chapter 26 sources of magnetic field. biot-savart law (p 614 ) 2 magnetic equivalent to c’s law by...

Chapter 26

Sources of Magnetic Field

Biot-Savart Law (P614)

2

Magnetic equivalent to C’s law by Biot & Savart

. P

Idl

r0

3

4

Idl rdB

r

Magnetic field due to an infinitesimal current:

Permeability of free space 70 4 10 /T m A

:r

position vector from to the field point PIdl

B due to full current

3

. P

Idl

r

03

4

Idl rdB

r

1) Magnitude: 02

sin

4

IdldB

r

2) Direction: right-hand rule

B

3) Total magnetic field due to a full current

03

4

Idl rB

r

Notice: directions

A straight current

4

Example1: Magnetic field of a straight current.

Pa

.

I

x

Idxr

o

x

Solution: Infinitesimal current

02

sin

4

IdxdB

r

/ sin r a

2

sin

addx

cot x a 0 sin

4

Id

a

2

1

0 sin4

IB d

a

1

2

01 2cos cos

4

I

a

5

Discussion:

0 2

IB

a

Pa.

I

01 2 cos cos

4

IB

a

1) Infinite straight current

0

2

Ea

2) Magnetic field at point A or B:

. A

. B

03

04

Idl rdB

r

0B dB

No field!

Square loop

6

Solution:

Example2: of a square loop at O, A and P.B

l

I.O

A

l

. P

01 2 cos cos

4

IB

a

o o0 cos 45 cos135 4 / 2O

IB

l

4

o o0 cos 45 cos90 24A

IB

l

BP can also be obtained by the formula .

Force between parallel wires (P605)

7

0 11

2

IB

d

Long parallel straight wires with current I1 and I2

I1 I2

d

Magnetic field due to I1 :

1 2F B I LAmpere force on I2 :

0 1 21 2 2

I IFB I

L d

F per unit length:

Operational definitions of Ampere & Coulomb

Circular current

8

Magnetic field of a circular current on the axis.

02

sin4

IdlB

r

o

R

P xx.

02

4

IdldB

r

r

Idl

dB

02

sin2

4

IR

r

20

32

IR

r

20

2 2 3/22( )

IR

R x

From the symmetry:B

9

Discussion:2I R

oR

P xx.

20

2 2 3/2

2( )

IRB

R x

1) Magnetic dipole moment

02 2 3/2

2 ( )

BR x

3

02

pE

x0

3 ( )

2x R

x

2) B at the center of a circular / arc current:

0 2

IB

R

2

l

R .

R

Io.

R

Io

Combined currents

10

Example3: Magnetic field at point O.

(a)0

2 2

IB

a

0

2( ) 2

I

a b

.a

b

I

o

(b) Uniform conductor ring

.I

R o 1 1 I l

2 2 I l

0 1 0 21 2 2 2 2 2

I Il lB

R R R R

01 1 2 22

( )4

I l I lR

0

1 21 2( )

l lI I

S S

Rotating charged ring

11

Question: A uniformly charged ring rotates about z axis. Determine the magnetic field at the center.

2

20

30

2 22

dr Q

BR

20

2 2 3/ 22( )

IRB

R x

2 202 0

sin8

Qd

R

0

8

Q

R

r

z

.o

R

Q

Magnetic flux

12

1) Uniform field:

ΦB : Magnetic flux through an area

∝ number of field lines passing that area

B B S

2) General case: B B dS

3) Closed surface:

B dS

→ net flux

Unit: Wb (Weber)

Gauss’s law for magnetic field

13

The total magnetic flux through any closed surface

is always zero.

→ Gauss’s law for magnetic field

0B dS

Closed field lines without beginning or end

0B

Ampere’s law (1)

14

0 inB dl I

1) Magnetic field is produced by all currents

Ampere’s law

The linear integral of B around any closed path is

equal to μ0 times the current passing through the

area enclosed by the chosen path.

2) The closed integral only depends on Iin

3) How to count the enclosed current?

15

The sign of enclosed current: right-hand rule

l

I1

I2

I31 2inI I I

l

I

I

2inI I I

1) Circular path around infinite straight current

Ampere’s law (2)

2B dl B r

0I 0

2

IB

r

r

16

0

2

IB

r

2) Any path around the current in same plane

dl

Bdr

0 cos2

IB dl dl

r

0

2

Ird

r

0

2

Id

Ampere’s law (3)

B

dr

dl

B

dl

3) Any path that encloses no current

17

Magnetic field is not a conservative field

Equations for E & B field

0 inB dl I

Maxwell equations for

steady magnetic field &

electrostatic field

0 0

0

/

0

0

E

E

B

B j

0 B j

Applications of Ampere’s law → symmetry

Cylindrical current

18

Example4: The current is uniformly distributed over the cylindrical conductor. Determine (a) magnetic field; (b) magnetic flux.

Solution: (a) Symmetry of system?

R

I

r

2B dl B r

0 inI

0 : 2

Ir R B

r

2 r

20 : 2

r R B j rr

0

2

jr

2( )

Ij

R

19

(b) Magnetic flux through the area:

R

I

2R

l

0 : ;2

Ir R B

r

0 02

: 2 2

jr Irr R B

R

B B dS

20 0

20

2 2

R R

R

Ir Ildr ldr

R r

ln24 2

o oIl Il

B

= BdS

Coaxial cable

20

Question: Coaxial cable: a wire surrounded by a cylindrical tube. They carry equal and opposite currents I distributed uniformly. What is B ?

×

×

×

×

I

·

··

R3

oI R1

R2·

01 2 :

2

IR r R B

r

3 : 0r R B

1 2 3 or ?r R R r R

21

Solenoid: a long coil of wire with many loops

Solenoid

0 in

abcd

B dl I

0 BL NI 0 0 N

B I nIL

→ Uniform

22

Toroid: solenoid bent into the shape of a circle

Toroid

N loops, current I

Outside: B = 0

Inside?

r

0

2

NIB

r

Nonuniform field

r becomes solenoid again

23

*B produced by moving charge

03

4

Idl rdB

r

F qv B

dF Idl B

Equations about Ampere force & Lorentz force:

03

4

qv rB

r

Magnetic field produced by a single moving charge

B created by rotating charge? .

Q

R

v

24

Increase the magnetic field by ferromagnetics

*Ferromagnetic

0 0B nI 0 mB K B nI

Km: relative permeability

μ : magnetic permeability

Hysteresis:

Electromagnet

“Hard / soft”