Sources of Magnetic Fields Chapter 30 Biot-Savart Law Lines of Magnetic Field Ampere’s Law...

Sources of Magnetic FieldsSources of Magnetic Fields

Chapter 30Chapter 30

Biot-Savart LawBiot-Savart Law

Lines of Magnetic FieldLines of Magnetic Field

Ampere’s LawAmpere’s Law

Solenoids and ToroidsSolenoids and Toroids

Sources of Magnetic Fields

• Magnetic fields exert forces on moving charges.• Something reciprocal happens: moving charges give rise

to magnetic fields (which can then exert a force on other moving charges).

• We will look at the easiest case: the magnetic field created by currents in wires.

• The magnetism of permanent magnets also comes from moving charges (the electrons in the atoms).

Magnetic Interaction

• A current generates a magnetic field.

• A magnetic field exerts a force on a current.

• Two conductors, carrying currents, will exert forces on each other.

Rather than discussing moving charges in general, restrict attention to currents in wires. Then:

Biot-Savart Law

• The mathematical description of the magnetic field B due to a current-carrying wire is called the Biot-Savart law. It gives B at a selected position.

• A current I is moving all through the wire. We need to add up the bits of magnetic field dB arising from each infinitesimal length dl.

€

dr B =

μ0

4π

Idr l × ˆ r

r2

Add up all the bits!

r

d l I

dB

Biot-Savart Law

• The mathematical description of the magnetic field B due to a current-carrying wire is called the Biot-Savart law. It gives B at a selected position.

• A current I is moving all through the wire. We need to add up the bits of magnetic field dB arising from each infinitesimal length dl.

€

dr B =

μ0

4π

Idr l × ˆ r

r2

Add up all the bits!

r

d l I

dB

€

rr = r ˆ r is the vector from dl to

the observation point

It turns out that and are related in a simple

way: (0)-1/2 = 3x10 8 m/s = c, the speed of light.

Why? Light is a wave of electric and magnetic fields.

The constant 0 = 4 x 10-7 T m/A is called the permeability of free space.

Biot-Savart Law

d

r B =

4Id

r l ×ˆ rr 2r

d l I

dB

Example: Magnetic field from a long wire

Consider a long straight wire carrying a current I.We want to find the magnetic field B at a point P, a distance R from the wire.

I

R

P


Consider a long straight wire carrying a current I.We want to find the magnetic field B at a point P, a distance R from the wire.

Break the wire into bits dl.To do that, choose coordinates:let the wire be along the x axis,and consider the little bit dx at aposition x.

The vector r = r r is from this bit to the point P.

dx

I

R

d l

x

x

0

r

P

^


dx

I

R

x

x

0

r

€

dr B =

μ0I

4π

dr l x ˆ r

r2

Direction of dB: into page.

+

€

ˆ r


dx

I

R

x

x

0

r

€

dr B =

μ0I

4π

dr l x ˆ r

r2

Direction of dB: into the page.This is true for every bit; so we don’t need to break into components, and B also points into the page.

+

€

ˆ r


dx

I

R

x

x

0

r

€

dr B =

μ0I

4π

dr l x ˆ r

r2

Direction of dB: into the page.This is true for every bit; so we don’t need to break into components, and B also points into the page.

Moreover, lines of B go around a long wire.

+

€

ˆ r


dx

I

R

x

x

0

r

Moreover, lines of B go around a long wire. Perspective:

+

€

ˆ r i

B

P

Another right-hand rule


dx

I

R

x

x

0

r

€

dr B =

μ0I

4π

dr l x ˆ r

r2

Direction of dB (or B): into page

€

dB =μ0I

4π

dx sinθ

r2

∴ B = dB =∫ μ0I

4π

sinθdx

r2x=−∞

x= +∞

∫

+

€

ˆ r


dx

I

R

x

x

0 +

€

∴B = dB =∫ μ0I

4π

sinθ dx

r2x=−∞

x= +∞

∫

r = x 2 + R2 , sinθ =R

r

B =μ0I

4π

Rdx

(x 2 + R2)3

2x=−∞

x= +∞

∫

=μ0I

4πR

x

(x 2 + R2)1

2x=−∞

x= +∞

=μ0I

2πR

r

€

ˆ r

Force between two current-carrying wires

I1

I2

B2

B1

Current 1 produces a magneticfield B1 =0I/ (2 d) at the positionof wire 2.

d

This produces a force on current 2:


I1

I2

B2

B1


d

This produces a force on current 2: F2 = I2L x B1


I1

I2

B2

B1


d


F2


I1

I2

B2

B1


d

This gives the force on a length L of wire 2 to be:

F2 =I2LB1 = I1I2L2 d

Direction: towards 1, if the currents are in the same direction.


