6/10/2014

1

Chapter 30 Sources of the Magnetic Field

CHAPTER OUTLINE

30.1 The Biot–Savart Law

30.2 The Magnetic Force Between

Two Parallel Conductors

30.3 Ampère’s Law

30.4 The Magnetic Field of a Solenoid

30.5 Magnetic Flux

30.6 Gauss’s Law in Magnetism

30.7 Displacement Current and the

General Form of Ampère’s Law

1

30.1 The Biot–Savart Law

Biot and Savart arrived at a mathematical expression that gives the magnetic field at some point in space in terms of the current that produces the field. That expression is based on the following experimental observations for the magnetic field dB at a point P associated with a length element ds of a wire carrying a steady current I :

2

6/10/2014

2

The total magnetic field B created at some point by a current of finite size

3

Magnetic Field on the Axis of a Circular Current Loop

4

6/10/2014

3

30.2 The Magnetic Force Between Two Parallel Conductors

parallel conductors carrying currents in the same direction attract each other, and parallel conductors carrying currents in opposite directions repel each other.

The force per unit length:

5

Example 4 Interaction between current-carrying wires

2 parallel wires A & B, carry the same size of

current.

(a) Draw a few magnetic field lines due to

the current through wire A.

A B

• •

6

6/10/2014

4

Example 4 Interaction between current-carrying wires

2 parallel wires A & B, carry the same size of

current.

(b) Mark the direction of this magnetic

field at wire B with an arrow.

A B

• •

field at B due to

current through A

7

Example 4 Interaction between current-carrying wires

2 parallel wires A & B, carry the same size of

current.

(c) Mark the direction of magnetic force acting

on B due to magnetic field of the current

through A.

A B

• •

field at B due to

current through A

magnetic force

8

6/10/2014

5

Example 4 Interaction between current-carrying wires

2 parallel wires A & B, carry the same size of

current.

(d) Deduce the direction of magnetic force

acting on A due to magnetic field of the

current through B.

A B

• •

field at B due to

current through A

magnetic force

field at A due to

current through B

9

Example 4 Interaction between current-carrying wires

2 parallel wires A & B, carry same size of

current.

(e) Do A & B attract or repel each other? A B

• •

field at B due to

current through A

magnetic force

field at A due to

current through B

They attract each other. 10

6/10/2014

6

The force between two parallel wires is used to define the ampere as follows:

The value 2 X 10-7 N/m is obtained from Equation 30.12 with I1 = I2 = 1 A and a = 1 m.

11

12

6/10/2014

7

30.3 Ampère’s Law

13

14

6/10/2014

8

Example 30.4 The Magnetic Field Created by a Long Current-Carrying Wire

15

We need to find I’

16

6/10/2014

9

A superconducting wire carries a current of 104 A. Find the magnetic field at a distance of 1.0 m from

the wire. ( µ 0 = 4∏ x 10-7 A-m/T)

1. 1. 2.0 x 10-3 T

2. 2. 8.0 x 10-3 T

3. 3. 1.6 x 10-2 T

4. 4. 3.2 x 10-2 T

17

A 2.0 m wire segment carrying a current of 0.60 A oriented parallel to a uniform magnetic field of 0.50 T experiences a force of what magnitude?

1. A. 6.7 N

2. B. 0.30 N

3. C. 0.15 N

4. D. zero

18

6/10/2014

10

A proton moving with a speed of 3 x 105 m/s perpendicular to a uniform magnetic field of 0.20 T will follow which of the paths described below? (qe = 1.6 x 10-19 C and mp = 1.67 x 10-27 kg)

1. 1. a straight line path

2. 2. a circular path of 1.5 cm radius

3. 3. a circular path of 3.0 cm radius

4. 4. a circular path of 0.75 cm radius 19

A proton which moves perpendicular to a magnetic field of 1.2 T in a circular path of 0.08 m radius, has what speed? (qp

= 1.6 x 10-19 C and mp = 1.67 x 10 -27 kg)

1. A) 3.4 x 106 m/s

2. B) 4.6 x 106 m/s

3. C) 9.6 x 106 m/s

4. D) 9.2 x 106 m/s

20

6/10/2014

11

A current carrying wire of length 50 cm is positioned perpendicular to a uniform magnetic field. If the current is 10.0 A and it is determined that there is a resultant force of 3.0 N on the wire due to the interaction of the current and

field, what is the magnetic field strength?

