• Certain objects and circuits produce magnetic fields• Magnetic fields, like electric fields, are vector fields

• They have a magnitude and a direction• Denoted by B, or B(r)

• They have no effect on charges at rest• They produce a force on moving charges given by

• Perpendicular to magnetic field• Perpendicular to velocity

• Magnetic field strengths are measured in units called a tesla, abbreviated T• A tesla is a large amount of magnetic field

B q F v B

sinF q vB

q

FB

v

N

C m/s N

TA m

Ch. 29

Warmup 12

• Vectors in the plane are easy to draw• Vectors perpendicular to the plane are hard

• Coming out of the plane• Going into the plane

The Right Hand RuleTo figure out the direction of magnetic force, use the following steps:1. Point your fingers straight out in direction of first vector v2. Twist your hand so when you curl your fingers, they point in the

direction of B3. Your thumb now points in the direction of v B4. If q is negative, change the sign

q F v B

v B

v B

ˆ ˆ ˆ

det x y z

x y z

v v v

B B B

i j k

v B

ˆ ˆ

ˆ

y z z y z x x z

x y y x

v B v B v B v B

v B v B

i j

k

Finding the direction

Ca+2 ion

Ca Bv

q F v BWhat is the direction of the force for each of the situations sketched?A) B) C) D) E) F) None of the above

Bproton

pv

B

v

electron

e

q F v B

sinF q vB

4. If q is negative, change the sign

JIT Quick Quiz 29.1

An electron moves in the plane of this paper toward the top of the page. A magnetic field is also in the plane of the page and directed toward the right. The direction of the magnetic force on the electron is described by which of the following?

(a) toward the top of the page (b) toward the left edge of the page (c) toward the bottom of the page (d) toward the right edge of the page (e) upward out of the page (f) downward into the page

Ans e

Ex- (Serway 29-11) A proton moves perpendicular to a uniform magnetic field B at 1.00 x 10 7m/s and experiences an acceleration of 2.00 x 1013 m/s2 in the positive x direction when its velocity is in the positive z direction. Determine the magnitude and direction of the field.

Solve on Board

Warmup 12

Warmup 13

Cyclotron MotionConsider a particle of mass m and charge q moving in a uniform magnetic field of strength B

B v

F

v

F

v v

F qF

• Motion is uniform circular motion• Centripetal force formula:

q F v B

2mvF

R qvB mv qRB

• Let’s find how long it takes to go around:

2 RT

v

2 mT

qB

2

T

qB

m

p qRB

CT –1 Cosmic rays (atomic nuclei stripped bare of their electrons) which come from all directions, would continuously bombard Earth’s surface if most of them were not deflected by Earth’s magnetic field. Given that Earth is, to an excellent approximation, a magnetic dipole, the intensity of cosmic rays bombarding its surface is greatest at the

A. poles. B. mid-latitudes. C. equator.

The Earth has magnetic field linesCharged particles from space follow themHit only at magnetic polesaurora borealisaurora australis

Concept QuestionTwo particles with the same mass are moving in the same magnetic field, but particle X is circling in less time than particle Y. What can account for this?A) Particle X is moving faster (only)B) Particle Y is moving faster (only)C) Particle X has more charge (only)D) Particle Y has more charge (only)E) A and C could both account for thisF) B and C could both account for this

mv qRB

B

X

Y• Moving faster doesn’t help

• Higher speed means bigger radius• Higher charge does help

• You turn corners faster

2 mT

qB

Velocity Selector / Mass Spectrometer• When we have both electric and magnetic fields, the force is

q F E v B

v

• Magnetic field produces a force on the charge• Add an electric field to counteract the magnetic force• Forces cancel if you have the right velocity

FB

FE

+–

0 E vB v E B

Now let it move into region with magnetic fields only• Particle bends due to cyclotron

motion• Measure final position• Allows you to determine m/q

mv qRB m RB

q v

2m RB

q E

detector

An electron has a velocity of 1.00 km/s (in the positive x direction) and an acceleration of 2.001012 m/s2 (in the positive z direction) in

uniform electric and magnetic fields. If the electric field has a magnitude of strength of 15.0 N/C (in the positive z direction),

determine the components of the magnetic field. If a component cannot be determined, enter 'undetermined'.

