1

Chapter 28

2

Gravitational and Electric Fields

rr

mGg ˆ

2

rr

qkE ˆ

2

We expect that B will1. have a 1/r2 dependence2. Directed along the radius

Recall F= iLxB and F=qE and F=mg Based on the pattern

rr

iB ˆ

2

3

Broken Symmetry

But those other fields diverge and B does not

Based on our experience with bar magnets, B must return to the magnet

4

Biot-Savart Law

Point P

X

X X

X

X

X

B into page

i ds

1. B goes into the page2. i ds and r are both

perpendicular to B3. B proportional to i ds x r4. But needs to be 1/r2 so

32 4

ˆ

4 r

rsdiBdor

r

rsdiBd oo

5

A new constant, 0

Called the permeability of free spaceValue = 4 x 10-7 T*m/AAnd yes, 0/ 410-7 T*m/A

6

Modified Biot-Savart

If a charged particle has a constant velocity, v, then I can modify Biot-Savart:

2

ˆ

4 r

rvqB o

7

Solving Biot-Savart Problems

Biot-Savart problems are typically one of geometry

You must integrate about the limits of the current loop while evaluating the cross-product

Sometimes, ds and r are parallel, which eliminates the contribution. You should examine the problem to see where this is true.

Sometimes, ds and r are perpendicular, which forces the cross-product to its maximum value.

Mostly, the ds and r are related only through sin . This means that you may have to create an integrand over the angle, not necessarily the length of the current loop.

2

ˆ

4 r

rsdiBd o

In conclusion, Biot-Savart problems are strongly dependent on creating the appropriate geometry over which to integrate.

8

Magnetic Field Due to a Long, Straight Wire

22 Rsr

Point P

ds

s

R

r

i

Direction of current

X

X

XX

X

Direction of B

22)sin(sin

Rs

R

9

Integrating

R

i

Rs

s

R

iB

Rs

Rdsi

Rs

ds

Rs

RiB

r

dsidBdBB

oo

oo

o

22

22

sin

22

02

122

0 23

220

2222

02

0

10

Magnetic Field due to Current of a Circular Arc of Wire

R

dsr

B out of page

r always perpendicular to ds so cross-product is ds

R

iB

dR

iB

RddswhereR

dsiB

r

dsidBB

o

o

o

o

4

4

4

sin

4

0

2

2

11

A Complete Loop

We can take the previous result and evaluate it at =2

The result is the same as your text on pg. 1077 Eq. 28-17

R

i

R

iB oo

24

2

aRwherea

iB o

2

12

Force Between Two Long Wires with Parallel Currents

zd

iB aoa ˆ

2

ia

ib

d

L

XX

XX

Force on b caused by the magnetic field of a

Fa on b =ibL x Ba

d

Lii

d

iLiF baoaobbona

22

By RH rule, Fa on b is towards a

13

By Symmetry

Fb on a is equal and opposite to Fa on b

So the two wires with parallel currents are attracted to one another

If I reverse the current on b (anti-parallel), then the forces generated by the wires will repel one another.

Parallel currents attract; anti-parallel currents repel.

14

Derivation of Ampere’s Law

IRR

IRBdsBsdB o

o )2(2

)2(

Consider a distance R from a wire carrying current I. Now consider all of the points which are a distance R from the wire. They form a circle of circumference 2R. Now we evaluate the closed loop integral at this point.

So the circle that we made is called an “Amperian Loop”.

enclosedoIsdB

15

Ampere’s Law

Ampere’s Law says that the magnitude of B is proportional to the net current enclosed within the Amperian Loop.

Amperian loops do not have to be circles. They could be rectangles but a circle is usually more convenient to integrate. The only rule is that they have to be closed!

enclosedoIsdB

16

What is the direction of the magnetic field?

We have an agreed convention:Curl your fingers around the Amperian loop

with your fingers curling in the direction of the integration. If the current is in the same direction as your thumb, then current is assigned a positive value. If your thumb is opposite the direction of the current, then the current is assigned a negative value.

