Chapter 30

Sources of the Magnetic Field

Amperes and Biot-Savart Laws

FB on a Charge Moving

in a Magnetic Field

Magnitude proportional to charge and speed of the particle

Direction depends on the velocity of the particle and the direction of the magnetic field It is perpendicular to both

FB = q v x B FB is the magnetic force

q is the charge

v is the velocity of the moving charge

B is the magnetic field

Direction given by

Right-Hand Rule

Fingers point in the

direction of v

(for positive charge;

opposite direction if

negative)

Curl fingers in the

direction of B

Then thumb points in the

direction of v x B; i.e. the

direction of FB

Hall Effect

When the charge carriers are negative, the upper edge of the conductor becomes negatively charged

When the charge carriers are positive, the upper edge becomes positively charged

Sign of Hall voltage, VH, gives the sign of the charges

Biot-Savart Law – Introduction

Biot and Savart conducted experiments

on the force exerted by an electric

current on a nearby magnet

They arrived at a mathematical

expression that gives the magnetic field

at some point in space due to a current

Biot-Savart Law – Set-Up

Measured magnetic field

dB at some point P,

distance r from wire

Wire carrying a steady

current of I

Length element is ds

Biot & Savart found that dB

is proportional to current, I,

length, ds and sin()

is inversely proportional to r2

Observations are summarized in the Biot-Savart law:

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

The magnetic field is due to the current-carrying conductor

Biot-Savart Law – Equation

d B

0

4

I d s ˆ r

r2

Total Magnetic Field

dB is the field created by the current in the length segment ds

To find total field, B, sum up contributions from all current elements I.ds

The integral is over the entire current distribution

B-S Law is also valid for a current consisting of charges flowing through space

B

0

4

I d s ˆ r

r2

Quick Quiz 30.1

Consider the current in the length of wire shown in the figure

below. Rank the points A, B, and C in terms of magnitude of

the magnetic field due to the current in the length element

shown, from greatest to least.

(a) A, B, C

(b) B, C, A

(c) C, B, A

(d) C, A, B

(e) An equal field applies at all these points.

Answer: (b). Point B is closest to the current element. Point

C is farther away and the field is further reduced by the sin θ

factor in the cross product . The field at A is zero because θ = 0.

Quick Quiz 30.1

d s x ˆ r

B Compared to E, 1

Distance

Magnitude of magnetic field varies as

inverse square of distance from source

Magnitude of electric field due to point

charge also varies as inverse square of

distance from the charge

B Compared to E, 2

Direction

Electric field created by a point charge is

radial in direction

Magnetic field created by a current element

is perpendicular to both the length element

ds and the unit vector

r̂

Source

Electric field is established by an isolated

electric charge

Current element producing a magnetic field

must be part of an extended current

distribution

Need to integrate over entire current

distribution

B Compared to E, 3

B Compared to E, 4

Ends of Field Lines

Magnetic field lines have no beginning and

no end

They form continuous circles

Electric field lines begin on positive

charges and end on negative charges

B for Long, Straight Conductor

Thin, straight wire carrying

constant current I

Integrating over all the current

elements gives (see ex 30.1)

d s ˆ r dx sin( ) ˆ k

B

0I

4 asin( ) d

1

2

0I

4 acos(

1) cos(

2)

B for a Long, Straight

Conductor, Special Case

For infinitely long,

straight wire

1= 0° and 2= 1

cos(11cos(21

The field becomes

B

0I

2 a

B for a Long, Straight

Conductor, Direction

Magnetic field lines are circles concentric with the wire

Field lines lie in planes perpendicular to to wire

Magnitude of B is constant on any circle of radius a

The right-hand rule for determining the direction of B is shown Grasp wire with thumb in

direction of current. Fingers wrap in direction of B.

B for a Curved Wire Segment

What is the field at point O due to the wire segment?

I and R are constants

ds is parallel to r along AA’ and CC’ so we only need to calculate the field from AC.

We find that

in radians

See Example 30.2

B

0I

4 R

B for a Circular Loop of Wire

Put = 2in previous result

This is the field at the centre of the loop

B

0I

4 R2

0I

2 R

Demo Ec10:

Forces between parallel wires

Wires can be made to

attract and repel one

another depending on

the direction of the

current flow through

each wire.

Magnetic Force Between Two

Parallel Conductors

Two parallel wires carrying

steady currents I1 and I2

Field B2 due to the current in

wire 2 exerts a force on wire 1

of F1 = B2 I1 ℓ

However B2 = 0 I2 / 2a, so

F1

0I

1I

2

2 al

Magnetic Force Between Two

Parallel Conductors, cont.

Parallel conductors carrying currents in

the same direction attract each other

Parallel conductors carrying current in

opposite directions repel each other

F1

0I

1I

2

2 al

Quick Quiz 30.2

For I1 = 2 A and I2 = 6 A in the figure below, which is true:

(a) F1 = 3F2

(b) F1 = F2/3

(c) F1 = F2

Answer: (c). F1 = F2 as required by Newton’s third law.

