Physics 100 Lecture 2Physics 212 Lecture 12, Slide 11 Three points are arranged in a uniform magnetic field. The B field points into the screen.1) A positively charged particle is

Physics 212 Lecture 12, Slide 1

Physics 212 Lecture 12

Today's Concept:

Magnetic Force on moving charges

F qv B

Physics 212 Lecture 2

Music Who is the Artist?

A) The Meters

B) The Neville Brothers

C) Trombone Shorty

D) Michael Franti

E) Radiators

Tribute to New Orleans and Mardi Gras


Your Comments

05

Good questions! But very deep!

“Magnets. How do they work?”

“Please explain what causes magnetic fields, and why they exist. Unless we learn that later or not at all, I just don’t understand.”

“What are we supposed to do if we have two left hands?”

“Could you go over this right hand rule again? There are so many of them...”

RHR confused many

A little at a time!

“We most definitely need examples. Also, every time you post an awful pun, a puppy dies. Just so you know.”

“I refuse to make an attempt at a bad pun.”

“How can a magnetic field point in a direction? What about the wire example in the prelecture where is shown to move in a counter-clockwise/clockwise direction?” “This stuff is really neat... It is fun to actually see the calculations for magnetism. However, since this is the first time I’ve really seen it, it is still a bit confusing. If you could go through different examples and go over the actual concepts more, that would be great.”


Today’s Plan:

1) Review of magnetism 2) Review of cross product 3) Example problem

Key Concepts:

1) The force on moving charges due to a magnetic field.

2) The cross product.

05


Magnetic Observations

N S

• Bar Magnets

• Compass Needles

S N N S

N S

N S

N S N S cut in half

07

N S

Magnetic Charge?


• Compass needle deflected by electric current

I

• Magnetic fields created by electric currents

• Direction? Right thumb along I, fingers curl in direction of B

• Magnetic fields exert forces on electric currents (charges in motion)

V

V

I

I F F

V

I

I

F

F


12


The magnetic field at P points

A. Case I: left, Case II: right B. Case I: left, Case II: left

C. Case I: right, Case II: left D. Case I: right, Case II: right


I (out of the screen)

P

Case I

I (into of the screen)

P

Case II

WHY? Direction of B: right thumb in direction of I, fingers curl in the direction of B


Magnetism & Moving Charges

All observations are explained by two equations:

F qv B Today

0

2

ˆ

4

I ds rdB

rNext Week

14


Cross Product Review

• Cross Product different from Dot Product

– A●B is a scalar; A x B is a vector

– A●B proportional to the component of B parallel to A

– A x B proportional to the component of B perpendicular to A

• Definition of A x B – Magnitude: ABsinq

– Direction: perpendicular to plane defined by A and B with sense given by right-hand-rule

16


Remembering Directions: The Right Hand Rule

F qv B x

y

z

B F

qv

17


Three points are arranged in a uniform magnetic field. The B field points into the screen.

1) A positively charged particle is located at point A and is stationary.

The direction of the magnetic force on the particle is:

Checkpoint 1a

18

A positively charged particle is located at point A and is stationary. The direction of the

magnetic force on the particle is

A. right B. left C. into the screen D. out of the screen

E. zero

“all are into the screen”

“Right hand rule says its pointing out of the screen”

“qvXB=0 if v=0.”





Checkpoint 1a

F qv B

The particle’s velocity is zero. There can be no magnetic force.

18

A positively charged particle is located at point A and is stationary. The direction of the

magnetic force on the particle is


E. zero





21

Checkpoint 1b

The positive charge moves from A toward B. The direction of the magnetic force on the

particle is


E. zero

“it should be diagonally outward”

“v is up, b is into screen, right hand rule says it should be left.”

“Because the magnetic force goes into the screen, so will the direction”





F qv B

21

X

qv

B F

Checkpoint 1b

The positive charge moves from A toward B. The direction of the magnetic force on the

particle is


E. zero


Cross Product Practice • protons (positive charge) coming out of screen (+z direction)

• Magnetic field pointing down

• What is direction of force on POSITIVE charge?

