Lecture 11 Magnetism of Matter: Maxwell’s Equations Chp. 32Wednesday Morning

• Cartoon -. Opening Demo -

• Warm-up problem

• Physlet

• Topics

– Finish up Mutual inductance– Ferromagnetism – Maxwell equations– Displacement current– Exam

• Demos

What is Mutual Inductance? M

When two circuits are near one another and both have currents changing, they can induce emfs in each other.

On circuit boards you have to be careful you do not put circuits near each other that have large mutual inductance.

They have to be oriented carefully and even shielded.

221111 IMILm +=φ

112222 IMILm +=φ

MMM == 2112

1 2I1

I2

71. Two coils, connected as shown, separately have inductances L1 and L2. Their mutual inductance is M.

(a) Show that this combination can be replaced by a single coil of equivalent inductance given by

.221 MLLLeq ++=

We assume that the current is changing at (nonzero) rate di/dt and calculate the total emf across both coils.

First consider coil 1. The magnetic field due to the current in that coil points to the left.

The magnetic field due to current in coil 2 also points to the left. When the current increases, both fields increase and both changes in flux contribute emf’s in the same direction.

.)()( 2211 dt

diMLand

dt

diML +−=+−= εε

Therefore, the total emf across both coils is

Thus, the induced emf’s are

dt

diMLL )2( 2121 ++−=+= εεε

which is exactly the emf that would be produced if the coils were replaced by a single coil with inductance .221 MLLLeq ++=

ε

(b) How could the coils in this figure be reconnected to yield an equivalent inductance of

?221 MLLLeq −+=

We imagine reversing the leads of coil 2 so the current enter at the back of the coil rather than front (as pictured in the diagram). Then the field produced by coil 2 at the site of coil 1 is opposite to the field produced by coil 1 itself.

The fluxes have opposite signs. An increasing current in coil 1 tends to increase the flux in that coil, but an increasing current in coil 2 tends to decrease it.

ε

The emf across coil 1 is

dt

diML )( 11 −−=ε

Similarly, the emf across coil 2 is

dt

diML )( 22 −−=ε

The total emf across both coils is

dt

diMLL )2( 21 −+−=ε

This is the same as the emf that would be produced by a single coil with inductance MLLLeq 221 −+=

75. A rectangular loop of N closely packed turns is positioned near a long, straight wire as shown in the figure.

(a) What is the mutual inductance M for the loop-wire combination?

(b) Evaluate M for N = 100, a = 1.0 cm, b = 8.0 cm, and l = 30 cm.

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Φ= Bwire ldr =μ0il

2πrdr =

a

a +b

∫a

a +b

∫ μ0il

2πln r a

a +b =μ0il

2πln(1+

b

a)

(a) The flux over the loop cross section due to the current i in the wire is given by

Thus,

€

M =NΦ

i=

Nμ0l

2πln 1+

b

a

⎛

⎝ ⎜

⎞

⎠ ⎟

(b) Evaluate M for N = 100, a = 1.0 cm, b = 8.0 cm, and l = 30 cm.

(b) From the formula for M obtained,

( ) ( )⎟⎠

⎞⎜⎝

⎛ +⋅×

=−

0.1

0.81ln

2

30.0104100 7

π

π mmHM

H5103.1 −×=

€

M =Nμ0l

2πln 1+

b

a

⎛

⎝ ⎜

⎞

⎠ ⎟

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

€

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:

€

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:

€

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?

(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)

(a) The number of iron atoms in the iron bar is

( )( )( )( ) ( ) .103.4

10022.6847.55

0.10.59.7 2323

23

×=×

=molmolg

cmcmcmgN

Thus, the dipole moment of the bar is

( )( ) .9.8103.4101.2 22323 mATJ ⋅=××= −μ

( )( ) mNTmABb o ⋅=⋅== 1357.19.890sin)( 2μτ

(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 m2

5 cm

A = 1 cm2

€

B =μ0μ

2πz3

€

B =4π ×10−7 N

A 2 8.9A.m2

π × 0.05m(.01m)2

€

B =4 ×10−78.9N / A.m

5 ×10−6

€

B = 0.71 T

€

BT =μ0μ

2πL

dz

z3.06

.01

∫ = (μ0μ

2πL)(−

2

z2) Evaluate at z = 0.01

€

BT = dB∫ =μ0

2π

dμ

z3∫

dμ =μ

ALAdz =

μ

Ldz

z. P

BigBite is a 50 ton electromagnet with a 25 cm by 100 cm

gap

B = 1 Tesla

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

classical electromagnetic phenomena.You already know two of them.