Another way to arrive at this answer is to realize that

gives the same result whether the multiplication of currents

is (2 A)(6 A) or (6 A)(2 A).

Quick Quiz 30.2

F1

0I

1I

2

2 al

Quick Quiz 30.3

A loose spiral spring is hung from the ceiling, and a large

current is sent through it. The coils move

(a) closer together

(b) farther apart

(c) they do not move at all

Answer: (a). The coils act like wires carrying parallel

currents in the same direction and hence attract one another.

Quick Quiz 30.3

Definition of the Ampere

The force between two parallel wires can be 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

F1

l

0I

1I

2

2 a with

0 4 10

7 T m A

-1

Definition of the Coulomb

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 s is 1 C

Magnetic Field of a Wire

Compass can be used to

detect the magnetic field

When there is no current

in the wire, there is no

field due to the current

Needle points towards the

Earth’s north pole

Magnetic Field of a Wire, 2

When wire carries a

current a compass

needle deflects in a

direction tangent to

the circle

Shows the direction

of the magnetic field

produced by the wire

Ampere’s Law

The line integral of B . ds around any

closed path equals µoI, where I is the

total steady current passing through any

surface bounded by the closed path.

B ds 0I

Ampere’s Law, cont.

Ampere’s law describes the creation of

magnetic fields by all continuous current

configurations

Direction: put thumb of right hand in the

direction of the current through loop.

Your fingers curl in the direction you

integrate around the loop.

Quick Quiz 30.4

Rank the magnitudes of for the closed paths in the

figure below, from least to greatest.

(a) a, b, c, d

(b) b, d, a, c

(c) c, d, b, a

(d) c, b, a, d

(e) d, c, a, b

B .d s

Answer: (b). Ampere’s law (above) indicates that the value

of the line integral depends only on the net current through

each closed path. Path b encloses 1 A, path d encloses 3 A,

path a encloses 4 A, and path c encloses 6 A.

Quick Quiz 30.4

B ds 0I

Quick Quiz 30.5

Rank the magnitudes of for the closed paths in the

figure below, from least to greatest.

(a) a, b, c, d

(b) b, c, d, a

(c) b, d, a, c

(d) d, c, a, b

(e) The criteria are

badly chosen in this case.

B .d s

Answer: (e). Ranked from least to greatest, first comes b,

then a = c = d. Paths a, c and d all give the same nonzero

value μ0I because the size and shape of the paths do not

matter. Path b does not enclose the current, and hence its line

integral is zero.

Quick Quiz 30.5

MFM04AN1: Ampere’s law for

magnetic field near a current-

carrying wire

Field Due to a Long Straight

Wire – from Ampere’s Law

Calculate the magnetic

field at a distance r from

the centre of a wire

carrying a steady

current I

The current is uniformly

distributed through the

cross section of the wire

Field Due to Long

Straight Wire

1. Outside of the wire, r > R

2. Inside the wire, current I' at radius r:

B ds B 2 r 0I

B

0I

2 r

I r

2

R2

I , so that

B ds B 2 r 0I

0

r2

R2

I

B

0I

2 R2

r

Field Due to a Long Straight Wire

– Results Summary

Field proportional to r

inside the wire

Field varies as 1/r

outside the wire

Both equations are

equal at r = R

Note: much easier to

calculate using

Ampere’s rather than

Biot-Savart law!

Magnetic Field of a Toroid

Find field at distance

r from the centre of

the toroid

Toroid has N turns

of wire

B ds B 2 r 0NI

B

0NI

2 r

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

Reasonably uniform magnetic field can be produced in its interior.

Here the field lines are approximately parallel

uniformly distributed

close together

Magnetic Field of a Solenoid

Magnetic Field of a Tightly

Wound Solenoid

Field distribution is

similar to a bar magnet

As the length of the

solenoid increases

the interior field becomes

more uniform

the exterior field

becomes weaker

i.e. an “ideal” solenoid

Ideal Solenoid –

Characteristics

An ideal solenoid is

approached when:

turns closely spaced

length much greater

than radius of turns

Field in interior

of a Solenoid

Apply Ampere’s law

Consider rectangle with side ℓ

parallel to the interior field and

side w perpendicular to the field

The side of length ℓ inside the

solenoid contributes to the field

This is path 1 in the diagram

BdsBdd

1path 1path

sBsB

Magnetic Field of Solenoid

Total current through the rectangular path equals the current through each turn multiplied by the number of turns

Ampere’s law then gives

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

valid only at points near centre of long solenoid

B 0

N

lI

0nI

NIBdo

sB

Quick Quiz 30.6

Consider a solenoid that is very long compared to the radius.

Of the following choices, the most effective way to increase

the magnetic field in the interior of the solenoid is:

(a) double its length, keeping the number of turns per unit

length constant

(b) reduce its radius by half, keeping the number of turns per

unit length constant

(c) overwrapping the entire solenoid with an additional layer

of current-carrying wire

Answer: (c). The magnetic field in a very long solenoid is

independent of its length or radius. Overwrapping with an

additional layer of wire increases the number of turns per

unit length.

