Our Goal:

Maxwell’s Equations

( ) :

I

⋅ =

⋅ =

Φ⋅ = −

Φ⋅ = +

∫

∫

∫

∫

Gauss's law electric

Gauss's law in (magnetic): 0

Faraday's law:

Ampere-Maxwell law

oS

S

B

Eo o

qdε

d

dddt

dd μ μdt

E A

B A

E s

B s

Electric Force and Field 1 2

12 2ˆq qk

r=F r

Magnetic Force and Field

q= ×F v B 2

ˆI 4

o dr

µπ

×= ∫

s rB

0

1E d dVρ

ε εΦ = ⋅ =∫ ∫

in

0

qE A =

0B dΦ = ⋅∫

B A = I⋅ = ⋅ =∫ ∫

s o od μ d μB J A

2ˆdqd k

r= =∫ ∫E E r

Continuous Charge Distribution

2ˆdqd k

r= =∫ ∫E E r

dq dxdq dAdq dV

λσρ

===

linear charge density

surface charge density

volume charge density

QLQAQV

λ

σ

ρ

≡

≡

≡

Note: For spheres, cylinders, infinite planes, rods, etc with high symmetry, the electric field is found using Gauss’s Law which we’ll do next chapter!

Gauss’s Law

E dε

Φ = ⋅∫

in

0

qE A =

The net electric flux through ANY closed surface is due to the net charge contained inside that surface (divided by ε0)!

• Empirical observations are summarized in the mathematical equation called the Biot-Savart law:

• The magnetic field described by the law is the field due to the current-carrying conductor – Don’t confuse this field with

a field external to the conductor

Biot-Savart Law

24oμ ddπ r

×=

s rB

ˆI

2

ˆI 4

o dr

µπ

×= ∫

s rB

Ampere’s Law Ampere’s law states that 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.

s Iod μ⋅ =∫ B

I⋅ = ⋅ =∫ ∫

s o od μ d μB J A

Gauss’ Law in Magnetism • The magnetic flux associated

with a magnetic field is defined in a way similar to electric flux

• The unit of magnetic flux is T.m2 = Wb is a weber

• 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:

0B dΦ = ⋅∫

B A =

B dΦ = ⋅∫B A

Maxwell’s Equations

( ) :

I

⋅ =

⋅ =

Φ⋅ = −

Φ⋅ = +

∫

∫

∫

∫

Gauss's law electric

Gauss's law in (magnetic): 0

Faraday's law:

Ampere-Maxwell law

oS

S

B

Eo o

qdε

d

dddt

dd μ μdt

E A

B A

E s

B s

Currents Produce Magnetic Fields

Smallest Magnet: Electron

How to Find the B Field?

If there is high symmetry: Use Ampere’s Law Otherwise use the Biot-Savart Law:

2

ˆI 4

o dr

µπ

×= ∫

s rB

s Iod μ⋅ =∫ B

Ampere’s Law Ampere’s law states that 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.

I⋅ =∫ s od μB

I⋅ = = ⋅∫ ∫

s o od μ μ dB J A

Field Due to a Long Straight Wire

• Outside of the wire, r > R

• Inside the wire, r < R

2

2

( ) I

Io

o

d B πr μ

μBπr

⋅ = =

=

∫ B s

2

2

2

2

2

( ) I ' I ' I

I

o

o

rd B πr μR

μB rπR

⋅ = = → =

=

∫ B s

The current is uniformly distributed through the cross section of the wire

Magnetic Fields due to Currents

0

2IBr

µπ

=

70 4 10 /x T m Aµ π −= ⋅

“mew not” : The permeability of free space. Permeability is a measure of a materials ability to be permeated by magnetic fields – ferromagnetic materials have a higher permeability.

Magnitude of the Field:

Currents Produce Magnetic Fields

Smallest Magnet: Electron

B for a Long, Straight Conductor • The magnetic field lines

are circles concentric with the wire

• The field lines lie in planes perpendicular to to wire

• The magnitude of B is constant on any circle of radius a

• The right-hand rule for determining the direction of B is shown

Direction for a loop

Right Hand Rules What is the direction of the current?

Electromagnets What’s the direction of the B field?

