Question

A particle (q= 5.0 nC, m = 3.0 mg) enters a region where the magnetic field has components By = 0, By = 0 and B, = 4.0 T. The initial velocity of the particle as it enters this region is given by Vx - 7.0 m/s, Vy = 6.0 m/s and vz = 8.0 m/s. (a.) What is the magnitude of the magnetic force and the magnitude of the acceleration of the particle (b.) What is the radius and pitch of the helical trajectory of this particle?

Answer 1

S02n :) Given that, charge of particle =9=5nc 9 =5810-9c mass of particle 3mg = 3x10-6 kg By=0 Bx=0 Be=UT Vx=7 ml Vy=.6mis V2=8 m/s z О, ЧТ V=8m/s Vx=7 m/s x Vy=omis 3 Force у Magnetic on a moving charge is given by, ² = 9(98) So, the force charge R = q (Vx XB²) + q (uA X8%) q (vx x B2) will be. on

G F = (5x109) 7 (4) SX10-976) (4) 1 F = QX10-8 (79-6î) TJ - IF 18.43 X 100 N (acceleration of particle) To = m -8 lal 6.14 x 10-2 m/s2 18.434101 3 x 10-6 | ਕੇ। - G.14 X 102 m/s2 (6) Radius of Helical Path is given by R= mv. 09 where Ve is the herhendicular velocity with the magnetic field. Ve=5v² + vy? 72 +62 z C 9.21 m mis

R= 3x10-6 X g. 21 4 X 5 X10-9 3 3x9:21 x 10 20 R = 1381. S m Pitch = Vzxt where t is time to complete one circle of radius R with Velocity 9.21 m/s sheed & time. QIR= 9.21 t Distance = = ta 2x 3.14 X 1381.5 9.21 t guz seconds Pitch Vzxt 8 X 942 Pitch. = 7536m