Question

<Chapter 27 Problem 27.74 9 of 9 > A Review | Constants Part C A particle with charge q and mass m is dropped at time t = 0 from rest at its origin in a region of constant magnetic field B that points horizontally. What happens? To answer, construct a Cartesian coordinate system with the y-axis pointing downward and the z-axis pointing in the direction of the magnetic field. At time t o the particle has velocity v = vzi+vy). The net force F= F,{+Fyj on the particle is the vector sum of its weight and the magnetic force. This result shows that vy is a simple harmonic oscillator. Use the initial conditions to determine vy(t). Write your answer in terms of the angular frequency w=qB/m. Note that (dv/dt)o = 9. Express your answer in terms of w. V ADO A O O ? vy(t) = Submit Request Answer Part D Substitute your result for vy(t) into your equation for dux/dt. Integrate using the initial conditions to determine vrt). Express your answer in terms of w. 50 ADD O O ?

Answer 1

charge =q, mass=m solution: At t=0, position ř= 0;topt ok B = B x Zanis yanis At time t, let the relocity v = valt lyjt vale E = Exit Fyj + FzR = a[ux B) + mgj F = $(-V 7 + V, Bí to) = fq =0 for all t.. since V₂ =0 initially and fz =0, V₂=0 for all t. = му в, г - avд в m due to a qBuy,. in dry ya q B Ux since as a w. dina - Wuy, do y = - Vx everything we doen het en die - We Vy. d²ly ata=-w2vu 'y Vy varies harmonical - 5

Vy = C Coswt + D sin wr At tao, Vy = 0 Đ C=0. Attro de los ang > DW = g > D- Hance Vy = q sinut Paute : Vy LE) = å sinet wuy = g swt jde = sig sinart at Valt) = 9(-oswt) = 3 (0-as Wit) Part D ve CH = (i-coswt).