17

Magnetic Field of Long, Straight Wire Revisited

R

IB

IRBRdBsdB

IsdB

o

o

enclosedo

2

)2(2

0

I

R

Integrate from 0 to 2

In this case, the enclosed current is defined to be in the negative direction.

18

Differential Version of Ampere’s Law

Recall sdvAdv

enclosedo

enclosedenclosed

enclosedo

JB

so

AdJI

and

IsdBAdB

A current density creates a steady magnetic field which circulates around the current density

19

Solenoid

Solenoid is a number of coils packed tightly together.

It solves an incredibly tough problem. Consider: It is easy to

make a uniform E-field. All you need are two parallel plates. But how do you make a uniform field out of a bunch of circles (i.e. a B-field).

20

Finding the magnetic field of a solenoid

iNB

iNLBL

so

iNLI

BbecausesdB

sdBsdB

BLsdB

sdBsdBsdBsdBsdB

o

o

enclosed

d

c

a

d

c

b

b

a

a

d

d

c

c

b

b

a

00

0

The solenoid is characterized by the number of turns per unit length, NThe magnetic field outside of the solenoid is so weak that it is consider equal to zero

21

Toroid– A circular solenoid

Consider a circular solenoid of radius, R

R

iNB

iNRB

so

iNI

RBsdB

o

o

enclosed

2

)2(

)2(

Current into page

Current out of pageAmperian

Loop

22

Bohr Magneton

In the last chapter, we discussed how atoms could be thought of as current loops

I=e/T where T=2r/v v=velocity of the electron r= radius of the electron’s orbit

If =IA then =((ev)/2r)*r2

=evr/2 Recall for an orbiting point, the angular

momentum, L, is equal to mvr So =eL/(2m)

23

Quantization of angular momentum

Each orbital is an integer value of Planck’s constant, h divided by 2 =h/ 2L=n

=e /2mCalled the “Bohr Magneton” Represents smallest amount of dipole

moment possible

24

Magnetic Materials

This represents any material. The arrows indicate the direction of the individual magnetic moments of the atoms. As you can see, their orientation is random and the vector sum of them is zero.

This represents a permanent magnet. The magnetic moments or dipoles have a net vector sum point to the left.

25

Magnetic Materials in an External Magnetic Field

Bexternal

Let’s turn on an external B which points to the left

Before

Possibility 1: The dipoles align with the external field

Possibility 2: The dipoles align to oppose the external field

26

The degree to which the magnet moments align or oppose the magnetic field determines their classification

Ferromagnetic—Strongly aligns with the magnetic field

Paramagnetic—Weakly aligns with an external magnetic field

Diamagnetic—Weakly opposes the external magnetic field

For any of these cases, we define the total magnetic field as B=Bexternal + B

27

B represents the magnetic field created by the alignment/opposition of the dipoles

Bm*Bexternal

mis called the magnetic “susceptibility” of the material

Paramagnetic materials and their susceptibilities Al 2.2 x 10-5

Cu 1.4 x 10-5

Air 3.6 x 10-7

Diamagnetic materials and their susceptibilities Bi -1.7 x 10-5

Ca -2.2 x 10-5

H2O -9.1 x 10-6

Ferromagnetic materials have susceptibilities from 103 to 106

28

--The Permeability of a Material

B=Bexternal + m*Bexternal

B=Bexternal (1+ m Relative permeability, Km is defined as

Km= 1+ m

Actually, we could replace with a new “” called the permeability of the material. = Km

29

Hysteresis

Let’s say you had a ferromagnet and you measured its magnetic field

You applied an external B-field and the dipoles are aligned with the field

Now you remove the field but some of the dipoles get “stuck” in their new position.

Now you measure the magnetic field of the ferromagnet and find that it is different.

You repeat the process and yet, you never get back to your original value as more or less of the dipoles stick or unstick in their new positions.

This behavior is called hysteresis.