What is the speed of the charged particle as it enters the region of the magnetic field?

Question

A particle of mass 1.1454X10^-25 kg. and charge of magnitude 4.8 x 10^-19 C is accelerated from rest in the plane of the page through a potential difference of 218 V between two parallel plates. The particle is injected through a hole in the right-hand plate into a region of space containing a uniform magnetic field of magnitude 0.138 T oriented perpendicular to the plane of the page. The particle curves in a semicircular path and strikes a detector.
What is the speed of the charged particle as it enters the region of the magnetic field?

Answer 1

This is just equating the electric field to the Kinetic Energy (as the particle is just entering the field).

q V = 1/2 m v^2

V - Potential
v - velocity

So we have-:

4.8 x 10^-19 * 218 = 1/2 * 1.1454 x 10^-25 * v^2

v^2 = 1827134625

v = 42744.99533

or v = 4.2744 x 10^4 metres per second

If you want to work out what happens to the particle inside the field use-:

Force = q v b (sin theta) = m v^2 / r

(you are not given an angle so assume sin theta = 90 -: thus = 1 and can be removed)

divide by v to give-:

q b = m v / r

You can now work out the radius of the particles circular motion (as you have the magnetic field strength and the other variables values apart from 'r')

Good Luck.......