Question

An alpha particle (charge of +2e) moving with a speed of 5.5E6 m/s enters a region of space where a uniform magnetic field of 5.0 T is directed right to left. If the angle in the image is 40°, what are the magnitude and direction of the magnetic force that acts on the alpha particle? B e V 5.7E-12 N, out of the screen 6.7E-12 N, out of the screen 3.5E7 N, into the screen 6.7E-12 N, into the screen 5.7E-12 N, into the screen

Answer 1

Force on the charge particle is given as

F= q( v×B)= qvB *sin( theta)

So, F = 2*1.6*10^-19*5.5 *10^6* 5 *sin(40°)

F= 56.56*10^-13

Or F=5.7 *10 ^-12 N.

Putting your right hand fingers along v and curl it towards the B, we get force Direction, so force is in to the screen.

So last option is correct.