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Unformatted text preview: 1/6 Ampere’s Law  Applications Consider the situation shown schematically in the diagram. Currents [1,12,13,14,I5 are ﬂowing
out of ()or (+) into the paper The corresponding B —f1elds swirl around their sources as shown.
The main point is that the B ﬁeld lines circulate around the currents Choose a closed loop (L)
Start atA, measure B, choose small stop Al along loop Calculate Dot product (component of Balong Al multiplied by AZ)
Kama 151m 0105 (E, M) .ILEUéf, Pagan; 0 p Rpeat this calculation at everystep as shown. BoAl
—> —> Write out the sum
‘23c BaoAl; c: closed loop.
This sum is called circulation of B around closed loop and Ampere says that it is determined . solely by currents threading through the surface on which the loop' 15 drawn & only currents
within the loop contribute, i e. exclude 15. The mathematical Equation 13: T—m EcBwAl=yoEInuo= 475x 10'7—— A 2J6 Qu
In words, circulation of B aroundA closed 100p is proportional to the algebraic sum of the currents
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Applications 1.) swam; Cu £12m Single Wire with current I, there IS cylindrical symmetry so B can be a function of r only & must
encircle]. EA, ’3 anyf (LE, m "Para/M “ID 63% Gaga—mad] Appropriate loop IS circle of radius r centered on the wire 32271? ,uOI
J44!“
30,13—2” ‘ .. ,, as claimed previously 2.) Next, we'begin by Showing that if current is outside the loop it contributes nothing to the '
circulation. Choose L MN 0 P Q with I at the center of the circles of radii r1 , and r2 rm in: 2/5 B( if” —
r1 2727”: ELM
103‘? ”‘3'
_#o .
2
#01 ’ ' #01 [The ﬁrst” (ﬁrm as #:ch
ggélzﬁzm+0+§g2ﬂé+0 l3; awdﬂﬂ (32442,
:L—)M—+N+N—)0+ OefaQ '+Q_,L $3,073,)??Y‘: fa m
=0 munch/b
3.)
y
x
Z
J At y=0 there is a Current sheet of thickness 2‘ carrying current density J = — J2. Looking? it end— on we SEC sources as B _) m and we see that y—components of g: cancel out. gun? survives. Let us take loop of width 1 and
height h. = 231 ’ T
= ,quz‘l '
so,
Jt Jt
B = 502—353 = L)? y > o «57.4 _ 4.) Hollow Cylindrical Conductor— Radius R, carries uniform current. We want B at a distance r
a a from its axis. Since there is a cylindrical symmetry we should use circles centered on the axis for our closed loop.
For r<R. use loop 1. B . 2727" : 0 , No Current threads through loop 1. fat «'2 >12 so, B. 27zr = ,u01,the entire current threadsloop 2.
#01 A B:—
SO’. a, 2”. . ZﬁRf 9
Note: if cylinder has wall thickness f: I = J . Wand ﬁeld at surface would be ,qut. . > i
Again ﬁeld would jump by yth on crossing a current sheet. 5.) SOLID CYLINDRICAL CONDUCTOR — with uniform current I
Deﬁne (I: ”R2
i‘IlIIFV
R
Now for r<R I: Jm‘2  ' J A
1;: egg R 6/5 _ . . ' [“01 A
For >R,'ent1re I contributes B = —
—> 2717” R x i r
6.) Solenoid: Tightly wound, smallradius, length much larger than radius:
N 1
Look at two neighboring turns , ' N turns, L longjn = =# of turns per meter. r—component cancels
By inside survives. Long—narrow solenoid
B m 0 just outside.
._) [:13 $bﬂll1’2éf “MM/St”
Came. out qui'DPleDl’
cﬁﬂouwd 61"”58 Bl = ,uonIl 42mm cuff Take loop as shown B : #011 I boa NW ”NE V For ease shown g = you]? be. Vic/lo ] ...
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 Fall '11
 Bhagat
 Physics

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