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Unformatted text preview: 1. (a) Ions accelerated through potential difference V.
vaKZ : qV
. v = [2q\r’;"m]l'r2 Ions move in circular path in the magnetic ﬁeld B, with magnetic force
providing the centripetal acceleration. mVZ/R = qu R = mquB x = 2R = 2m [zqwm] ”2qu 2 (2/B)[2m‘Wq] ”2 03) Mass of (:0 ion = 28(1.67x10'27) = 4.63;:10‘26 kg
Mass ofOz ion = 32(1.67x10'”) = 5.34x10‘2" kg
Singly—ionized, so q = e = 1.6x10"9 C
v=1000v,andn=0.2r
x(CO) = (2f0.2)[2(4.68x1026)(1000)/1.6x10_19)]m
= 0.24 m x(02) : (2/0.2)[2(5.34x10'25)(1000M.6x10*‘9)] “2
= 0.26 m 2. (a) Let the components of velocity parallel and perpendicular to the ﬁeld be v“ and vl. The perpendicular component is responsible for the magnetic
force, and hence the centripetal acceleration. Let the radius of the circular
path in the plane at rightangles to E be R. mvffR= qviB
vi=qBRfm ..............................(1) The electrons take time t = 21:ij L to execute a single turn. During that
time they move a distance vnt along the ﬁeld lines. If the pitch is p
p = v” (21tquB) ........................... (2)
q = 1.6x10"9 (3, m = 9.1x10'3‘ kg, B = 0.005 T, R = 0.02 m, p = 0.06 m.
. vi= 1.76:;107 mfs
v“ = 3.4::10" mfs
Electron speed v = [vi2 + “[2]” = 1.953(10? mfs (b) For zero acceleration, the electric force and ﬁeld must be equal (and
opposite to the magnetic force. quB = qE
E = viB = (1.76x107}(0.005) = 88000 Vr'm = 88 kV/m (a) If V is the accelerating potential mvzfz : eV
v = [2(1.6x10"9)(3x104)/9.lx1{Turn
= 1.033008 m/s
(b) in the region of the magnetic ﬁeld the electrons follow a circular
path of radius R given by
Invz/R = evB
R = mv feB = 0.090 m
In the ﬁeld they are deﬂected a distance d1 = R  R c059
where sin 0: 003520.090, e = 23°
d. = 0.007 m The electrons subsequently travel in a straight line, and are deﬂected
an additional distance
d2 = 0.22 tame = 0.093 m
Total deﬂection = d1+ d2 = 0.100 m = 10 cm “‘3 "‘3 N
(a) (a)
(a) When B is in the vertical direction, the force F is horizontal.
N = mg F = HR = MN = mg
B = tumg/il = 0.6(0.05)(9.81)f[10(1)] = 0.294 T (b) Magnetic force F acts at angle 0 to the vertical, at rightangles to the ﬁeld B.
N = mg  Fcose
F sit10 = .uSN = us{mg  FcosB) F = usmg/(sine + tLScosB) ....................... (1)
To ﬁnd the minimum of this, set d(sin0 + moose) = 0
0050  itssinB = 0
tame = H3 = 0.6
6 = 310 , angle B makes with the vertical is 59° sine + ugcosE) = 2 sine From (1) F = 0.6(0.05)(9.81)/(0.5145+0.5145) = 0.286N F = ilB B = 0.29/[10(l)] = 0.0286 T 5. E B The circuit can be regarded as the sum of two loops, as shown, each carrying a current i. The area of the ﬁrst is 0.04 m2, and the area of the
second is 0.02 m2. The total magnetic moment is therefore a = 0.06i z 1.2 Am:
(a) Force is zero, as it is for any closed circuit.
(1)) Torque 3 = u x B, and by righthand rule g is in the z direction.
1: = as sin(40°) = (1.2)(o.5)sin(40°) = 0.39 Nm Torque I = g x E, where y. = iA is the magnetic moment of the loop.
I = i ABsine = i azBsinE}
where a is the length of a side of the square.
Using 1: 2101,
I dzeldtz = —i azB sine e i 3230
dZBIdtZ =  (i azBi'I) e
= — (1)1,2 9 Each side of the loop has mass Mi'4.
Moment of inertia of the loop about the axis 00' is
I = 2 (M/4)a"/12 + 2 (M/4)(af2)2
= Mazfﬁ
Period of SHM is T = 21m.)o
= 21: Wa2/(6iazBHm
= 2n: [M/(éiB)]”2 7. Since 3 = (31+ 4i)x105m/s and E = 0.00513, the magnetic force is
Em = qu g =(1.6x10"9)(—o.0151+ 0.0201)t105
= (3.21; 2.41)t10“6 N Sum of magnetic and electric forces is 5x10'mi, so electric force is
&=(1.81+ 2.4j)x10"6 = qg
g: (131+ 2.41)x10"6t1.6x10"9 =11251+15001
g has magnitude [(1125)24(15002]”2 = 1375 Wm, and is in the xy plane
making an angle of tan'1(1500f1125) = 53° with the x axis. ...
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 Fall '10
 ProfessorSomi

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