An alpha particle 4He nucleus has mass M and charge q There

An alpha -particle (4^He nucleus) has mass M and charge q There is a uniform magnet is field 13 pointing in the +z direction. The a -particle is moving in the x-y plane with speed v M = 6.G4 middot 10^-27 kg q = 3.20 middot 10^19 C B = 0.100 T u = 2.00 10^6 m/s What is the magnitude of the magnetic force acting on the particle If there are no other forces on the particle, what is the of its acceleration?

Solution

a)

Magnetic force is given by :

F = q*v* B = 3.2*10^-19*2*10^6*0.1

= 6.4*10^-14 N <-----answer

b)

F = m*a

So, a = F/m

= 6.4*10^-14/(6.64*10^-27)

= 9.64*10^12 m/s2

c)

For circular motion,

the centripetal force is provided by the magetic force

So, mv^2/R = qvB

So, R = mv/qB

= 6.64*10^-27*2*10^6/(3.2*10^-19*0.1)

= 0.415 m