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Lecture 12 of Matter: Maxwell’s Equations Chp. 32

• Cartoon • Warm-up problem • Opening Demo • Topics – Finish up Mutual – – Maxwell equations –

• Demos What is the magnetic stored in a or coil

i i dU di U B 2 B dU = Lidi dU = Lidi U = Lidi = 1 Li = Li B !0 B ! B ! 2 dt dt 0 0

U = 1 Li 2 For an inductor L Now define the energy per unit volume B 2 U u = B B Al 1 2 2 L 2 Li L i 2 u = = = µ0n A B Al l 2A l 2 Area A 1 Li u = 2 = 1 µ n 2i2 B = µ0ni l B Al 2 0 2 The energy formula B E 2 uB = is valid in general uE = 2µ0 2!0

What is Mutual Inductance? M

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

I 1 2 2 I1

!m1 = L1I1 + M 21I2

!m2 = L2 I2 + M12 I1

M12 = M 21 = M

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.

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.

(a) The flux over the loop cross section due to the current i in the wire is given by a +b a +b µ0il µ0il a +b µ0il b ! = # Bwireldr = # dr = lnr a = ln(1+ ) a a 2"r 2" 2" a N! M = i N! Nµ l # b& M = = 0 ln%1 + ( i 2" $ a'

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

Nµ l " b% M = 0 ln$1 + ' 2! # a&

(b) From the formula for M obtained, 100(4* (10)7 H m)'(0.30m) & 8.0 # M = ln$1+ ! 2* % 1.0 "

=1.3"10!5 H Ferromagnetism

Iron, cobalt, nickel, and rare alloys exhibit ferromagnetism. The so called exchange coupling causes magnetic moments of one to align with of other . This alignment produces magnetism. Whole groups of atoms align and form domains.

(See Figure 32-12 on page 756)

A material becomes a when the domains line up adding all the magnetic 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 or . What is the atomic origin of magnetism?

Electron spinning on its axis Electron orbiting around the nucleus Magnetic of the Electron ! e ! µ = ! S m e =1.6 "10!9 Coulombs !31 m = 9.11"10 kg S is the due to the electron’s spin. It has units kg.m2/s. µ has units of A.m2 - current times area Recall for a current loop, the moment = current times area of loop

In the quantum theory of the electron, S can not be measured. Only it’s component along the z axis can be measured. In quantum physics, 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: eh µ = ± Corresponding to the two values of z 4!m the electron +1/2 and -1/2

The above quantity is called the and is equal to: eh µ = µ = = 9.27 "10#24 A.m2 B z 4!m

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

7.9 g cm3 5.0cm 1.0cm2 ( )( )( ) 23 N = = 4.3 ! 10 . (55.847 g mol) (6.022 ! 1023 mol)

Thus, the dipole moment of the bar is µ = (2.1! 10"23 J T )(4.3 ! 1023 ) = 9.03A # m2 . (b) What 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) ! = µBsin90o = (9.03A " m2 )(1.5T ) = 13.5N " m (C) Use the dipole formula to find the magnitude and direction of the 1cm from the end of the bar magnet on its central axis at P.

5 cm P . z

A = 1 cm2 µ0µ µ= 8.9 !.m2 B = 2!z3 µ dµ B = dB = 0 T ! 2" ! z3 µ µ dµ = Adz = dz AL L 0.01 µ µ .01 dz µ µ 2 µ µ 2 B = 0 = ( 0 )(# ) = ( 0 )(# ) T " 3 2 2 2! L .06 z 2! L z 0.06 2! L (0.01)

#7 N 2 4! "10 2 8.9A.m B = A ! " 0.05m(.01m)2 4 !10"78.9N / A.m B = B = 0.71 T 5 !10"6 BigBite is a 50 ton with a 25 cm by 100 cm gap

B = 1

Maxwells Equations: In 1873 he wrote down 4 equations which govern all classical electromagnetic phenomena. You already know two of them. ! ! 1. ! = E .dA = q /# 2. ! = B .dA = 0 E " enc 0 B "

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

! ! d# 3. ! E .ds = " B dt of the electric field around the wire equals the change of through the area Bounded by the loop Electric lines curl around changing magnetic field lines

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

.

! ! d#E ! B .ds = µ0"0 dt

Maxwell’s law of induction

! ! d#E ! B .ds = µ0"0 dt "# current ever actually flows through the Consider the charging of our circular capacitor plate capacitor B field also induced at point 2.

When capacitor stops charging B 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 < R Flux within ! ! d#E the loop of ! B .ds = µ0"0 radius r dt 0 ! ! E B ! B .ds = (B)(2"r) since B parallel ds r d"E d(AE) dE R µ0!0 = µ0!0 = µ0!0 A dt dt dt r< R r > R dE µ ! r dE µ ! R2 dE (B)(2!r) = µ " !r2 B = 0 0 B = 0 0 0 0 dt 2 dt 2r dt

-Maxwell’s Law

! ! ! d#E ! ! B .ds = µ0"0 ! B .ds = µ0i dt

Maxwell combined the above This term has units of current two equations to form one equation ! ! d#E 4. ! B .ds = µ0"0 + µ0ienc dt

How do we interpret this equation?

What is the displacement current? ! ! d#E ! B .ds = µ0"0 + µ0ienc dt

d"E This is called the !0 = id dt displacement current id ! ! B .ds = µ i + µ i ! 0 d 0 enc The term is really is a transfer of electric and 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. Show that the displacement current in the gap of the two capacitor plates is equal to the real current outside the gap

d" ! E = i 0 dt d

! ! d#E ! B .ds = µ0"0 dt

Can I detect the magnetic field associated with displacement current? Calculation of id First find the real current i ! q/ A For the field inside a parallel plate capacitor E = = "0 "0 Solving for q q = !0 AE dq dE i = = ! A This is the real current i charging the capacitor. dt 0 dt Next find the displacement current d" i = ! E d 0 dt displacement current = real current. No charge d(AE) dE id = !0 = !0 A actually moves across dt dt 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.

µ i B = ( 0 d )r Inside the capacitor 2!R2 µ i B = 0 d Outside the capacitor 2!r

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 Equations Integral form ! 1. ! = E .dA = q /# E " ! enc 0 2. ! = B .dA = 0 B " ! ! d# 3. ! E .ds = " B dt ! ! d" 4. B.ds = µ ! E + µ i # 0 0 dt 0 enc