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Unformatted text preview: (A) i/s ' 1. What is aZEAR Mag et? 2. GAUSS’ LAW FOR B  FIELD Any object which has a non—zero magnetic moment (,u) can be designated as a BAR
7 a  MAGNET because it will experience a torque r = [ ,ux B]
—> W, ~> . When placed in a B ﬂeld.
“7 MA 6N5 T. 7 Wininacuﬂkgvur rrszeg Vwé; 1,021}? 15 ’7 "3M , , To begin, let us recall that a current carrying Icop of area A experiences a torque ’Z' = [Ari x B _.
a , , —> —> —> Because it has a magnetic moment ,u = IAﬁ. Incidentally, it also has potential energy
—> 11,, = — ,LNB. So that ﬁwants to line up along B.
a —> » ' —) Note, the similarity to an Electric Dipole of moment p = q I placed in E .
_, —> —> Torque is T = [px E]
~ > —) _, —) Potential Energy is" ,, ' Next, let us take a compass or a store bought bar magnet. If you suspend it in Earth’smagnetic ~ ﬁeld, the compass or bar will line up along the Earth’s B —ﬁeld. Again, this happens because of the torque on the compass (B.M.) To understand this let us pretend that the EM. can be imagined as consisting of magnetic
“charges” im, separated by]. 3/5 And we ﬁnd that it experiences a Torque: T = —mclB SinHz” or r = [ax B] with y = me!
a —> _> —) . _> _..) Question: Is the pretense justiﬁed? Answer: A FIRM NO! 190
Why: If you break the B. M inﬂate two parts you will n0_t separate +mc and— —.mc You will get two bar magnets. Keep breaking and you get more and more bar magnets.
19 29 39 49 89 ........ 9 Single atom—) ELECTRON Ultimately, you will discover that a SINGLE Electron is a complete bar magnet because it has a 7millimagneticjipolemoment 7 7 ,, 7 7 7 , _ , ,, ,, p6 = 9.27x 10—1"i"N—;—"—2 Which is called a Bohr magneton (,uB ). Thanks to quantum mechanics we now know that in addition to charge 6 (= 1.6 x 10”19 C) and mass m (= 9 x lO‘ﬁlkg) an electron has an intrinsic
property called spin s (=l/2) which endows it with a magnetic moment 261‘:
W (“B H Em—S
whereﬂ h: 10H34J— s Jib: (7:!) ca“, QJIWbm is «p/MM “o a.
O '1Tm/Xbﬁ9—aﬁi “04 <19 "MMM 1134,.49/3) bf: Witt MPW’éokm
‘ Indeed, even nuclei have a magnetic moment but since mass of proton IS about 2000 me, 415T? w};
, nuclear moments are much smaller. Ci} 2 aﬁ/EM So 110W that We have the elementary building blocks let us start‘ ‘constructing” a bar magnet
begimﬁng with electrons. The next constituent is the atom. Here, to get a nonzero ,uA we need unpaired Electrons. That is, if the eleCtrons are arranged in pairs so that they look like was; and that atom is no good to make a B. M. Iron, on the other hand, is a “good” atom. F 83 J“ has
ﬁve unpaired electrons.  Alwyn }'\ i; i 3/5 at the simplest level.
Next, put these atoms into a solid. At high temperatures, the thermal energies make these atomic moments wobble rapidly so that the time average
. We 0
and all we get is a paramagnet. There is m net moment, so no bar magnet.
However, [recall the dielectric in an E ﬁeld] if we put the material in an applied magnetic B it
—> —) will cause all the moments to line up and the solid will vauire a magnetic moment #30,: We can
deﬁne the Magnetization O , ,, 7 7 , 115%....— ,,
, M WVaﬂumc.
A—m2 A B 'It has units of 3 or — which are the same as units of — m m ,uo Simple Experiment: Take a Solenoid. Pass a current I. You will get B0 = port I = no H This can be applied to our paramagnet. We will ﬁnd that the B ﬁeld enhances to B: ,uOnI+ ,uOM
= #0(H+ M) = #0 170+ rm)
M
Which deﬁnes the magnetic susceptibility gm = E M H
In the paramagnet, )5", is a constant, independent of H. M is linear'in H. However, gm is a function of temperature. On reducing T, gm Increases accordlng to gm = ? and this Wlll hold 1f the atomic moments do not “talk” to one another and we will never get a bar magnet. , Fortunately, again thanks to quantum mechanics, in some materials (Fe and Ni are most
familiar) the atomic moments interact with one another and produce an internal ﬁeld
proportional to M. In such systems one can write M = anH + M] where A. is a constant. . _ E ' I __ £ _ C
Agam M — T(H+ AM) and we should ﬁnd gm — H — T— 1c
reduced until T = 2.0 , 95;, will blow up. That is, the material will have a spontaneous magnetic
moment M. We have succeeded in making a FERROMAGNET. Remember that in class we did an experiment to show that M becomes a ferromagnet at T S 630K _ and we note that if 773 4/5 At lower temperatures the system can be imagined as consisting of “domains” each of which
contains billions of atoms with their ,uAl's aligned. These domains Ferromagnet (Only few domains
are drawn) 7 have giant magnetic moments #1) and the response to an applied ﬁeld is greatly enhanced 7 [Experiment in class: nickel was attracted strongly'when it was cold] The last step in constructing a BM. is to recognize that we need a material in which there is a
preferred direction. That is, the ,uD ’s prefer to lie along some axis. Let us assume that for our bar this is along 33. If we apply a EH)? all the ,uD ’S will align along 3? and We get and every domain has its ,uD along direction where it prefers to stay. So next if we remove B ,
x 7 —> the ,uD ’s will stay put, the Bar has a “permanent” magnetic moment. Indeed, we ﬁssile have." “constructed” a _
BAR MAGNET
Which has a B ﬁeld looking like
—> Q C) ( 345' 2. GAUSS’S LAW FOR B FACT: A single electron is a complete magnet with a dipole moment ,uB .
' CONSEQUENCE: Elementary source of B is a DIPOLE with effectively ”zero” size.
—>  ‘ “sources” and “sinks” are coincident. When we use lines to map B —ﬁelds the lines must
close on Themselves. There is no “beginning” and no “end” to a B ﬁeld line.
PROFOUND IMPLICATION: Total ﬂux of B through any closed surface must be —> M. always equal to zero: Every line that comes into enclosed volume must go out as B —ﬁeld lines do not stop or
—> start anywhere. NOTE: The law tells you that total 1.1% of B is zero. It says nothing about B. , ~ 073463 be. (simmer! ll ”
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 Fall '11
 Bhagat
 Physics

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