A) Left B) Right C) UP D) Down E) Zero

F qv B

x

y

z B 24

- x + x + y - y


Motion of Charge q in Uniform B Field

x x x x x x x

x x x x x x x

x x x x x x x

x x x x x x x

x x x x x x x

x x x x x x x

Uniform B into page

• Force is perpendicular to v – Speed does not change

– Uniform Circular Motion

• Solve for R:

2

v

aR

2

v

qvB mR qB

mvR

q v

F q

v

F

q v

F

q

v

F

R

F qv B F qvB

30


mv pR

qB qB

LHC 17 miles diameter


The drawing below shows the top view of two interconnected chambers. Each chamber has a unique magnetic field. A positively charged particle is fired into chamber 1, and observed to follow the

dashed path shown in the figure.

Checkpoint 2a

34

What is the direction of the magnetic field in chamber 1?

A. up B. down C. into the page D. out of the page

“Since the direction of the magnetic field has to be perpendicular to the charged

particle’s direction, it will point up..”

“Having a magnetic field going into the screen indicates counterclockwise motion.”

“Thumb points to the left because particle is deflected to the left and fingers point

up because that is velocity vector. Fingers curl out of the screen.”




Checkpoint 2a

34

F qv B qv

F B

.

What is the direction of the magnetic field in chamber 1?

A. up B. down C. into the page D. out of the page




36

Checkpoint 2b

Compare the magnitude of the magnetic field in chamber 1 to the magnitude of the

magnetic field in chamber 2

A. |B1| > |B2| B. |B1| = |B2| C. |B1| < |B2|

“because the curve is much sharper inside box 1”

“It equals out.”

“Because the particle changed direction more slowly in the second chamber.”




Observation: R2 > R1

36

qB

mvR 21 BB

Checkpoint 2b

Compare the magnitude of the magnetic field in chamber 1 to the magnitude of the

magnetic field in chamber 2

A. |B1| > |B2| B. |B1| = |B2| C. |B1| < |B2|


Calculation A particle of charge q and mass m is accelerated from rest by an electric field E through a distance d and enters and exits a region containing a constant magnetic field B at the points shown. Assume q,m,E,d, and x0 are known. What is B?

X X X X X X X X X

X X X X X X X X X

X X X X X X X X X

X X X X X X X X X

B

B

x0

x0/2

exits here

enters here

d

q,m

– Absolutely ! We need to use the definitions of V and E and either

conservation of energy or Newton’s Laws to understand the motion of the particle before it enters the B field.

– We need to use the Lorentz Force Law (and Newton’s Laws) to determine what happens in the magnetic field.

• Conceptual Analysis – What do we need to know to solve this problem?

(A) Lorentz Force Law (B) E field definition (C) V definition (D) Conservation of Energy/Newton’s Laws (E) All of the above

)( EqBvqF

E



• Strategic Analysis – Calculate v, the velocity of the particle as it enters the magnetic

field – Use Lorentz Force equation to determine the path in the field as a

function of B – Apply the entrance-exit information to determine B

X X X X X X X X X

X X X X X X X X X

X X X X X X X X X

X X X X X X X X X

B

B

x0/2

exits here

enters here

d

q,m

Let’s Do It !!

x0 E



X X X X X X X X X

X X X X X X X X X

X X X X X X X X X

X X X X X X X X X

B

B

x0/2

exits here

enters here

d

q,m

• Why?? – Conservation of Energy

• Initial: Energy = U = qV = qEd • Final: Energy = KE = ½ mv0

2

– Newton’s Laws • a = F/m = qE/m • v0

2 = 2ad

x0 E

• What is v0, the speed of the particle as it enters the magnetic field ?

(A) (B) (C) (D) (E) m

Evo

2

m

qEdvo

2 advo 2

md

qEvo

2

m

qEdvo

qEdmvo 2

21

m

qEdvo

2

dm

qEvo 22

m

qEdvo

2



X X X X X X X X X

X X X X X X X X X

X X X X X X X X X

X X X X X X X X X

B

B

x0/2

exits here

enters here

d

q,m

• Why?? – Path is circle !

• Force is perpendicular to the velocity • Force produces centripetal acceleration • Particle moves with uniform circular motion

x0

X X X X X X X X X

X X X X X X X X X

X X X X X X X X X

X X X X X X X X X

• What is the path of the particle as it moves through the magnetic field?

