Question

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Problem 5-15pts A magnetic field B points in the x-direction and electric field E points in the 2-direction. A positive particle at rest is released from the origin. a) Starting with general solution y(t) = 6 cos wt + C sin wt + + + C3; z(t) = C2 cos wt - Csin wt + C4 and using initial conditions, determine the constants G,C2, Cand C, and write the exact solution (10pts) b) Find the trajectory of the particle (5pts)

Answer 1

Page RANKA (a) As ylt) = Coswt + C2 Sinut +/Elt +62 Ř Ilt) = C2 Coswt -ci sinut tch : 40) -- As bouticle relfesed from the origin - thus C + C3=0 is 2/0=0 Particle stouts foom ouest is (0) = 0 :(0)=0 įſt)= -c, w sinut +6w Coswt +18) ZH) = -(2 wsinut - Cow Coswt

B Ý (0) = GW+ t = 0 [C2 = - 1 - 2 ; (0) = -4W = 0 [C=0 faran û legio] funca en

Page RANKA yt) = - E Sinut tet Bu Zit)= E BW Coswt t = sinut -E BW ye) - + = -E sinut Z lt) = E cosut - Squaring and adding both sides (4€) – +)* + [200]= (sw 1° {cos intense ning and ad si Trajectory of particle. This is equation of cycloid motion.