€

1. ΦE =r E .dA∫ = qenc /ε0

€

2. ΦB =r B .dA∫ = 0

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

€

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

B is increasing inmagnitude

Note that induced E field is in such a direction that the B fieldit produces opposes the original B field.

€

3. r E .d

r s ∫ = −

dΦB

dtFaraday’s Law

Note there is no electric potentialassociated with the electric field induced by Faraday’s Law

Can a changing electric field with time produce an magnetic field:

.

Yes it can and it is called

Maxwell’s law of induction

€

rB .d

r s ∫ = μ0ε0

dΦE

dt

Maxwell’s law of induction

€

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.

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

charging circular parallel plate capacitor

€

r B .d

r s ∫ = μ0ε0

dΦE

dt

0

BE

R

r

€

rB .d

r s ∫ = (B)(2πr) since B parallel ds

Flux withinthe loop of radius r

r < R

€

μ0ε0

dΦE

dt= μ0ε0

d(AE)

dt= μ0ε0A

dE

dt

€

(B)(2πr) = μ0ε0πr2 dE

dt

€

B =μ0ε0r

2

dE

dt

r< R r > R

€

B =μ0ε0R

2

2r

dE

dt

Ampere-Maxwell’s Law

€

4. r B .d

r s ∫ = μ0ε0

dΦE

dt+ μ0ienc

€

rB .d

r s ∫ = μ0ε0

dΦE

dt

€

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?

€

r B .d

r s ∫ = μ0ε0

dΦE

dt+ μ0ienc

€

ε0

dΦE

dt= id

This is called the displacement current id

€

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?

€

ε0

dΦE

dt= id

€

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

€

E =σ

ε0

=q / A

ε0

€

q = ε0AE

For the field inside a parallel plate capacitor

€

i =dq

dt= ε0A

dE

dtThis is the real current i charging the capacitor.

Solving for q

First find the real current i

€

id = ε0

dΦE

dt

Next find the displacement current

€

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.

€

B = (μ0id

2πR2)r Inside the capacitor

€

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?

At any instant the displacement current id in the gap between the plates equals the conduction current i in the wires. Thus, id = i = 3.0 A.

37. A parallel-plate capacitor has square plates 1.0 m on a side as shown in the figure. A current of 2.0 A charges the capacitor, producing a uniform electric field between the plates, with perpendicular to the plates.

Er E

r

(b) What is dE/dt in this region?

The rate of change of the electric field is

( )( ) smV

mmF

A

A

i

dt

d

Adt

dE dE

⋅×=×

==⎟⎠

⎞⎜⎝

⎛ Φ=

−

11212

00

0

103.20.11085.8

0.21

εε

ε

(a) What is the displacement current id through the region between the plates?

The displacement current through the indicated path is

( ) Am

mA

plateeachofarea

pathbyenclosedareaii dd 50.0

0.1

50.00.2

2

=⎟⎠

⎞⎜⎝

⎛=

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛×=′

(c) What is the displacement current through the square, dashed path between the plates?

(d) What is around this square, dashed path?∫ ⋅ sdBrr

( )( ) mTAmHisdB d ⋅×=×=′=⋅ −−∫ 760 103.650.01026.1μ

rrThe integral field around the indicated path is

(e) What is the value of B on this path?????

Summary of Maxwell EquationsIntegral form

€

4. r B .d

r s ∫ = μ0ε0

dΦE

dt+ μ0ienc

€

3. r E .d

r s ∫ = −

dΦB

dt

€

1. ΦE =r E .dA∫ = qenc /ε0

€

2. ΦB =r B .dA∫ = 0

Warm up set 10 Due 8:00 am Tuesday

1. HRW6 31.TB.02. [120186] Suppose this page is perpendicular to a uniform magnetic field and the magnetic flux through it is 5 Wb. If the page is turned by 30° around an edge the flux through it will be: 4.3 Wb 10 Wb 5.8 Wb 2.5 Wb 5 Wb

2. HRW6 31.TB.08. [120192] Faraday's law states that an induced emf is proportional to: the rate of change of the electric field the rate of change of the magnetic field zero the rate of change of the magnetic flux the rate of change of the electric flux

3. HRW6 31.TB.09. [120193] The emf that appears in Faraday's law is: around a conducting circuit perpendicular to the surface used to compute the magnetic flux throughout the surface used to compute the magnetic flux none of these around the boundary of the surface used to compute the magnetic flux