Force Acting on a Straight Piece of Current-Carrying Conductor

A

q + dv

B

bF

L

BvqF db ×=

:charge single aon Force

( ) ( ) ( )

BLIF

LIL

BLIBLAnqvnALBvqnVBvqNBvqF

b

ddddb

×=

×=×=×=×=×=

. magnitudebut asdirection sameh Vector wit :

:egray volumin thecharges allon force Total

Physics for Scientists and Engineers II , Summer Semester 200912

…Small Segment of Current-Carrying Conductor

B

BsIdFd b ×=

:dssegment short wire aon Force

∫ ×=b

a

b BsdIF

: wireoflength aon Force

IsdBFd

Physics for Scientists and Engineers II , Summer Semester 200913

Example (a variation on Example 29.4 – different direction of B)

B

I

I

r

x

y

What is the net force on this loop?

Physics for Scientists and Engineers II , Summer Semester 200914

Net force on the straight section

B

jrBIBLIF straightB ˆ2

:2rlength ofsegment irestraight w on the Force

, =×=

I

straightBF ,

x

y

r2

Physics for Scientists and Engineers II , Summer Semester 200915

Force on the semicircle

B

( )jiBrdΘIFd

BrdΘIBdsIdF

B

B

ˆsinˆcos

:dssegment short wire aon Force

Θ−Θ−=

==

sd

semiBFd ,

x

y

Θ

( ) [ ]( ) ( )[ ] jIrBjiBrI

jiBrIdΘjiBrIF semiB

ˆ2ˆ0coscosˆ0sinsin

ˆcosˆsinˆsinˆcos

:semicircle entireon Force

00

,

−=−++−=

Θ+Θ−=Θ−Θ−= ∫ππ

ππ

j

Physics for Scientists and Engineers II , Summer Semester 200916

Total Force

0ˆ2ˆ2

: wireentireon Force

,, =+−=+= jIrBjIrBFFF straightBsemiBB

Physics for Scientists and Engineers II , Summer Semester 200917

Torque on a Current Loop in a Uniform Magnetic Field

B

I

a

b

1

2

3

4

The forces on sections 1 and 3 are equaland opposite.Similarly, the forces on 2 and 4 are equaland opposite.

Bottom view:

2 4

2F

4FO

2b

( ) ( )loop) theof (area

2222

,0

max

42max

abAwhereIAB

IabBbIaBbIaBbFbF

butF net

==

=+=+=

=

τ

τ

Physics for Scientists and Engineers II , Summer Semester 200918

…and rotated by 90 degrees….

Bottom view:

2

4

2F

4F

O

0,0

==

τandF net

Physics for Scientists and Engineers II , Summer Semester 200919

…at some arbitrary angle

2F

4F

A

Θb sin2

Θ

( ) ( )

BAI

BIAIabBbIaBbIaBbFbF

×=

Θ=Θ=Θ+Θ=Θ+Θ=

τ

τ sinsinsin2

sin2

sin2

sin2 42

Θ

Physics for Scientists and Engineers II , Summer Semester 200920

Right hand rule to determine direction of A

A

I

I

A

I

Physics for Scientists and Engineers II , Summer Semester 200921

Magnetic Dipole Moment of the Loop

A

I

IA

I

µ µ

B

B

AIN

AI

coil

•=

×=⇒

==

≡

µ

µτ

µ

µ

- U:is dipole magnetic

theofenergy potential thedipole, electric analogy toIn

coil theof windingsofnumber N where,

:Definition

Physics for Scientists and Engineers II , Summer Semester 200922

Magnetic Dipole Moment of the Loop

)180,vs.direction opposite in the points( BU

)0,asdirection same in the points(B-U

- U:is dipole magnetic

theofenergy potentialthedipole,electric analogy toIn

max

min

°=Θ+=

°=Θ=

•=

B

B

B

µµ

µµ

µ

BEp

BEp

•=•=

×=×=

µ

µττ

- U - UEnergy Potential

- - Torque

DipoleMagneticDipoleElectric