Question

Problem 3 A particle with charge of 2 (uc), mass of 2X10-20 (kg) and velocity of 200 (m/s) enters into a magnetic field of 0.06 (T). The particle experiences a force of 1.01x10-5 (N) in the magnetic field. A) Calculate the angle between velocity and magnetic field directions. B) Calculate the acceleration of charged particle in the magnetic field. C) Draw a diagram showing direction of velocity, magnetic field, force and the angle between velocity and magnetic field, using RHR-1 D) In order to maximize the force, what should be the angle between velocity vector and magnetic field direction? What would be the value of the maximum magnetic force at this angle?

Answer 1

1. 
   ③ 9 = 2x106c 1189 me 2 x 10-20 Geiven Deeta. V- 200 mis B-0.06T F = lope 105N Part II As we know magnetic magnetic force f = q VB sin e is Sine o qVB Sintl 1101x10.5 2x10-6x200x 0.06 © - Sin-' (42083) 10 = 24.89°/> angle Hence, bu velocity & magnetic fic field direction.
   
   Part (B) As we also known r = ma 101x10-5 - 2x10-20% a Acceleration las 5.05x 1014 m/s² Y Parte) 7B 241890 x n z Pout (d) for maxime 'zing the force the angle should be 90°, at that value of sing will me maximum ire. l. Sin 90°) Fmax = q VB Sin 90' = 2x1067 200x0.06 Fmax = 2,4x10-5 N