(A) (B) (C)

X X X X X X X X X

X X X X X X X X X

X X X X X X X X X

X X X X X X X X X

X X X X X X X X X

X X X X X X X X X

X X X X X X X X X

X X X X X X X X X

m

qEdvo

2

E



X X X X X X X X X

X X X X X X X X X

X X X X X X X X X

X X X X X X X X X

B

B

x0/2

exits here

enters here

d

q,m

• Why??

x0

X X X X X X X X X

X X X X X X X X X

X X X X X X X X X

X X X X X X X X X

R xo/2

• What is the radius of path of particle?

(A) (B) (C) (D) (E)

oxR oxR 2oxR

2

1

qB

mvR o

a

vR o

2

m

qEdvo

2

E



X X X X X X X X X

X X X X X X X X X

X X X X X X X X X

X X X X X X X X X

B

B

x0/2

exits here

enters here

d

q,m

• Why??

x0

m

qEdvo

2

(A) (B) (C) (D) (E)

q

mEd

xB

o

22

q

mEd

xB

o

21

qEd

mEB

2

v

EB

o

o

qx

mvB

amF

R

vmBqv o

o

2

R

v

q

mB o

021 xR

m

qEd

xq

mB

o

22

q

mEd

xB

o

22

E


Follow-Up A particle of charge q and mass m is accelerated from rest by an electric field E through a distance d and enters and exits a region containing a constant magnetic field B at the points shown. Assume q,m,E,d, and x0 are known. What is B?

X X X X X X X X X

X X X X X X X X X

X X X X X X X X X

X X X X X X X X X

B

B

x0/2

exits here

enters here

d

q,m x0

• Suppose the charge of the particle is doubled (Q = 2q),while keeping the mass constant. How does the path of the particle change?

(A) (B) (C)

X X X X X X X X XX X X X X X X X X












q

mEd

xB

o

22

No slam dunk.. As Expected ! Several things going on here

1. q changes v changes

2. q & v change F changes

3. v & F change R changes

E



X X X X X X X X X

X X X X X X X X X

X X X X X X X X X

X X X X X X X X X

B

B

x0/2

exits here

enters here

d

q,m x0


How does v, the new velocity at the entrance, compare to the original velocity v0?

• Why??

q

mEd

xB

o

22

2

212

21 22 omvqEdQEdmv

(A) (B) (C) (D) (E) 2

ovv ovv ovv 2 ovv 2

2

ovv

22 2 ovv ovv 2

E



X X X X X X X X X

X X X X X X X X X

X X X X X X X X X

X X X X X X X X X

B

B

x0/2

exits here

enters here

d

q,m x0


How does F, the magnitude of the new force at the entrance, compare to F0, the magnitude of the original force?

• Why??

q

mEd

xB

o

22

ovv 2

(A) (B) (C) (D) (E) 2

oFF oFF oFF 2 oFF 2

oFF 22

BvqQvBF o 22 oFF 22

E



X X X X X X X X X

X X X X X X X X X

X X X X X X X X X

X X X X X X X X X

B

B

x0/2

exits here

enters here

d

q,m x0


How does R, the radius of curvature of the path, compare to R0, the radius of curvature of the original path?

• Why??

q

mEd

xB

o

22

ovv 2 oFF 22

(A) (B) (C) (D) (E) 2

oRR

2

oRR oRR oRR 2

oRR 2

R

vmF

2

F

vmR

2

2222

2 22o

o

o

o

o R

F

vm

F

vmR

E



X X X X X X X X X

X X X X X X X X X

X X X X X X X X X

X X X X X X X X X

B

B

x0/2

exits here

enters here

d

q,m x0


(A) (B) (C)













E

q

mEd

xB

o

22

2

oRR

A Check: (Exercise for Student) Given our result for B (above), can you show:

Q

mEd

BR

21

2

oRR

E

Documents

Physics 100 Lecture 2Physics 212 Lecture 12, Slide 11 Three points are arranged in a uniform magnetic field. The B field points into the screen.1) A positively charged particle is