Chapter 30

Sources of the Magnetic Field

A whole picture helps

Charge q as source Current I as source

Electric field E Magnetic field B

Gauss’s Law Ampere’s Law

Force on q in the fieldForce on or I in the filed

qv

Ampere-Maxwell Law

Faraday’s Law

Summarized in

Maxwell equations

E qF E

B qF v B

q qF E v B

Lorentz force

Math -- review

C A BVector cross product:

θ

A

B

C A B

Vector cross product:

D B A C

θ

A

B

D A B

sin

C AB θC

Determine the direction. If

Magnitude of the vector :C

BF v B

The right-hand rule:1. Four fingers follow

the first vector.2. Bend towards the

second vector.3. Thumb points to the

resultant vector.

Sources of Electric field, magnetic field

From Coulomb's Law, a point charge dq generates electric field distance away from the source:

r

2

0

1

4

dq ˆdr

E rr

dE

dq

P

From Biot-Savart's Law, a point current generates magnetic field distance away from the source:

Ids

r dB

Ids

024

Id ˆdr

sB r

Difference:1. A current segment , not a point charge dq,

hence a vector.2. Cross product of two vectors, is determined

by the right-hand rule.

Ids

dB

Total Magnetic Field

is the field created by the current in the length segment , a vector that takes the direction of the current.

To find the total field, sum up the contributions from all the current elements

The integral is over the entire current distribution

o = 4 x 10-7 T. m / A , is the constant o is called the permeability of free space

dB

24

oμ Id

π r

s rB

Ids

ds

Example: from a Long, Straight Conductor with current I The thin, straight wire is carrying a

constant current I Constructing the coordinate

system and place the wire along the x-axis, and point P in the X-Y plane.

Integrating over all the current

elements gives

Icos

I ˆsin sin

2

1

1 2

4

, or 4

θo

θ

o

μB θ dθ

πaμ

θ θ Bπa

B k

ˆ ˆˆ sin cos

2πd dx θ dx θs r k k

B

The math part

Icos ˆ ˆˆ cos

cos , tan ,cos

Icos

Isin sin

ˆ

2

1

3

2 2

1 2

4 4 4

2

4

4

o o o

θo

θ

o

μ μ I μId θd dx θdθ

π r π a πaa adθ

θ x a θ dxr θ

μB θ dθ

πaμ

θ θπa

B

sB r k k

B k

ˆ ˆˆ sin cos

2πd dx θ dx θs r k k

Now make the wire infinitely long

The statement “the conductor is an infinitely long, straight wire” is translated into:

= /2 and = -/2 Then the magnitude of the field

becomes

The direction of the field is determined by the right-hand rule. Also see the next slide.

2

IoμB

πa

For a long, straight conductor the magnetic field goes in circles

The magnetic field lines are circles concentric with the wire

The field lines lie in planes perpendicular to to wire

The magnitude of the field is constant on any circle of radius a

A different and more convenient right-hand rule for determining the direction of the field is shown

B

Example: at the center of a circular loop of wire with current I

From Biot-Savart Law, the field at O from is

B

I I I 2 2

24 4 2

o o o

full circle

μ μ μB ds πr

πr πr r

0 02 24 4

Id Ids ˆˆdr r

sB r k

Ids

Ids

O

Y

Xr

Off center points are not so easy to calculate.

I ˆ

or 2oμ

B Br

B k

This is the field at the center of the loop

How about a stack of loops?

Along the axis, yes, the formula isI

2oμ

B Nr

N is the number of turns.

When the loop is sufficiently long, what can we say about the field inside the body of this electric magnet?

We need Gauss’s Law Oops, typo.

We need Ampere’s Law. Sorry, Mr. Ampere.

Ampere’s LawConnects B with I

Ampere’s law states that the line integral of around any closed path equals oI where I is the total steady current passing through any surface bounded by the closed path:

s od μ IB

dB s

This is a line integral over a vector field

from a long, straight conductorre-calculated using Ampere’s LawB

Choose the Gauss’s Surface, oops, not again!

