Download The problem states

Please review my solution to the problem and explain in detail what I may be doing wrong and what concepts I
may not be applying correctly.
The problem states:
An alpha particle has a charge of +3.2 x 10 -19 C and a mass of 6.6 x 10-27 kg. The alpha particle travels at a
velocity v of magnitude 550 m/s through a uniform magnetic field B of magnitude 0.045T. The angle between v
and B is 52º.
a) What is the magnitude of the force FB acting on the particle due to the field?
b) What is the acceleration of the particle due to FB?
c) Does the speed of the particle increase, decrease, or remain the same?
We know:
V = 550 m/s
B = 0.045T
Q = +3.2 x 10-19 C
Mass = 6.6 x 10-27 kg
Φ = 52º
Solution:
a)
B = 0.045T = 0.045 N/Cm/s
FB = qvB
FB = (+3.2 x 10-19 C)( 550 m/s)( 0.045 N/Cm/s)
FB = 7.92 x 10-18 N
FB =q (v X B) = qvB*sin
FB = (+3.2 x 10-19 C)( 550 m/s)( 0.045 N/Cm/s)(sin520)
FB = 6.24 x 10-18 N
(the force acting on the charge particles in magnetic field only if they have velocity perpendicular to the
magnetic field. If the velocity of the particle is in the direction of the field, no force will act on it. .Hence due to
the component of the velocity parallel to the field no force will be there. It is only due to the component of the
velocity perpendicular to the field and this is v*sin For v perpendicular to field  is 90 and sin is 1 then F=qvB
b)
FB = │q│vBsinΦ
sinΦ = sin(52º) = .7880
v = 7.92 x 10-18 N / [(3.2 x 10-19 C)( 0.045 N/Cm/s)( .7880)
v = 6.9797 x 102 m/s
Acceleration of a particle is given by Newton’s law of motion a = F/m
And will be,
a = (6.24 x 10-18 N)/( 6.6 x 10-27 kg) =9.45*108 m/s2
c)
The speed stays the same because FB cannot change the particles speed, just direction.
Because the force is always perpendicular to the direction of motion.
Correct,