Question

A particle with charge of 2 (HC), 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% (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

Solution a given С q= 2llc = 2x10 m = 2x1020 kg 29 = 200 m18 B = 0.06 T ( Let the angle between veelocity and magnetic field direction is co?: Magnetic force actieng on a charged particle is given by E = q (0 xB) F=_qub sino fra = xÊ magnitude of force is fa que B sino. given E- 1:01 x 105 N 1.01x10.5 = qu B. sino the but thus sino = 1.01x105 19 В 1.OLX105 2x 106 x 200x0.06 -5 1.01 x 10 4x6 x 106 10.1 24 sino. -0.42

sin' (0.42) c = 24.83 Answer: (B it charged bastidie móvės a circular are when it goes through the magnetic field. Thus for circulas notion , F- mar {van=923 Thus acceleration of charged particle ara E q VB sino こ mo2= 71 - m m 2x106 x 200 x0:06 x0.42 a ar = m 5 a= 1:01 x 10 2x2020 a = 6.05 xł014 m 82 Required answer

B q m there 2= ve ceo sino 2 + coso BG and os then force F = 9.10 x (10 XB² ) = qv Csino i teosgo ŷ IX Bý f'= 920 B sino Thus force acts on the charged particle in Ê direction. (lie perpendicular to the blane of the paper). F=90B sino こ 1.01 XLO thus. F' = 1:01 x1 K N. This this the force actiøng on the charged particle instanceusly when charged particle enters in the magnetic field.

D from part (A) the force actieng on the particle when it enters enters in the magnetic field. ei F = 9,20 B sino. here 66.90B is constant . Thus for maximum force fmar, sino sino 'should also be maxeimum. (sin o) Thus =1 max Ⓡ = 90 answer Thus velocity vector of charged barticle should be perpendicular to magneti'o field maximin force the charged particle is vector and acting on Fmax =9UB sin(go) = 923 Fmax こ 2x106 X200 X 0.06 fmax = 2:4 x 105 N. Answer