Chapter 20 The Production and Properties of Magnetic Fields

Main points

•Biot–Savart law:

•Ampère’s law:

•Gauss’ law for magnetism:

What is the source of magnetic fields, if there is no magnetic charge?

Answer: electric charge in motion!

bar magnet

current in wire

moving point charge

Therefore, understanding source of field generated by bar magnet lies in understanding currents at atomic level within bulk matter.

20-1 The Biot-Savart Law

The electric field:The magnetic field ：

20

ˆ

4 r

rlIB

d

d

Permeability of Free Space 27

0 AN104

?

the magnetic field of a straight section of wire is given by the Biot–Savart law:

It is a defined value

The magnitude of the field ：

The direction of the field is given by the cross product.

B

P

lI

dr

Example ：

lI

d

lI

d

P

r B

r B

r

B lI

d r

0B

20

ˆ

4 r

rlIB

d

d

Using the Biot-Savart Law

lI

d r B

a

IFind the magnetic field at a point

P located a distance a from a

straight wire carrying a current

I. P

Field around a current-carrying wire:

we integrated dl over the whole wire

20

ˆ

4 r

rlIB

d

d

?

I

alI

d r

P

l

1

2

O

Magnitude:

Direction: determined by a right-hand rule

Field around a current-carrying wire

The magnetic field lines are circles around the wire

a right-hand rule

(1) For an infinite long wire

)cos(cos4 21

0

aI

B

aI

B

2

0B

(2) A very long wire carrying a current I is bent into the shape shown in figure

P

a

I

1

2

B

r

Discussion

I

1

2

PThe direction of the field is given by the right-hand rule

I 1

2

3R

O

Example Consider the wire shown in the figure below ， find the magnetic field at a point O .

Solution

The magnetic field at point O is the sum of three fields

They are directed out of page

lId

ACTA current I flows in the +y direction in an infinite wire; a current I also flows in the loop as shown in the diagram.

– What is Fx, the net force on the loop in the x-direction?

(a) Fx < 0 (b) Fx = 0 (c) Fx > 0

I

I

o

y

x

You may have remembered that the net force on a current loop in a uniform B-field is zero, but the B-field produced by an infinite wire is not uniform!

Example Two parallel, infinitely long wires, separated by a distance a, carry parallel currents I1 and I2 , find the force exerted by one long, straight current on another.

a

2I1I

1B21f

Solution :

The magnetic field of wire 1 at the position of wire 2 is

12f

The force on per unit length of wire 2 is

Similarly, the force on per unit length of wire 1 is

attract

(1)The force between two currents is in accordance with Newton’s third law

Discussion

Currents in the same direction attractCurrents in the opposite directions repel

(2) Defining the ampere

One ampere is the current which, following through two infinitely long, thin, straight wires placed one meter apart in a vacuum, produces a force of newtons on each meter length of wire

This definition gives

(3) The force between two current element

22dlI

11dlI

12r12f

21f

Similarly,

The force between two current element doesn’t abide by Newton’s third law.

(4) Two infinite line charges of linear charge density 1 and 2 move along the length direction with velocity of v1 and v2. Find the forces between the two wires. 1v

2v

1

2

a The current of the two wires is

The electrical force on per unit length of the wire is

00

1

c

Commonly, the magnetic force is much less than the electrical force.

The magnetic force on per unit length of the wire is

Example The figure below shows an infinite straight wire and a current loop. Find the torque on the loop.

x

Solution 1

Use

x

Solution 2

Use

I

b

P

r

x

y

O

xd 1

Field around a long current-carrying flat :

Break the flat into strips of width dx

(1) (2) (3)

y

b

b

IBp 2

tan 10

(1)

(2)

i021

magnetic shield

i i

Discussion

Infinite Straight Wire

Infinite Current Sheet

P

x

R

xO

Find the magnetic field at a point P located a distance x from the center along the axis of a circular current loop of radius R and current I.

lI

dB

d

B

d

B

r

Px

I

ixR

IRB

2/322

20

)(2

The direction of the field is given by the right-hand rule.

Field due to a current loop :

(1) At the center of the circular loop

If the circular loop has N turns

Discussion

(2) The magnetic field at the center of the arc segment

I

ixR

IRB

2/322

20

)(2

B

I

(3)

2/322

20

)(2 xRIR

B

A

nnIA

Define

30

2 xB

Magnetic dipole moment

Compare with electric dipole

The magnetic field of a magnetic dipole

The electric field of an electric dipole

The magnetic field lines of a circular loop of a current and a bar magnet

Solution

Example: A nonconducting disk of radius R has charge q and rotates with angular velocity about its axis, find the magnetic field at a point on the axis of the rotating disk.

x

OR

q

P

r

B

d

At the center of the disk

Example A hydrogen atom may be described as consisting of an electron that moves in a circular orbit around a proton. The force that gives rise to the motion is the Coulomb attraction between the proton and the electron, which have charges respectively, where .The motion is further constrained by the requirement that the angular momentum has the value where n is an integer and , Planck’s constant. Calculate the magnitude and direction of the magnetic field at the location of the proton. What is the magnetic moment of the current loop?