Choose the Ampere’s loop, as a circle with radius a.

Ampere’s Law says

s od μ IB

a

a

2

2

o

o

B ds B π μ I

μ IB

π

is parallel with , soB

ds

When the wire has a size: with radius R

Outside of the wire, r > R

Inside the wire, we need I’, the current inside the ampere’s circle

22

oo

μd B πr μ B

πr B s

I( ) I

2

2

2

2

2

o

o

rd B πr μ

Rμ

B rπR

B s

( ) I ' I ' I

I

Plot the results

The field is proportional to r inside the wire

The field varies as 1/r outside the wire

Both equations are equal at r = R

Magnetic Field of a Toroid

The toroid has N turns of wire

Find the field at a point at distance r from the center of the toroid (loop 1)

There is no field outside the coil (see loop 2)

2

2

o

o

d B πr μ N

μ NB

πr

B s

( ) I

I

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 the wire

The field lines in the interior are nearly parallel to each other uniformly distributed

The interior of the solenoid field people use.

Magnetic Field of a Tightly Wound Solenoid

The field distribution is similar to that of a bar magnet

As the length of the solenoid increases the interior field becomes

more uniform the exterior field

becomes weaker

Ideal (infinitely long) Solenoid An ideal solenoid is

approached when: the turns are closely spaced the length is much greater than

the radius of the turns

Apply Ampere’s Law to loop 2:

1 1path path

d d B ds BB s B s

od B NI B s

I Io o

NB μ μ n

n = N / ℓ is the number of turns per unit length

Magnetic Force Between Two Parallel Conductors

Two parallel wires each carry steady currents

The field due to the current in wire 2 exerts a force on wire 1 of F1 = I1ℓ B2

Substituting the equation for gives

Check with right-hand rule: same direction currents attract

each other opposite directions currents

repel each other

2B

PLAYACTIVE FIGURE

I I

a 1 2

1 2oμ

Fπ

The force per unit length on the wire isI I

a

1 2

2oB μF

πAnd this formula defines the current unit Ampere.

Definition of the Ampere and the Coulomb The force between two parallel wires is used to

define the ampere When the magnitude of the force per unit length

between two long, parallel wires that carry identical currents and are separated by 1 m is 2 x 10-7 N/m, the current in each wire is defined to be 1 A

The SI unit of charge, the coulomb, is defined in terms of the ampere

When a conductor carries a steady current of 1 A, the quantity of charge that flows through a cross section of the conductor in 1 second is 1 C

Magnetic Flux, defined the same way as any other vector flux, like the electric flux, the water (velocity) flux

The magnetic flux over a surface area associated with a magnetic field is defined as

B d B A

The unit of magnetic flux is T.m2 = Wb (weber)

Magnetic Flux Through a Plane

A special case is when a plane of area A, its direction makes an angle with

The magnetic flux is

B = BA cos When the field is perpendicular

to the plane, = 0 (figure a) When the field is parallel to the

plane, = BA (figure b)

B

PLAYACTIVE FIGURE

dA

Gauss’ Law in Magnetism

Yes, that is Gauss’s Law, but in Magnetism. Magnetic fields do not begin or end at any

point The number of lines entering a surface equals the

number of lines leaving the surface Gauss’ law in magnetism says the

magnetic flux through any closed surface is always zero:

0d B A

Example problemsA long, straight wire carries current I. A right-angle bend is made in the middle of the wire. The bend forms an arc of a circle of radius r. Determine the magnetic filed at the center of the arc.

Formula to use: Biot-Savart’s Law, or more specifically the results from the discussed two examples:

I Isin 2

04 4

πo oμ μ

B θπr πr

For the straight section

For the arcI I I

14

2 2

2

4 4 4 8o o o

circle

μ μ μπrB ds

πr πr r

The final answer:I I I I

2 1

4 8 4 4 2o o o oμ μ μ μ

Bπr r πr r π

magnitude

direction pointing into the page.