F2


I1

I2

B2

B1

Current I1 produces a magneticfield B1 =0I/ (2 d) at the positionof the current I2.

d

Thus, the force on a length L of the conductor 2 is given by:

F2 =I2LB1 = I1I2L2 d

The magnetic force between two parallel wires carrying currents in the same direction is attractive .

This produces a force on current I2:

F2 = I2L x B1

F2

[Direction: towards I1]

What is the force on wire 1? What happens if one current is reversed?

Magnetic field from a circular current loop

I

dBzdBperp

B

r z

R

dl

€

dr B =

μ0

4π

Idr l × ˆ r

r2

dB =μ0

4π

Idl

r2

dBz = dBcosα =μ0

4π

Idlcosα

r2

r = R2 + z2 ,cos = RR2 + z2

Only z componentis nonzero.

along the axis only!

I

dBzdBperp

B

r z

R

dl

At the center of the loop

At distance z on axis from the loop, z>>R

Magnetic field from a circular current loop

€

B = dBz∫ =μ0

4π∫ IR

(R2 + z2)3

2dl

B =μ0

4π

IR

(R2 + z2)3

2dl∫ =

B =μ0

4π

IR

(R2 + z2)3

22πR =

μ0

2

IR2

(R2 + z2)3

2

€

B =μ0I

2R

€

B =μ0IR

2

2z3

along the axis only!

The magnetic dipole moment of the loop is defined as = IA =IR2.

The direction is given by the right hand rule: with fingers closed in the direction of the current flow, the thumb points along .

Magnetic field in terms of dipole moment

I

B

z

R

Far away on the axis,

€

B =μ0IR

2

2z3

In terms of , the magnetic field on axis (far fromthe loop) is therefore

This also works for a loop with N turns. Far from the loop the same expression is true with the dipole moment given by =NIA = IR2

B =2z3

Magnetic field in terms of dipole moment

Dipole Equations

Electric Dipole

= p x E

U = - p · E

Eax = (20 )-1 p/z3

Ebis = (40 )-1 p/x3

Magnetic Dipole

= x B

U = - · B

Bax = ( 0/2 /z3

Bbis = (0/4) /x3

Electric fields

Coulomb’s law gives E directly (as some integral).

Gauss’s law is always true. It is seldom useful. But when it is, it is an easy way to get E.

Gauss’s law is a surface integral over someGaussian surface.

Ampere’s Law

Magnetic fields

Biot-Savart law gives B directly (as some integral).

Ampere’s law is always true. It is seldom useful. But when it is, it is an easy way to get B.

Ampere’s law is a line integral around some

Amperian loop.

Draw an “Amperian loop” around the sources of current.

The line integral of the tangential component of B around this loop is equal to oIenc:

Ampere’s Law

€

rB ∫ • d

r l = μ0Ienc

I2

I3

Ampere’s law is to the Biot-Savart law exactly

as Gauss’s law is to Coulomb’s law.

Draw an “Amperian loop” around the sources of current.

The line integral of the tangential component of B around this loop is equal to oIenc:

Ampere’s Law

€

rB ∫ • d

r l = μ0Ienc

I2

I3

Ampere’s law is to the Biot-Savart law exactly

as Gauss’s law is to Coulomb’s law.

The sign of Ienc comes from another RH rule.

Ampere’s Law - a line integral

blue - into figurered - out of figure

I1

I3

I2

a

c

b

d

€

rB ⋅d

r l

a

∫ =

r B ⋅d

r l

b

∫ =

r B ⋅d

r l

c

∫ =

r B ⋅d

r l

d

∫ =


€

rB ⋅d

r l

a

∫ = μ0I1

r B ⋅d

r l

b

∫ =

r B ⋅d

r l

c

∫ =

r B ⋅d

r l

d

∫ =


I1

I3

I2

a

c

b

d


€

rB ⋅d

r l

a

∫ = μ0I1

r B ⋅d

r l

b

∫ = μ0I2

r B ⋅d

r l

c

∫ =

r B ⋅d

r l

d

∫ =


I1

I3

I2

a

c

b

d


€

rB ⋅d

r l

a

∫ = μ0I1

r B ⋅d

r l

b

∫ = μ0I2

r B ⋅d

r l

c

∫ = μ0 −I3( )

r B ⋅d

r l

d

∫ =


I1

I3

I2

a

c

b

d


€

rB ⋅d

r l

a

∫ = μ0I1

r B ⋅d

r l

b

∫ = μ0I2

r B ⋅d

r l

c

∫ = μ0 −I3( )

r B ⋅d

r l

d

∫ = μ0 I1 − I3( )


I1

I3

I2

a

c

b

d

Ampere’s Law on a Wire

i

What is magnetic fieldat point P ?

