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Unformatted text preview: Problem 6.12 z
 . Eh
(a)M=ksi;Jb=VxM=—k¢;K5=Mxﬁ=kR¢. "
B is in the z direction (this is essentialiy a superposition of solenoids). So
Use the amperian loop shown (shaded)—inner side at radius 3: I f3 ' d1 = B! = saIenc = no ”Jo da + Kai] = #11 [—kl(B.  s) + le] = pickle. (b) By symmetry, 11 points in the z direction. That same amperian loop gives fII . d] = H: = #01:.“ = 0: since there is no free current here. So  and hence Outside M = 0, so B = 0; inmde M = ksﬁ, so B = poksi. 13 me _ _ __ ink, ate—a,
=J°=me:3:s( 27s)” KFM“ Total enclosed current, for an amperian 100p between the cylinders: Hoa + Xmﬂﬁ ( I+X L12'1”»:(1+)(m)l', sodel: 1.1013“: po(1+xm)I=:B= 2“ 21m l’roblem 7.1 (a) Let Q be the charge on the inner shell. Then 13— _
“KE'dr=—4—1renQLurJ5dr=rarq$(%_% _ . _o . = __a41reo(an4)_1W(%—%) —%r in the space between them, and (Vn— Vb)— — 4—1565 _V..—v,,_ 1 1 1
{b)R_ I _47rcr(a b)‘ (c) For large b (b :s a), the second term is negligible, and R—  1/41raa. Essentially all of the resistance' 15 in
the region right around the' mner sphere. Successive shells, as you go out, contribute lee and lessi because the
crosssectional area (4in2) gets larger and larger. For the two submerged spheres. R—  = n(one R as the current leaves the ﬁrst, one R as it converges on the second). Therefore I: H‘V/R — Problem 7. 2
(a) V: QIC‘: IR Because positive I means the charge on the capacitor is decreasing, i314 =—1—«esoom=ooe m Bucoa= 000 =cvo.so Q(t)= warm. one on 2 00
(b) W = The energy delivered to the resistor is [O Pd: 2 fa IaRdt = V—"j eqclfﬂcdt =
u Visa RC —2:/nc 00H 1 2
E(—Te )0 5015. (
(c) VI, = Q/0+IR. This time positive 1 means Q is increasing: ﬂ = I—  1 —(C‘Vo — Q) =>— d—Q
cit RC Q— Ova: 'Rl—Ctdt => 511“?  CVo) = "kl—Ct +constant =:~ Q(t) = CV0 + kc"/Rc. But 62(0) = 0 => k = ~C'Va, so at) =cvo (1 —e““‘") 1(t)= $9 = 6% (R—Ge W”) = . _ “124,ch _ V_u2 _ —£/RC °° _ V_0’ _ 2
(:1) Energy from battery. ["00 VDI dt — E 0 e dt — R ( RCe )io — R RC_ Since I(t) is the same as in (a), the energt delivered to the resistor is again The ﬁnal energy in the capacitor is also so the energy from the battery goa to the capacitor, and the other half
to the resistor. Problem7.5
5 2 63R dP 2[ 1 ZR]
= _. : =__= _ =0: R=2R R=.
I r+R’P IR (r+R)2’dR ‘9 (7'+R)2 (H.313 r+ :’ T Problem 7.8 s+a
. . upoI _ _ml 1 _ ,uoIa 8+0.
(3) Theﬁeld oflong w1re1sB——2m¢,so(I>—[B da—— s(ads)— 21r ln s . _ dII>_ poled 3+0, ds_ hole 1 ds lds _ #01621!
(ma dt_ 21: dz'“( s),anddt—1J,so 21r (8+adt salt) The ﬁeld points out of the page, so the force on a charge in the nearby side of the square is to the tight. In
the far side it’s also to the right, but here the ﬁeld is weaker, so the current ﬂows (c) This time the ﬂux is constant, so Problem 7. 11 5 = Btu = IR => I = ﬁlo => upward magnetic force = NB 2 $3. This opposes the gravitational force
downward: 3212 _ do_ do_ _leg _ _9 MR
mg— R v_mE,E—y—ow,wherea=m. g—a‘vg—Oévg—E= sz‘ d
U =dt=>—lln(g—cw)=t+oonst.=>g—au=Ae‘°‘; att=0, v20, soil=9.
9—01: a At 90% 0f terminal velocity, 11/11; = 0.9 : 1 — 6““ :> 3““ = 1  0.9 = 0.1; 111011) = —at; In 10 = at; t: 5:111 10, or 190% = %lu 10. Now the numbers: 171 = 417111, where r: is the mass density of aluminum, A is the crosssectional area, and
l is the length of a side. R = 41an, where a is the conductivity of aluminum. So p=2.8 x10‘aﬂm _ 4nAlg4l _ 16719 __ 16911.0 and g = 9.8 mls2
”‘ ‘ Aaezlz ' 0'33 _ B2 ’ n = 2.7x milks/m3
B = IT I _
So u: = $1519.st rxioﬂlgzsmo 1 ___ 190% = 1.235130 2 1n{10) =
If the loop were cut, it would fall freely, with acceleration g. \ {mace
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 Summer '11
 Agshe
 Magnetism

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