A cross-sectional view of a coaxial cable is shown. The center conductor is surrounded by a rubber layer, which is surrounded by an outer conductor, which is surrounded by another rubber layer. In a particular application, the current in the inner conductor is 1.00 A out of the page and the current in the outer conductor is 3.00 A into the page. Determine the magnitude and direction of the magnetic field at points a and b.

Adding B Fields The wires carry a current of 8.0A. Find the magnitude of the net B field at A and B.

0 1 0 2

1 22 2netAI IBr r

µ µπ π

= −

74 10 / (8 ) 1 1( )2 .03 .15

x Tm A Am m

ππ

−

= −

0

1 2

1 12

Ir r

µπ

= −

First, define positive directions. Up out of plane is positive.

I1

I2

54.27 10netAB x T−=

Is there any place along the line where the fields cancel?

Current Affair

An electric current is flowing through two parallel wires in the same direction. Do the wires tend to

a) repel each other b) Attract each other c) Exert no force on each other d) Twist at right angles to each other e) Spin

F

Current Affair

F F

Magnetic Force between two Current Carrying Wires ON Each Other:

Opposite CURRENTS repel! Like CURRENTS attract!

Magnetic Force between two Current Carrying Wires ON Each Other:

2 2 1 2 sinon atF I LB θ=

0 12 ( )sin 90

2II Lr

µπ

=

sinF ILB θ=

0 1 22 12on on

I I LF Fr

µπ

= = −

F I= ×L B

Problem The wires carry a current of 8.0A. What is the force per meter on each wire? What direction?

I1

I2

0 1 22 12on on

I I LF Fr

µπ

= = −

2 0 1 2

2onF I IL r

µπ

=

They repel each other.

7 24 10 / (8 )2 (.12 )

x Tm A Am

ππ

−

=

41.07 10 /x N m−=

NTA m

=⋅

Questions Is the net B field greater, less or the same at A and B?

Do the wires exert forces on each other? Do the wires exert torques on each other?

If so, what directions?

Question Is the wire attracted or repelled by the coil?

Electric Force and Field

1 212 2

ˆq qkr

=F r

Magnetic Force and Field

q= ×F v B 2

ˆI 4

o dr

µπ

×= ∫

s rB

0

1E d dVρ

ε εΦ = ⋅ =∫ ∫

in

0

qE A =

0B dΦ = ⋅∫

B A = I⋅ = ⋅ =∫ ∫

s o od μ d μB J A

2ˆdqd k

r= =∫ ∫E E r

Compare E & B

Compare E & B Symmetry

I

( ) I

⋅ =

=∫

2o

o

d μ

B πr μ

B s⋅ =∫0

inqE dAε

( ) =0

2 λlE πrlε

=02

λEπε r

I=

2oμBπr

Use Gauss’s Law Use Ampere’s Law

Compare E & B: Infinite Current Sheet

0 2Jµ=B

02E σ

ε=

Infinite Charged Plane Use Gauss’s Law

Use Ampere’s Law

Circular Charge Distribution

( )3/ 22 2ringkxQE

x a=

+ ( )2

32 2 22

Iox

μ RBx R

=+

3ringkxQEa

= I=

2

32o

xμ RB

x

Far from ring: Dipole Field

Compare Finite Rods

2IoμB

πa=

02E

yλ

πε=

Compare Gauss’s Law Flux for Dipoles

0B dΦ = ⋅∫

B A = E dε

Φ = ⋅∫

in

0

qE A =

Compare Dipoles

, , -IA Uµ ≡ = = ⋅τ μ×E μ B 2 , , -p aq U≡ = × = ⋅τ p E p E

Compare Dipoles

0net indE E E= −0net indB B B= +

Reduces External Field Increases External Field

E0 & B0: External Field

(ferromagnetic)

How to Find the B Field?

If there is high symmetry: Use Ampere’s Law Otherwise use the Biot-Savart Law:

2

ˆI 4

o dr

µπ

×= ∫

s rB

s Iod μ⋅ =∫ B

Ampere’s Law Ampere’s law states that 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.

s Iod μ⋅ =∫ B

I⋅ = ⋅ =∫ ∫

s o od μ d μB J A

A long cylindrical conductor of radius R carries a current I as shown. The current density J, however, is not uniform over the cross section of the conductor but is a function of the radius according to J = br, where b is a constant. Find an expression for the magnetic field B (a) at a distance r1 < R and (b) at a distance r2 > R, measured from the axis.