SolutionThe Coulomb attraction provides the centripetal force

The constraint on the angular momentum

The magnetic field at the location of the proton

Perpendical to the orbit

The magnetic moment

Perpendical to the orbit

I

PR

A solenoid of radius R has n turns in per unit length, with each turn carrying a current I .

ldl

B

d

r

Magnetic field in a solenoid :

A solenoid is a length of wire coiled into a cylinder

PR

l

B

d

r 21

(1) For a long solenoid

nIB 0

Discussion

(2) If the origin is at one end of the solenoid

nIB 02

1

20-2 Gauss’ Law for Magnetism

Electrostatic field ：Magnetic field ：

Magnetic flux

In the Magnetic field, the number of Magnetic field lines crossing a curved surface S is defined as magnetic flux.

ABB

dd

For a curved surface

If the magnetic field lines enter the surface

For a closed surface

If the magnetic field lines leave the surface

SI: weber (Wb)

• No magnetic monopoles (single magnetic charge) have ever been observed

• Magnetic field lines must be continuous

• The magnetic “charge” inside any closed surface must always be zero.

Gauss’s Law For Magnetism

0d SB AB

The magnetic field lines must be continuous

20-3 Ampere’s Law

Ampere’s Law

Electrostatic field ： conservative field

Magnetic field:

Take a line integral over the magnetic field along a circular path C

Path C circles a current-carrying wire

ld

Current I is through the surface whose edge is defined by the path

Path C’ consists of C1 and C2

Take a line integral over the magnetic field along a closed path C’(C1+C2)

In this case, there is no current through the surface whose edge is defined by the path

In the static magnetic field, the path integral of the total magnetic field around a closed path is equal to times the algebraic sum of the currents threading the path.

Ampere’s law enclosedLIμlB 0d

Integral around a closed path Current “enclosed” by that path

(2) The enclosed current can be positive or negative contribution.

Discussion

Curl your right hand around the closed path, with the fingers pointing in the direction of integration. A current through the path in the general direction of your outstretched thumb is assigned a plus sign, and a current generally in the opposite direction is assigned a minus sign

(1)Ampere’s Law applies to an arbitrary closed path The direction of the closed path is arbitrary

(3) Ampere’s law is valid only if the currents are steady.

aI

L

A finite current segment, if

1

2

For example

The integral of the magnetic field around the closed path L is

Using Ampere’s law to find the magnetic field

Ampere's Law can simplify the calculation if there is symmetry of the current! (and the magnetic field )

• Choose a closed path with direction

• Evaluate line integral in Ampere’s Law (The integral is simple)

• Calculate the current enclosed by the path

• Apply Ampere’s Law

Example: A long, hollow, cylindrical surface of radius R carries a current I that is uniformly distributed over the surface. Find the magnetic field both outside and inside the cylindrical surface.

R

I

rP

LSolution Because of the high degree of symmetry, we know that is tangent to the circle of radius r about the axis and constant in magnitude everywhere on the circle.

PId'dI

B

d 'dB

B

rI

B

2

0

0B

ACT For a long cylinder that current is uniformly distributed over the cross-sectional area, find the magnetic field.

R

I

Example Find the magnetic field of an infinite sheet of current, the current per unit length (along the direction which is perpendicular to the current) is i .

Solution i

B

'B

P ab

c d

If the thickness of the sheet is dj is the current density

jd• outside the sheet

• inside the sheet

x

Choose the rectangle abcd as the integral path.

Example Find the the magnetic field of a long ideal solenoid carrying a current i.

Solution

The field inside is parallel to the cylinder axis

The field outside is insignificant

1l

2l

S

 The Magnetic field lines for a real solenoid of finite length. The field is strong and uniform at interior points such as P1 but relatively weak at external points such as P2.

o

I

N

Solution

• If the cross-sectional width is much less than the radius of the toroidal solenoid ,

r

• outside the torus

Example Find the magnetic field of a toroidal solenoid

The Magnetic field within the torus must be parallel to the walls

o

I

N

h

1R

2R A

rd

If the cross-sectional surface is rectangle with width h.Find the magnetic flux in it.

Example Figure below shows a cross section of a long cylindrical conductor containing a long cylindrical hole .The axes of the cylinder and the hole are parallel and are a distance d apart. The current is uniformly distributed in the conductor. The current density is Find the magnetic field in the hole.

O1 O2

d

Solution

Regard the cylindrical hole as resulting from the superposition of a complete cylinder (no hole) carrying a current in one direction and a small cylinder carrying a current in the opposite direction, both cylinders having the same current density .

OPr

The magnetic field in a cylinder

O1 O2

d1r

2rUse superposition

0d SB AB

0 B

Ampere’s law

enclosedLIμlB 0d

Gauss’s Law For Magnetism

Bj

0

1

Gauss equation

Stokes equation

20-4 The Magnetic Field of Moving Point Charges

lI

d

P

r

lI

d+q

A

—— The magnetic field of a moving charges

30

4 r

rqB

v

The Electric and Magnetic Field of Moving Point Charges

•q

v

E

Br

Ec

EB

vv 200

1

•

Oa

b

Example A closed circuit consists of two semicircles of radii a and b that are connected by straight segments as shown in the figure, the linear charge density is (>0) ， and rotates with angular velocity about O. Find the magnetic field at point O .

1

2

34

qd

d

Segment 1 ：

v

Solution

Segment 2 ：

Oa

b

1

2

34

qd

dvv

B

d

Segment 3 ：

Segment 4 ：

qd