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Exam electromagnetism Oct 27,2017
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2 g Given : potential outside spherical shell
A-swu
= M¥4 sing g
Use superposition to get potential outside
uniformly charged ball by adding mfruste -
sand shells :
A ball = {Ahsan = § Aiwa = µ=gdR]=
= mg sniff Igpidr = mYfI sniff
⇐ntostaUniformly charged ball ⇒ g = }¥a3
" Aim = might sniff g
b, By definition B = Tx I
,A- = A
, f
⇒ Is =T×I = muffed [ instnotfolimosnnd) . ago)
+ a Haake o . ÷ Its 'm¥)) + go ]
=mYoQ÷ lived + a sing )
g we recognize that the frdd on b,
is a
pure dipole freld in the z . Imation.
For a magnetic dipole in = MI we have
B = many Ii 20¥ + § snftst ]
Comparing with b, we see that
we = wQa25
Smee this is a pure dipole field there
are no higher multiples
^ I3 a , -a
-
Is=nIy^f( s . b)
5 J J = . nIf8( s - a)
|
iffiF¥¥lf
"
if.ru#B.ut
.
..
- -
.
.
,^
,
tying.
Ampere 's law,
o,
noE- field
Ix B = MEE+ no J
Symmetry + mhnitely long solenoid ⇒ D= BE
wtegrate over surface is, iiand iii respectively
and uomg Holes'
theorem 9msright-hand rule
§ B. ill = µ . f 5. Is,
DJ !. cfdsdz
i,
B = BE ⇒ only controls from tz. part to LHS
Bout l - Bml = 0 ⇒ Bout = Bm
Bout = 0 for s → a ⇒ 13=0 for all s > b
>=O
ii
¥b
l -
Basil= m.JJs.is =
l Smaa
= - Monty .ie f ) SCS - b) dsdz = - Month0 Smrh
⇒Baab= Mont ,Ba<s< b = µ°nIE
iii
Bauble-
B.al= µ . JJ b.
is = + µ .uIl
⇒ Bs<a
= Ba<s< b
- µ°nI = 0
b,
I = I. sin #,we ÷
quasi . state approximation B=µuIosin@HE-
Faraday 's law : B.
EXE = - FEB
tutegrate over area + Stoke 's theorem
⇒ §E. at = - dftf B. Is
B m I . direction + symmetry gives E m
§ - dmectiou.
Choose C to be a omde with
radius s in xy . plane sign from right-hand rate
it = if sdy dJ=tI sdsdp
⇒ Esfdy =- It ( §Bids 'dy )
÷TS
i,
s< a 13=0 ⇒ E= 0
ii,
a<s< b,
B = Bosinfwt)
E = . at fetlsinlwtt )
Bo§ s
'd÷it ( s
'. a
'
)
= - w cos ( wt ) Bo s±a'
ZS
iii,
s > b :
E = - has oft ( imlwtt ) B.
§ sididp =
= - co coscwt ) Bo b±a'
ZS
E = Eig
g Poywtmg vector
§ = µt ( Ex B)
a < s < b : ( vacuum ⇒ µ =µ . )
§ = ¥ ( - B. waoslwtt she'd ) × ( B. sin Cwttz )
= - to Bo'
w tz sin ( zwt ) s÷g2g×I[ Bo =
non I ]5
= - M¥7 sinczwt ) s¥'
g
d, Energy density m EM fndds
u= zt e. E '
+ tz µ÷ 182=
= se Borut wicwt ) ( Kf )'
+ fµ B.
'
sin' Cwt ) =
= ±µBilw'go.ae#Icos4wH+sin4wH]
a < s < b,
Waz < < 1
,had < < 1
⇒ u = ÷µ.
Bi sin Ywtl
local conservation of EM energy
¥ u = -T.j.gg#70noohmreheahugfor a < scb
Tzu = F. Bi sinlwt ) ascwt ) = To B ! sinkwt )
8.5 = 5 . ftp.B?wsinkwtltjds ) =
= - ytu.
Bolw sinkwt ) ts Is ( s'
- a' ) =
⇒µ
Biwsinfrwt )= 2
. : Tzu = -8.5a
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