Question

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=vi+vj. The net force F=Fi +Fyj on the particle is the vector sum of its weight and the magnetic force.

A) Using Newton's second law, write equations for a_x and a_y, where a⃗ =axi^+ayj^ is the acceleration of the particle. Express your answer in terms of the variables q, B, v_x, v_y, and m. Enter your answers separated by a comma.

B) Differentiate the second of these equations with respect to time. Then substitute your expression for a_x=dv_x/dt to determine an equation for dv²_y/dt² in terms of v_y. Express your answer in terms of the variables q, B, vy, and m.

This is all the info that appears on the question.

Answer 1

P = mgh t q(WXB) 2 (B) Vy (mg) (ux ît vg ) x (BR) = (-VeВ f t vyBP) | P- m 2 + %. ř - GV y B) ↑ + (mg qu x B) ja ma au = (avy B) ay = (mg qV x B) VU dt V - Vụ t (Atto, Vezo 2. cy the truth aya melihat duyz = g - q²vy get a m m2 e m gf vy to countered vete Of not, olum = 2 2 2 m2