An α-particle has a charge of +2e and a mass of 6.64 × 10-27 kg. It is accelerated from rest through a potential difference that has a value of 1.20 × 106 V and then enters a uniform magnetic field whose magnitude is 2.14 T. The α-particle moves perpendicular to the magnetic field at all times. What is (a) the speed of the α-particle, (b) the magnitude of the magnetic force on it, and (c) the radius of its circular path?

the velocity you are looking for is made up of two parts: horizontally velocity, vh, and its vertical velocity, vv. the actual velocity of the ball will be these two parts added together, and because velocity is a vector then they will need to be added vectorially.  

neglecting air resistance etc etc.,  

the horizontal velocity component will be constant (because there is no accelerating force in the horizontal direction.  

vh = 8 ms-1  

the vertical velocity component has an accelerating force: 'g'.  

using v = u +at where a = g = 9.81 ms-2.  

the vertical component of velocity, vv = 0 + gt = vv =9.81 x 2  

vv = 19.62 ms-1 . the zero appears because the virtical component of velocity at time t=0 is zero. remember, the ball was thrown horizontally.  

how fast it will be moving (it actual path (being a parabola )) will be the vector sum of these two.  

adding vectorally gives: -  

v = sqr root (8^2 + 19.62^2) = sqr root(64+384.9) = 21.18 or 21.2 ms-1

b

explanation:

need to see circuit to get time constant

grav forces affect all objects magnetic forces affect magnetic objects and electric currents