A charge moving at right angles to a uniform B (Into the page)field moves in a circle at constant speedbecause F and i are always perpendicular toyeach other.(A)(В)Figure 2:2. A proton moves perpendicular to a uniform magnetic field B at a speed of 1.00 × 107m/s and experiences an acceleration of 2.00 x 1013m/s²in the negative x-direction when its velocity is in the positive y-direction[see figure 2(A)].(a) The radius r, The cyclotron angular velocity wc, the magnitude and direction of the field can be found as follows[mp1.66×10-27kg1.60 × 10¬1°C]. Since ac =5m, and wc == 2 x 106 Rad/sqBand from the second law: qutBsin90° = mpɑcВ —трасqut= 0.021T and the direction of the field is into the page as indicated in the figure 2A(True, False).The acceleration, the velocity and the position can be written as[see figure 2(B)][4= w. -$ = wct + ¢o= wct and ro is a constantvector]ār = -ac(cos pi + sin øj) = ac(-ô)(b)āt = at(- sin pî + cos ĵ = atô = t = 0d = är + đt + āz = ac(-ê) (True, False)dzdtJ = Ut + Uz = vt(- sin øi + cos oj) + vz kvrô + Ok7 = ro + S ūdt = 7o + 2 (cos oî + sin øî)= 7o + r(cos pi + sin ø3)(True, False)K = }mū · i = m[vt(– sin pî + coss¢3) + vzk] = }m[v? sin² ø + v? cos² + + v?] = }mv?φ0)+ υ . [υ (-sin φί + cos2= constantVz%3Dūdt(d) The work done by the magnetic field is zero since dWFB· dr = (qữ x B) · dř(qữ x B) • vdt = 0 This implies that the magneticfield can not change the speed[or kinetic energy] of the particle(True, False)

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2. A proton moves perpendicular to a uniform magnetic field B at a speed of 1.00 x 107 m/s and experiences an acceleration of 2.00 x 1013 m/s? a the negative x-direction when its velocity is in the positive y-direction[see figure 2(A)]. (a) The radius r, The cyclotron angular velocity we, the magnitude and direction of the field can be found as follows[m, = 1.66x 10-27kg, q = 60 x 10-19C). Since ac = = 5m, and w. = t = 2 x 106 Rad/s = gB and from the second law: qv, Bsin90° = m,a. B = ac m puc = 0.021T and the direction of the field is into the page as indicated in the figure 2A (True, False). qut The acceleration, the velocity and the position can be written as[see figure 2(B)]| = we - d = wet + do = wet and fo is a constant ector] är = -ac(cos pi + sin oj) = ac(-p) (b) ät = at(- sin oi + cos oj = a,d = duL = 0 - ā = är + ất + ãz = ac(-p) (True, False) ä, = duz k = 0 i = it + = vi(-sin pi + cos oj) + vzk = vi6 + ok = ved 7 = 7o + S ūdt = ro + " (cos oi + sin oj) = ro + r(cos oi + sin oj) (True, False) K = }mũ i = m[vt(-sin oi + cos oj) + vz k] [vr(-sin oi + cos oi) + vz k] = }m[v? sin? o+ v? cos² + v?] = } mv; = constant

2. A proton moves perpendicular to a uniform magnetic field B at a speed of 1.00 x 107 m/s and experiences an acceleration of 2.00 x 1013 m/s? a the negative x-direction when its velocity is in the positive y-direction[see figure 2(A)]. (a) The radius r, The cyclotron angular velocity we, the magnitude and direction of the field can be found as follows[m, = 1.66x 10-27kg, q = 60 x 10-19C). Since ac = = 5m, and w. = t = 2 x 106 Rad/s = gB and from the second law: qv, Bsin90° = m,a. B = ac m puc = 0.021T and the direction of the field is into the page as indicated in the figure 2A (True, False). qut The acceleration, the velocity and the position can be written as[see figure 2(B)]| = we - d = wet + do = wet and fo is a constant ector] är = -ac(cos pi + sin oj) = ac(-p) (b) ät = at(- sin oi + cos oj = a,d = duL = 0 - ā = är + ất + ãz = ac(-p) (True, False) ä, = duz k = 0 i = it + = vi(-sin pi + cos oj) + vzk = vi6 + ok = ved 7 = 7o + S ūdt = ro + " (cos oi + sin oj) = ro + r(cos oi + sin oj) (True, False) K = }mũ i = m[vt(-sin oi + cos oj) + vz k] [vr(-sin oi + cos oi) + vz k] = }m[v? sin? o+ v? cos² + v?] = } mv; = constant