Lecture 12 Magnetism of Matter: Maxwell’s Equations Ch. 32

• Cartoon • Opening Demo• Topics

– Ferromagnetism – Maxwell equations– Displacement current

• Demos• Elmo• Polling

Ferromagnetism

Iron, cobalt, nickel, and rare earth alloys exhibit ferromagnetism.The so called exchange coupling causes electron magnetic momentsof one atom to align with electrons of other atoms. This alignment producesmagnetism. Whole groups of atoms align and form domains.

(See Figure 32-12 on page 756)

A material becomes a magnet when the domains line up adding all themagnetic moments.You can actually hear the domains shifting by bringing up an magnet and hear the induced currents in the coil. Barkhausen Effect

Two other types of magnetic behavior are paramagnetism or diamagnetism.

What is the atomic origin of magnetism?

Electron spinning on its axis Electron orbiting around the nucleus

Spin Magnetic Dipole Moment of the Electron

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rμ =−

e

m

r S

e =1.6 ×10−9Coulombs

m = 9.11×10−31kg

S is the angular momentum due to the electron’s spin. It has units kg.m2/s. μ has units of A.m2 - current times areaRecall for a current loop, the magnetic dipole moment = current times area of loop

In the quantum field theory of the electron, S can not be measured. Only it’s component along the z axis can be measured. In quantumphysics, there are only two values of the z component of the electron spin.

Therefore, only the z component of μ can be measured.Its two possible values are:

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μz = ±eh

4πmCorresponding to the two values of the electron spin quantum number +1/2and -1/2

The above quantity is called the Bohr magneton and is equal to:

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μB = μ z =eh

4πm= 9.27 ×10−24 A.m2

The magnetic moment of the electron is the prime origin of ferromagnetism in materials.

22. The dipole moment associated with an atom of iron in an iron bar is 2.1x10-23 J/T. Assume that all the atoms in the bar, which is 5.0 cm long and has a cross-sectional area of 1.0 cm2, have their dipole moments aligned.

(a) What is the dipole moment of the bar?

The number of iron atoms in the iron bar is

N =7.9g cm3( ) 5.0cm( ) 1.0cm2( )

55.847g mol( ) 6.022 ×1023mol( )=4.3×1023.

Thus, the dipole moment of the bar is

μ = 2.1 ×10−23 J T( ) 4.3 ×1023( ) = 9.03A ⋅m2 .

(b) What torque must be exerted to hold this magnet perpendicular to an external field of 1.5 T? (The density of iron is 7.9 g/cm3)

(b) τ =μBsin90o = 9.03A⋅m2( ) 1.5T( ) =13.5N⋅m

(C) Use the dipole formula to find the magnitude and direction of the magnetic field 1cm from the end of the bar magnet on its central axis at P.

μ 89 m2

5 cm

A = 1 cm2

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B =μ0μ

2πz3

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B =4π ×10−7 N

A 2 8.9A.m2

π × 0.05m(.01m)2

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B =4 ×10−78.9N / A.m

5 ×10−6

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B = 0.71 T

BT =μ0μ2πL

dzz3

.06

.01

∫ =(μ0μ2πL

)(−2z2

0.06

0.01

) = (μ0μ2πL

)(−2

(0.01)2)

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BT = dB∫ =μ0

2π

dμ

z3∫

dμ =μ

ALAdz =

μ

Ldz

z. P

Maxwells Equations:In 1873 he wrote down 4 equations which govern all

classical electromagnetic phenomena.You already know two of them.

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1. ΦE =r E .dA∫ = qenc /ε0

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2. ΦB =r B .dA∫ = 0

A magnetic field changing with time can produce an electric field: Faraday’s law

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3. r E .d

r s ∫ = −

dΦB

dt

Electric lines curl around changing magnetic field lines

Line integral of the electric fieldaround the wire equals the change of Magnetic flux through the areaBounded by the loop

Example

New Question: Can a changing electric field with time produce an magnetic field?

.

Yes it can and it is called

Maxwell’s law of induction

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rB .d

r s ∫ = μ0ε0

dΦE

dt

Maxwell’s law of induction

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rB .d

r s ∫ = μ0ε0

dΦE

dt

Consider the charging of our circular plate capacitor

B field also induced at point 2.