• Empirical observations are summarized in the mathematical equation called the Biot-Savart law:

• The magnetic field described by the law is the field due to the current-carrying conductor – Don’t confuse this field with

a field external to the conductor

Biot-Savart Law

24oμ ddπ r

×=

s rB

ˆI

2

ˆI 4

o dr

µπ

×= ∫

s rB

B for a Long, Straight Conductor • The thin, straight wire is

carrying a constant current

• If the conductor is an infinitely long, straight wire, θ1 = 0 and θ2 = π

( )

2

1

1 2

4

4

I sin

I cos cos

θo

θ

o

μB θ dθπaμ θ θπa

=

= −

∫

( )ˆ ˆ ˆ ˆˆ (cos )× = × + =sin sin d dx θ θ dx θs r i i j k

2IoμB

πa=

ˆI ×=

24oμ ddπ r

s rB

, cos cot , cscsin

= = = = − 2ar x r θ a θ dx a θdθθ

ˆ sinˆ ˆsin csc sin×= − =

22

2 21d θθk a θdθ θdθk

r a as r

Same result with Ampere’s Law which is way easier

• Outside of the wire, r > R

• Inside the wire, r < R

2

2

( ) I

Io

o

d B πr μ

μBπr

⋅ = =

=

∫ B s

2

2

2

2

2

( ) I ' I ' I

I

o

o

rd B πr μR

μB rπR

⋅ = = → =

=

∫ B s

The segment of wire carries a current of I = 5.00 A, where the radius of the circular arc is R = 3.00 cm. Determine the magnitude and direction of the magnetic field at the origin.

B for a Circular Current Loop • The loop has a radius

of R and carries a steady current of I

• Find B at point P

• At x = 0:

( )2

32 2 22

Iox

μ RBx R

=+

( )I I

= =+

2

32 2 2 22o o

xμ R μB

Rx R

24oμ ddπ r

×=

s rB

ˆI

Magnetic Field of a Coil (Not a Solenoid)

0

2IB N

Rµ

=

Magnetic Field at the center of the loop:

Magnetic Field of a Solenoid • A solenoid is a long

wire wound in the form of a helix

• 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

The Field Outside is nearly zero because the fields from different

loops cancel each others

Ideal Solenoid

BdsBdd1path 1path

=∫ ∫ ∫=⋅=⋅ sBsB

NIBd oµ==∫ ⋅ sB

I Io oNB μ μ n= =

The field inside is uniform the field outside is ~zero.

What current is required in the windings of a long solenoid that has 1 000 turns uniformly distributed over a length of 0.400 m, to produce at the center of the solenoid a magnetic field of magnitude 1.00 × 10–4 T?

I Io oNB μ μ n= =

Magnetic Field of a Toroid

• Find the field at a point at distance r from the center of the toroid

• The toroid has N turns of wire 2

2

o

o

d B πr μ N

μ NBπr

⋅ = =

=

∫ B s

( ) I

I

A magnetic field of 1.30 T is to be set up in an iron-core toroid. The toroid has a mean radius of 10.0 cm, and magnetic permeability of 5 000 μ0. What current is required if the winding has 470 turns of wire? The thickness of the iron ring is small compared to 10 cm, so the field in the material is nearly uniform. I

=2oμ NBπr

Coil, Solenoid, Toroid

0

2IB N

Rµ

=Coil:

I=

2oμ NBπr

Toroid:

Solenoid: I Io oNB μ μ n= =

Magnetic Dipoles The magnetic dipole moment of a current loop enclosing an area A is defined as

The SI units of the magnetic dipole moment are A m2. The on-axis field of a magnetic dipole is

Smallest Magnet: Electron

Classical Atom • The electrons move in circular orbits

• The orbiting electron constitutes a tiny current loop

• The magnetic moment of the electron is associated with this orbital motion

• is the angular momentum

• is magnetic moment

L

µI

/= = =

∆ 2q e et T πr v

12

Iμ A evr= =

In Bohr’s 1913 model of the hydrogen atom, the electron is in a circular orbit of radius 5.29 × 10–11 m and its speed is 2.19 × 106 m/s. (a) What is the magnitude of the magnetic moment due to the electron’s motion? (b) If the electron moves in a horizontal circle, counterclockwise as seen from above, what is the direction of this magnetic moment vector?