Quick Quiz 30.6

Ampere’s vs. Gauss’s Law

Integrals around closed path vs. closed surface. i.e. 2D vs. 3D geometrical figures

Integrals related to fundamental constant x source of the field.

Concept of “Flux” – the flow of field lines through a surface.

B ds 0I

E dA q

0

MFM05AN1: Magnetic Flux

Magnetic Flux

Magnetic flux defined in a similar way to electric flux

Consider an area element dA on an arbitrarily shaped surface, and magnetic field B

The magnetic flux FB is

Units T.m2 = Wb Wb is a weber

FB B .d A

Magnetic Flux

through a Plane

Suppose a plane, of area A,

makes an angle with dA

Magnetic flux FB = BA cos

a) When field parallel to plane,

and FB = 0

b) When field perpendicular to

plane, and FB = BA

maximum value

Gauss’ Law in Magnetism

Magnetic fields do not begin or end at any point i.e. they form closed loops, with the number of

lines entering a surface equaling the number of lines leaving that surface

Gauss’ law in magnetism says:

Would not be true if magnetic monopoles were found!

FB B .d A 0

MFA05AN2: Gauss’s law for

magnetism, FB=0

Earth’s Magnetic Field

The Earth’s magnetic field resembles that of a huge bar magnet buried deep in its interior South magnetic pole is

located near the north geographic pole

North magnetic pole is located near the south geographic pole

Dip Angle of

Earth’s Magnetic Field

A compass needle free to rotate vertically as

well as horizontally will point towards the

Earth’s surface

Angle between horizontal and direction of the

magnetic field is called the dip angle

The closer the compass is moved to the magnetic

poles, the farther from horizontal its needle will be

Needle horizontal at equator, with dip angle = 0°

Needle points straight down at south magnetic pole, with dip

angle = 90°

Earth’s Magnetic Poles

Dip angle of 90° near Hudson Bay in Canada

Earth’s south magnetic pole

Dip angle of -90° near Dumont Durville, Antarctica

Earth’s north magnetic pole

The magnetic and geographic poles are not at the

same location

The difference between true north and magnetic north is

called the magnetic declination

Declination varies by location on the Earth’s surface

Source of the Earth’s

Magnetic Field

There cannot be large masses of permanently magnetized materials in Earth’s core since the high temperatures there prevent materials from retaining permanent magnetization

Most likely source of the Earth’s magnetic field is believed to be convection currents in the liquid part of the core It is also likely related to rate of the

Earth’s rotation: dynamo action

QuickTime™ and a

TIFF (Uncompressed) decompressor

are needed to see this picture.

Reversals of the Earth’s

Magnetic Field

The direction of the Earth’s magnetic

field reverses every few million years

Evidence of these reversals are found in

basalts resulting from volcanic activity

The origin of the reversals is not

understood

Demo Ec19: Magnetic Flux

An Earth inductor compass. Coil is rotated at constant speed, with its axis orientated either parallel or anti-parallel to the Earth’s magnetic field. The galvanometer will show a different deflection, depending on the direction the axis is pointing.

Puzzle: try to work out the direction of the Earth’s magnetic field?

End of Chapter

Quick Quiz 30.7

In an RC circuit, the capacitor begins to discharge. During

the discharge, in the region of space between the plates of

the capacitor, there is

(a) conduction current but no displacement current

(b) displacement current but no conduction current

(c) both conduction and displacement current

(d) no current of any type

Answer: (b). There can be no conduction current because

there is no conductor between the plates. There is a time-

varying electric field because of the decreasing charge on

the plates, and the time-varying electric flux represents a

displacement current.

Quick Quiz 30.7

Quick Quiz 30.8

The capacitor in an RC circuit begins to discharge. During

the discharge, in the region of space between the plates of

the capacitor, there is

(a) an electric field but no magnetic field

(b) a magnetic field but no electric field

(c) both electric and magnetic fields

(d) no fields of any type

Answer: (c). There is a time-varying electric field because of

the decreasing charge on the plates. This time-varying

electric field produces a magnetic field.

Quick Quiz 30.8

Quick Quiz 30.9

Which material would make a better permanent magnet?

-one whose hysteresis loop looks like Figure (a)

-one whose hysteresis loop looks like Figure (b)

Answer: (a). The loop that looks like Figure 30.32a is better

because the remnant magnetization at the point

corresponding to point b in Figure 30.31 is greater.

Quick Quiz 30.9

Quick Quiz 30.10

If we wanted to cancel the Earth’s magnetic field by running

an enormous current loop around the equator, the current

loop would be directed:

(a) east to west

(b) west to east

Answer: (b). The lines of the Earth’s magnetic field enter the

planet in Hudson Bay and emerge from Antarctica; thus, the

field lines resulting from the current would have to go in the

opposite direction. Compare Figure 30.7a with Figure 30.36.

Quick Quiz 30.10