P


What is magnetic fieldat point P? Draw Amperian loop through P around current source and integrate B · dl around loop

i

BP

dl

TAKE ADVANTAGE OF SYMMETRY!!!!


i

B

What is magnetic fieldat point P? Draw Amperian loop through P around current source and integrate B · dl around loop P

Then dl

€

rB ∫ • d

r l = Bdl∫ = B(2π r) = μ0I

TAKE ADVANTAGE OF SYMMETRY!!!!

B = I2r

Ampere’s Law for a Wire

What is the magnetic field at point P?

Draw an Amperian loop through P, around the current source, and integrate B · dl around the loop. Then:

i

BP

dl

€

rB ∫ • d

r l = Bdl∫ = B(2π r) = μ0I

A Solenoid.. is a closely wound coil having n turns per unit length.

current flowsout of plane

current flowsinto plane

A Solenoid.. is a closely wound coil having n turns per unit length.



What direction is the magnetic field?

A Solenoid

Consider longer and longer solenoids.

Fields get weaker and weaker outside.

Apply Ampere’s Law to the loop shown.Is there a net enclosed current?In what direction does the field point?What is the magnetic field inside the solenoid?



Apply Ampere’s Law to the loop shown.Is there a net enclosed current?In what direction does the field point?What is the magnetic field inside the solenoid?



€

BL = μ0NI ⇒ B = nμ0I

where n = N /L

L

Gauss’s Law for Magnetism

For electric charges

Gauss’s Law is:

because there are single electric charges. On the other hand, we have never detected a single magnetic charge, only dipoles. Since there are no magnetic monopoles there is no place for magnetic field lines to begin or end. €

E∫ • dA = qε0

€

B∫ • dA = 0Thus, Gauss’s Law for magnetic charges must be:

Laws of Electromagnetism

We have now 2.5 of Maxwell’s 4 fundamental

laws of electromagnetism. They are:

Gauss’s law for electric charges

Gauss’s law for magnetic charges

Ampere’s law (it is still incomplete as it only

applies to steady currents in its present form.

Therefore, the 0.5 of a law.)

Magnetic Materials

The phenomenon of magnetism is due mainly to the orbital motion of electrons inside materials, as well as to the intrinsic magnetic moment of electrons (spin).

There are three types of magnetic behavior in bulkmatter:

FerromagnetismParamagnetismDiamagnetism

Magnetic Materials

Because of the configuration of electron orbits in atoms, and due to the intrinsic magnetic properties of electrons and protons (called “spin”), materials can enhance or diminish applied magnetic fields:

rBapplied

Magnetic Materials


€

rB applied

€

rB internal

Magnetic Materials


€

rB int = κ M

r B app

€

rB applied

€

rB internal

Magnetic Materials

M is the relative permeability

(the magnetic equivalent of E )

Usually M is very close to 1.

- if M > 1, material is “paramagnetic” - e.g. O2

- if M < 1, material is “diamagnetic” - e.g. Cu

Because M is close to 1, we define the

magnetic susceptibility M= M - 1

r rB BM appint =

Magnetic Materials

Hence: For paramagnetic materials M is positive

- so Bint > Bapp For diamagnetic materials M is negative

- so Bint < Bapp

Typically, M ~ +10-5 for paramagnetics,

M ~ -10-6 for diamagnetics.

(For both M is very close to 1)

€

rB int = κ M

r B app = (1+ χ M )

r B app

Magnetic Materials

Ferromagnetic Materials: These are the stuff permanent magnets are

made of.

These materials can have huge susceptibilities:

M as big as +104

Magnetic Materials


made of.

These materials can have huge susceptibilities:

M as big as +104

But ferromagnets have “memory” - when you turn off the Bapp, the internal field, Bint , remains!

Magnetic Materials


made of.These materials can have huge susceptibilities:

M as big as +104

But ferromagnets have “memory” - when you turn off the Bapp, the internal field, Bint , remains!

permanent magnets!

Sources of Magnetic Fields Chapter 30 Biot-Savart Law Lines of Magnetic Field Ampere’s Law...

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