Domains

• All ferromagnetic materials are made up of microscopic regions called domains – The domain is an area within which all

magnetic moments are aligned • The boundaries between various domains

having different orientations are called domain walls

Domains, Unmagnetized Material

• The magnetic moments in the domains are randomly aligned

• The net magnetic moment is zero

Domains, External Field Applied

• A sample is placed in an external magnetic field

• The size of the domains with magnetic moments aligned with the field grows

• The sample is magnetized

Domains, External Field Applied

• The material is placed in a stronger field

• The domains not aligned with the field become very small

• When the external field is removed, the material may retain a net magnetization in the direction of the original field

Ferromagnetism

• Some substances exhibit strong magnetic effects called ferromagnetism

• Some examples of ferromagnetic materials are: – iron – cobalt – nickel – gadolinium – dysprosium

• They contain permanent atomic magnetic moments that tend to align parallel to each other even in a weak external magnetic field

Paramagnetism • Paramagnetic substances have small but

positive magnetism • It results from the presence of atoms that

have permanent magnetic moments – These moments interact weakly with each other

• When placed in an external magnetic field, its atomic moments tend to line up with the field – The alignment process competes with thermal

motion which randomizes the moment orientations

Diamagnetism

• When an external magnetic field is applied to a diamagnetic substance, a weak magnetic moment is induced in the direction opposite the applied field

• Diamagnetic substances are weakly repelled by a magnet – Weak, so only present when ferromagnetism or

paramagnetism do not exist

Curie Temperature • The Curie temperature is

the critical temperature above which a ferromagnetic material loses its residual magnetism – The material will

become paramagnetic • Above the Curie

temperature, the thermal agitation is great enough to cause a random orientation of the moments

Meissner Effect

• Certain types of superconductors also exhibit perfect diamagnetism in the superconducting state – This is called the

Meissner effect • If a permanent magnet is

brought near a superconductor, the two objects repel each other

Earth’s Magnetic Field

• The Earth’s magnetic field resembles that achieved by burying a huge bar magnet deep in the Earth’s interior

• The Earth’s south magnetic pole is located near the north geographic pole

• The Earth’s north magnetic pole is located near the south geographic pole

Earth’s interior consists of a rocky mantle and an iron rich core

The Earth’s magnetic field is caused by dynamo movements in Earth’s core

Earth’s magnetic field flips poles. Consecutive reversals are spaced 5 thousand years to 50 million years apart. The last reversal happened 740,000 years ago. Some researchers think our planet is overdue for another one, but nobody knows exactly when the next reversal might occur.

The Sun's north magnetic pole is pointing through the Sun's southern hemisphere, until the year 2012 when they will reverse. This transition happens, as far as we know, at the peak of every 11-year sunspot cycle -- like clockwork.

Babcock’s magnetic dynamo is one possible explanation of the sunspot cycle where magnetic field lines become complexly

entangled after many solar rotations

The Sun’s magnetic fields create sunspots

Zeeman effect - spectral lines split in regions of high magnetic fields

Magnetic field lines connect sunspots on the Sun’s photosphere

Solar magnetic fields also create other atmospheric phenomena

• plages • filaments

Solar magnetic fields also create other atmospheric phenomena

• plages • filaments • prominences

Solar magnetic fields also create other atmospheric phenomena

• plages • filaments • prominences • solar flares

Solar magnetic fields also create other atmospheric phenomena

• plages • filaments • prominences • solar flares • coronal holes

Solar magnetic fields also create other atmospheric phenomena

• plages • filaments • prominences • solar flares

• coronal holes • coronal mass

ejections (CMEs)

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 rocks provide a timeline for the periodic

reversals of the field • The rocks are dated by other means to determine the

timeline

This Week’s Lab Using a Tangent Galvanometer to

Measure the Earth’s Magnetic Field

0

2IB Nr

µ=

Magnetic Field at the center of the loop:

Horizontal Earth Field

Total Field