PHYS 2421 - Fields and Waves

Consider a charge moving in a magnetic field

Since F is

perpendicular

to v, the force

leads to a

circular motion

B field

into plane

F=ma acceleration change of direction of velocity

0

2

cos90mv

RqB

F qvB qvB

vF m

RCentripetal

Radius of

circular motion

Take F as

centripetal force:

Angular velocity:q Bv

R m

Homework :

Problem 27.15 and 27.21 (both 11th and 12th Eds.)

Radiation is produced by oscillating electrons

Angular frequency of electrons = radiation frequency

22

q B fmf B

m q

6 31

19

2 2450 10 9.11 1020.0877

1.6 10

fmB

q

Hz kg T

C

Summary of Section 27.4

Hmwk Sect. 27.4: Probls 27.15 and 27.21 (11th and 12th Eds.)

mvR

qB

q Bv

R m

Charge in a magnetic field move in circular path

Radius:

Angular frequency:

Charges pass by an area

with a B and an E crossed

The E field

deflects +

charges down

The B field

deflects +

charges up

Particles will go through undeflected if Fe and Fm balance each other

m eF F qvB qE

Ev

B

Only those particles with v = E/B will go through undeflected

Crossed B and E fields act as a velocity selector

2

2

2

2

1 1

2 2

2

EKE eV mv m

B

e E

m VB

Hmwk : Probl. 27.27 (11th Ed.) or 27.25 (12th Ed.)

Velocity selectors are used in

analog TV screens

He received the Nobel prize for discovering that the electron was a particle

J.J. Thomson

used a velocity

selector to study

the electron:

• Accelerated electrons through a potential V

• Sent them through a v-selector with E and B

adjusted to let them to go undeflected

• Calculated the ratio of the charge to the mass

of the electron

His son received it for discovering that it was a wave . . .

Homework : Problem 27.31 (both 11th or 12th Eds.)

2

21 1;

2 2

E EKE eV mv m v

B B

231 62

19

9.1 10 6 100.6825 0.826

2 2 1.6 10 150

mEB

eV

kg N/C T

C V

Summary of Section 27.5

Hmwk Sect. 27.5: Problems 27.27 and 27.31 (11th Ed.) or 27.25 and 27.31 (12th Ed.)

Ev

BVelocity selector

2

22

e E

m VBRatio of charge to mass of electron

Consider charges moving inside a conductor

placed in a magnetic field B

Each charge feels a forcedF qv B

Magnitude:

Direction: to the left

Number of charges

in length of wire nlA= n = density of charges

A = cross sectional area

Now obtain the total force on the whole length of wire

Total force in

length of wire d dF nAl qv B nqv A lB IlB

dF qv B

Vd = drift velocity

Force is to the left

For cases in which B is not to perpendicular to wire

F Il B sinF IlB

Some examples

For a segment of a wire

dF Idl B

Hmwk: Probl. 27.34 (11th Ed.) or 27.36 (12th Ed.)

Solution: a) b) and c): 7.06x 10-3 N

Force on wire?0sin 50 1 1.2 sin45

42.4

F IlB A m T

N

Again using vectors:0 0

0

0 0

ˆ ˆ ˆcos45 sin 45

ˆˆ ˆ

ˆ ˆ0 0 sin 45 42.4

cos45 sin 45 0

F Il B I li B i B j

i j k

I l IlB k k

B B

N

Force pointing out of the plane

Find the force on the

three segments of wire

Segment 1 Segment 2

Segment 3: length

L into the plane

Segment 1:

Segment 2 is more interesting . . .

0 ˆ ˆsin90F ILB j ILBj

Segment 3: 0sin180 0F IlB

Homework : Problem 27.35 and 27.39 or

27.37 and 27.39 (12th Ed.)

Segment 3:

• x components of force cancel

• net force will be in the y direction

0 0

0 0

0

0

180 180

0 0

180

0

sin

sin 2

y yF dF IB dl

IBR d IRB0 0

0 0

0

0

180 180

0 0

180

0

cos

cos 0

x xF dF IB dl

IBR d

Total force on all three segments: ˆ ˆ ˆ2 2F ILBj IRBj IB L R j

Summary of Section 27.6

Hmwk Sect. 27.6: Probls 27.34, 27.35 and 27.39 (11th Ed.) or

27.36, 27.37 and 27.39 (12th Ed.)

Forces on current carrying conductors

F Il B dF Idl B

Add forces

Consider a current

loop in a B field

perpendicular to

current

• x components of force cancel

• y components of force cancel

• net force on loop will be zero

Now consider a loop tilted respect to the B field

Again

• x and y components of force cancel

• net force on loop will be zero

• but there will be a torque on loop

Forces in x and y cancel

Forces in x form a couple

F Il B F IaB

Magnetic dipole moment is:

Loop tilted respect to the B field

Segments in y direction feel forces:

in ± x

sinbSeparation between these forces is

and torque is then:

sin sin sin sinFb IaB b IAB B

IA

Or in vector form: B

r F

Valid for all geometries

Some examples

What are the directions of and ?IA B

A current loop in a B field has potential energy:

cosU B B

Torque, method 1:0

1.18 1.2 sin 90 1.41sin 2

Am T NmB

2

30 5 0.05 1.182

Total A m AmNIA

Torque, method 2:2 0

30 5 1.2 0.05 sin 90 1.41sin A T m NmNIBA

Torque tends to rotate

Magnetic moment:

Homework:

Probls 27.42 and 27.44 (both 11th and 12th Eds.)

Soln. 27.42: a), b) and c): 7.06x 10-3 N

Homework : Problem 27.47 (both 11th and 12th eds.)

Initial potential energy: 0cos90 0InitialU BFinal potential energy:

0cos0 1.18 1.2 1.412

Final Am T JU B

Change in potential energy:

1.41Final Initial JU U

If for any loop in a uniform B field

• The forces in x and y cancel

• Forces form a couple that produces only a torque

Then, how do magnets attract or repel?

• Magnets affect atomic loops in materials

• B field of magnets is not uniform

• Forces produced are not zero

How do magnets work?

Summary of Section 27.7

Hmwk Sect. 27.7: Probls 27.42, 27.44 and 27.47

(both 11th Ed and 12th Eds.)

In loops :

•Forces in x and y cancel

•Forces in x form a couple

•Loops feel a torque

Magnetic dipole moment is: IA

B

cosU B BPotential energy:

Magnets work by exerting forces on atomic loops

in materials thanks to their not uniform B field

PHYS 2421 - Fields and Waves

PHYS 2421 - Fields and Waves