q F E v B

31 12 2

19

9.1094 10 kg 2.00 10 m/sˆ

1.602 10 C

E v B k

ˆ ˆ15.0 11.37 k v B k

q

FE v B

m

q

a

ˆ26.37 v B k

3

ˆ ˆ ˆ

ˆ26.37 det 10 0 0

x y zB B B

i j k

k 3 ˆˆ10 y zB B k j

.02637 TyB

0zB

undeterminedxB

The Hall effect• Consider a current carrying wire in a magnetic field

• Let’s assume it’s actually electrons this time, because it usually is I

• Electrons are moving at an average velocity of vd

• To the left for electrons• Because of magnetic field, electrons feel a force upwards• Electrons accumulate on top surface, positive charge on bottom• Eventually, electric field develops that counters magnetic force

vd

FB

0 H dq F E v B H d E v B• This can be experimentally measured as a voltage

t

H HV E d

d

I AJ dtd nqvd

Iv

tdnq

IBd tdnqH

IBV

tnq

V

dv Bd

Force on a Current-Carrying Wire• Suppose current I is flowing through a wire of cross sectional area A

and length L • Think of length as a vector L in the direction of current• Think of current as charge carriers with charge q and drift velocity

vd

BdNq F v B dVnq v B

AL J B I L B

I

LI F L B

F

A wire carries current in the plane of this paper toward the top of the page. The wire experiences a magnetic force toward the right edge of the page. The direction of the magnetic field causing this force is which of the following?

JIT Quick Quiz 29.3

(a) in the plane of the page and toward the left edge (b) in the plane of the page and toward the bottom edge (c) upward out of the page (d) downward into the page

Ans c

Ex- (Serway 29-39) A wire having a mass per unit length of 0.500 g/cm carries a 2.00 A current horizontally to the south. What are the direction and magnitude of the minimum magnetic field needed to lift this wire vertically upward?

Solve on Board

Warmup 13

Force/Torque on a Loop• Suppose we have a current carrying loop in a constant magnetic field

• To make it simple, rectangular loop size L W

I

L

W

I F L B

• Left and right side have no force at all, because cross-product vanishes

• Top and bottom have forcesˆ

t ILBF k

Ft

BFb

ˆb ILBF k

• Total force is zero• This generalizes to general geometry

I d F s B I d s B 0

• There is, however, a torque on this loop τ r F ˆ

tW j F ˆ ˆILWB j k ˆIAB i

t

IAB

Torque on a Loop (2)

This formula generalizes to other shapes besides rectangles?• It is true for circular loops, or oddly shaped loops of current

• What if the loop were oriented differently?• Torque is proportional to separation of forces

F

F

n̂

WWsi

n

sinFW sinBIWL

Edge-on view of Loop

sinBIA

ˆIA τ n B

B

RI

Torque and Energy for a Loop• Define A to be a vector perpendicular to the

loop with area A and in the direction of n-hat• Determined by right-hand rule by current

• Curl fingers in direction current is flowing• Thumb points in direction of A

• Define magnetic dipole moment of the loop as

ˆIA τ n B

I τ A B

Iμ A τ μ B

AˆAA n

• Torque is like an angular force• It does work, and therefore there is energy associated with it

sinB

U d sinB d cosB

U μ B

• Loop likes to make A parallel to B

Edge-on view of Loop

A

B

JIT Quick Quiz 29.4(i) Rank the magnitudes of the torques acting on the rectangular loops shown below, from highest to lowest. All loops are identical and carry the same current. (ii) Rank the magnitudes of the net forces acting on the rectangular loops shown below, from highest to lowest.

(i) c>b>a (ii) a=b=c

CT – 2 A rectangular loop is placed in a uniform magnetic field with the plane of the loop perpendicular to the direction of the field. If a current is made to flow through the loop in the sense shown by the arrows, the field exerts on the loop:

A. a net force. B. a net torque. C. a net force and a net torque. D. neither a net force nor a net torque.

Ans D

CT - 3 A rectangular loop is placed in a uniform magnetic field with the plane of the loop parallel to the direction of the field. If a cur-rent is made to flow through the loop in the sense shown by the arrows, the field exerts on the loop:

A. a net force. B. a net torque. C. a net force and a net torque. D. neither a net force nor a net torque.

Ans B

+–How to make an electric motor

• Have a background source of magnetic fields, like permanent magnets• Add a loop of wire, supported so it can spin on one axis• Add “commutators” that connect the rotating loop to outside wires• Add a battery, connected to the commutators

A

• Current flows in the loop• There is a torque on the current loop• Loop flips up to align with B-field• Current reverses when it gets there

I τ A B

F

F

• To improve it, make the loop repeat many times

IN τ A B