Sources of Magnetic Field Chapter 28

• Study the magnetic field generated by a moving charge

• Consider magnetic field of a current-carrying conductor

• Examine the magnetic field of a long, straight, current-carrying conductor

• Study the magnetic force between current-carrying conductors

• Consider the magnetic field of a current loop• Examine and use Ampere’s Law

1

The magnetic field of a moving charge

• A moving charge will generate a magnetic field relative to the velocity of the charge.

2

3

Magnetic Field of a Moving Charge)

20

sin

4 r

vqB

(28-1)

70 104

70 104 x

Permeability of free space

Direction of B determined by rxv

Magnitude of B

The vector form:

Force between two moving protons• Two protons moving at the same velocity

(much less than speed of light) in opposite directions.

• The electric force FE is repulsive.

• The right-hand rule indicates the magnetic force FM is repulsive. (i x k=-j)

• Find the ratio of the magnitude of the forces.

2 2

02

2

2

4B

B

E

q vF

r

F v

F c

The ratio of the two forces. Where c=speed of light. Therefore:FE>>FB

5

Magnetic Field of a Current Element

Figure 28-3

Let dQ = charge in wire segment dl

Let A = cross section area of wire segment dl

Let n = charge density in wire segment dl

dQ = nqAdl

20

20 sin

4

sin

4 r

Adlnqv

r

dQvdB dd

I = nqvdA

20 sin

4 r

IdldB

(28-

5)Biot-Savart Law

Direction of dB determined by rxlId

Vector form of Biot-Savart Law

Total magnetic field of several moving charges = vector sum of fields caused by individual charges

Magnetic field of a straight current-carrying conductor

• Biot and Savart contributed to finding the magnetic field produced by a single current-carrying conductor.

6

7

Magnetic Field of a Current-Carrying Conductor

Figure 28-5

22)sin(sin

yx

x

dydl

20 sin

4 r

IdldB

2/3220

22220

)(4)(4 yx

dyIx

yx

x

yx

IdydB

a

a yx

dyIxB 2/322

0

4

22

0

2220 2

4

1

4 ax

a

x

I

yx

y

x

IxB

a

a

If a x

x

I

a

a

x

IB

2

2

40

2

0

Based upon symmetry around

the y-axis the field will be a circle

8

Magnetic Field of a Current-Carrying Conductor

Figure 28-6

where r = perpendicular distance from the current-carrying wire.

9

Force between Parallel Conductors

Each conductor lies in the field set up by the other conductor

' ' ' 0

'0

'0

sin2

2

2

IF I LB I LB I L

r

IIF L

r

IIF

L r

BLIF

'Note: If I and I’ are in the samedirection, the wires attract. If I and I’ are in opposite directions, the wires repel.

See Example 28.5 Page 966

Only field due to I shown

Substitute for B

10

Magnetic Field of a Circular Current Loop

Figure 28-12

20 sin

4 r

IdldB

190sinsin 222 axr

)(4 220

ax

dlIdB

)2()(4)(4 2/322

02/322

0 aax

Iadl

ax

aIBx

By= 0

02 2 2 2 1/2

cos4 ( ) ( )x

I dl adB dB

x a x a

cos

11

Magnetic Field of a Circular Current Loop

2/322

20

)(2 ax

NIaBx

(on the axis of N circular loops) (x=0)

Figure 28-13

x

Figure 28-14

Ampere’s Law I—specific then general

Similar to electric fields if symmetry exists it is easier to use Gauss’s law

12

Ampere’s Law II

• The line integral equals the total enclosed current

• The integral is the sum of the tangential B to line path

14

Ampere’s Law (Chapter 28, Sec 6)

Figure 28-15

Irr

IrBdlB 0

0 )2(2

)2(

r

IB

2

0 rdl 2

dlBBdldlBdlB cos

IdlB 00 dlB

For Figure 28-15a

For Figure 28-15b

For Figure 28-15c

d

c

a

d

b

a

c

b

dlBdlBdlBdlBdlB

0)(2

0)(2 2

2

01

1

0

rr

Ir

r

IdlB

d

c

a

d

b

a

c

b

dldlBdldlBdlB 00 21

15

Ampere’s Law

Figure 28-16

dlBdlB cos

rddl cos

d

Ird

r

IdlB

2)(

200

r

IB

2

0

2dI

IdlB 0

0 )2(2

0 d

0 dlB

16

Applications of Ampere’s Law Example 28-9Field of a Solenoid (magnetic field is concentrated in side the coil)

Figure 28-20

Figure 28-21nLIIdlB encl 00

n = turns/meter

d

c

a

d

b

a

c

b

dlBdlBdlBdlBdlB

BLdldldldlBdlBd

c

a

d

b

a

c

b

000

nLIBL 0 Il

NnIB 00

l

Nn

where N = total coil turns l = total coil length

turns/meter

(28-23)

17

Applications of Ampere’s LawExample 28-9 Field of a Solenoid

Figure 28-22

l

Nn

where N = total coil turns l = 4a = total coil length

turns/meter

18

Applications of Ampere’s Law Example 28-10Field of a Toroidal Solenoid – (field is inside the toroid)

Figure 28-23

00 enclIdlB

00 enclIdlB

NIIdlB encl 00

Path 1

Path 3

Path 2

N turns

)2( rBdlBdlB NIrB 0)2(

r

NIB

20 (28-24)

0B

0BCurrent cancels

No current enclosed

Magnetic materials • The Bohr magneton will determine how to

classify material. • Ferromagnetic – can be magnetized and

retain magnetism• Paramagnetic – will have a weak response

to an external magnetic field and will not retain any magnetism

• Diamagnetic – shows a weak repulsion to an external magnetic field

19

Bohr Magneton- In atoms electron spin creates current a loop, which produce magnetic their own field

Ferromagnetism and Hysteresis loops

• The larger the loops the more energy that is lost magnetizing and de-magnetizing.

• Soft iron produce small loops and are used for transformers, electromagnets, motors, and generators

• Material that produces large loops are used for permanent magnet applications

0

BM