# soln6 - 1(a Ions accelerated through potential difference V...

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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(4.68x10-26)(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 right-angles 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 right-angles to the ﬁeld B. N = mg - Fcose F sit-10 = .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 right-hand 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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