When capacitor stops chargingB field disappears.

current ever actually flows through the capacitor

Find the expression for the induced magnetic field B that circulates around the electric field lines of a

charging circular parallel plate capacitor

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r B .d

r s ∫ = μ0ε0

dΦE

dt

BE

R

r

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rB .d

r s ∫ = (B)(2πr) since B parallel ds

Flux withinthe loop of radius r

r < R

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μ0ε0

dΦE

dt= μ0ε0

d(AE)

dt= μ0ε0A

dE

dt

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(B)(2πr) = μ0ε0πr2 dE

dt

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B =μ0ε0r

2

dE

dt

r< R r > R

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B =μ0ε0R

2

2r

dE

dt

Ampere-Maxwell’s Law

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4. r B .d

r s ∫ = μ0ε0

dΦE

dt+ μ0ienc

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rB .d

r s ∫ = μ0ε0

dΦE

dt

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rB .d

r s ∫ = μ0i

Maxwell combined the above two equations to form oneequation

How do we interpret this equation?

This term has units of current

What is the displacement current?

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r B .d

r s ∫ = μ0ε0

dΦE

dt+ μ0ienc

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ε0

dΦE

dt= id

This is called the displacement current id

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r B .d

r s ∫ = μ0id + μ0ienc

The term is really is a transfer of electric and magnetic energy from one plate to the other while the plates are being charged or discharged. When charging stops, this term goes to zero. Note it is time dependent.

Can I detect the magnetic field associated with displacement current?

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ε0

dΦE

dt= id

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rB .d

r s ∫ = μ0ε0

dΦE

dt

Show that the displacement current in the gap of the two capacitor plates is equal to the real current outside the gap

Calculation of id

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E =σ

ε0

=q / A

ε0

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q = ε0AE

For the field inside a parallel plate capacitor

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i =dq

dt= ε0A

dE

dtThis is the real current i charging the capacitor.

Solving for q

First find the real current i

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id = ε0

dΦE

dt

Next find the displacement current

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id = ε0

d(AE)

dt= ε0A

dE

dt

displacement current =real current. No charge actually moves across the gap.

Calculate Magnetic field due to displacement current

Current is uniformly spread over the circular plates of the capacitor.Imagine it to be just a large wire of diameter R. Then use the formula for the magnetic field inside a wire.

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B = (μ0id

2πR2)r Inside the capacitor

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B =μ0id

2πrOutside the capacitor

Question 11: A circular capacitor of radius R is being charged through a wire of radius R0. Which of the

points a, b, c, and d correspond to points 1, 2, and 3 on the graph

Where is the radius R0 and R on the graph?

Summary of Maxwell EquationsIntegral form

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4. r B .d

r s ∫ = μ0ε0

dΦE

dt+ μ0ienc

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3. r E .d

r s ∫ = −

dΦB

dt

€

1. ΦE =r E .dA∫ = qenc /ε0

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2. ΦB =r B .dA∫ = 0

Chapter 32 Problem 61 In Figure 32-39, the capacitor with circular plates of radius R = 16.0 cm is connected to a source of emf script e = script em sin ωt, where script em = 220 V and ω = 120 rad/s. The maximum value of the displacement current is id = 7.60 µA. Neglect fringing of the electric field at the edges of the plates.

(a) What is the maximum value of the current i in the circuit?(b) What is the maximum value of dE/dt, where E is the electric flux through the region between the plates?(c) What is the separation d between the plates?(d) Find the maximum value of the magnitude of B between the plates at a distance r = 11.0 cm from the center.

Chapter 32 Problem 56Suppose that 4 are the limits to the values of mc for an

electron in an atom. (a) How many different values of the z component µorb,z of the

electron’s orbital magnetic dipole moment are possible?(b) What is the greatest magnitude of those possible values?

Next suppose that the atom is in a magnetic field of magnitude 0.250T, in the positive direction of the z axis.

(c)What are the maximum potential energy(d)What are the minimum potential energy associated with

those possible values of µorb,z?

±