a) 0.6 T

b) 1.50 T

c) 1.85 x 103 T

d) 6.7 x 103 T

21

A horizontal wire of length 3.0 m carries a current of 6.0 A and is oriented

so that the current direction is 50° S of W. The horizontal component of

the earth's magnetic field is due north at this point and has a strength of 0.14 x 10-4 T. What is the size of the force on the wire?

a)0.28 x 10-4 N

b)2.5 x 10-4 N

c)1.9 x 10-4 N

d)1.6 x 10-4 N

22

6/10/2014

12

30.4 The Magnetic Field of a Solenoid

A solenoid is a long wire wound in the form of a helix

A reasonably uniform magnetic field can be produced in the space surrounded

by the turns of wire—which we shall call the interior of the solenoid—when the

solenoid carries a current. When the turns are closely spaced, each can be

approximated as a circular loop, and the net magnetic field is the vector sum of

the fields resulting from all the turns.

23

•The contributions from sides 2 and 4 are both zero, again

because B is perpendicular to ds along these paths

An ideal solenoid is approached when the turns are

closely spaced and the length is much greater than the

radius of the turns.

We can use Ampère’s law to obtain a quantitative expression for the

interior magnetic field in an ideal solenoid.

•The contribution along side 3 is zero because the magnetic

field lines are perpendicular to the path in this region.

•Side 1 gives a contribution to the integral because along

this path B is uniform and parallel to ds

24

6/10/2014

13

If N is the number of turns in the length !, the total current

through the rectangle is NI.Therefore, Ampère’s law applied

to this path gives

25

A solenoid with 500 turns, 0.10 m long, carrying a current of 4.0 A and with

a radius of 10-2 m will have what strength magnetic field at its center?

(magnetic permeability in empty space µ 0 = 4∏ x 10 -7 T-m/A)

1. A) 31 x 10-4 T

2. B) 62 x 10-4 T

3. C) 125 x 10-4 T

4. D) 250 x 10-4 T

26

6/10/2014

14

30.5 Magnetic Flux

Consider an element of area dA on an arbitrarily shaped surface, as shown in

Figure . If the magnetic field at this element is B, the magnetic flux through the

element is B. dA, where dA is a vector that is perpendicular to the surface and

has a magnitude equal to the area dA. Therefore, the total magnetic flux ΦB

through the surface is:

ΦB=0 27

28

6/10/2014

15

Magnetic Flux

Graphical Interpretation of Magnetic Flux

The magnetic flux is proportional to the number of magnetic flux lines

passing through the area.

29

Magnetic Flux

30

6/10/2014

16

A General Expression for Magnetic

Flux

ACosBAB )(

The SI unit of magnetic flux is the weber (Wb), named after the German Physicist

W.E. Weber (1804-1891). 1 Wb = 1 T.m2.

31

EXAMPLE : Magnetic Flux

A rectangular coil of wire is situated in a constant magnetic field whose

magnitude is 0.50 T. The coil has an area of 2.0 m2 . Determine the magnetic

flux for the three orientations, f = 0°, 60.0°, and 90.0°, shown below.

32

6/10/2014

17

Example 30.8 Magnetic Flux Through a Rectangular Loop

33

So

34

6/10/2014

18

30.6 Gauss’s Law in Magnetism

The magnetic field lines of a bar magnet

form closed loops. Note that the net

magnetic flux through a closed surface

surrounding one of the poles (or any other

closed surface) is zero. (The dashed line

represents the intersection of the surface

with the page.)

35

The electric field lines surrounding an

electric dipole begin on the positive charge

and terminate on the negative charge. The

electric flux through a closed surface

surrounding one of the charges is not zero.

36

6/10/2014

19

30.7 Displacement Current and the General

Form of Ampère’s Law

When a conduction current is present, the charge on the positive plate changes

but no conduction current exists in the gap between the plates.

For surface S1

For surface S2

because no conduction current passes

through S2

37

Thus, we have a contradictory situation that arises from the

discontinuity of the current! Maxwell solved this problem by

postulating an additional term on the right side

of Equation 30.13, which includes a factor called the displacement

current Id , defined as

we can express the general form of Ampère’s law (sometimes called the

Ampère–Maxwell law) as

38

6/10/2014

20

The electric flux through S2 is simply

Hence, the displacement current through S2 is

39

Example 30.9 Displacement Current in a Capacitor

40