# A particle with charge q and kinetic energy K travels in a uniform magnetic field of magnitude B. If the particle moves in a circular path of radius R, find expressions for each of the following

A particle with charge q and kinetic energy K travels in a uniform magnetic field of magnitude B. If the particle moves in a circular path of radius R, find expressions for each of the following. (Use any variable or symbol stated above as necessary.)

(a) its speed
v =

(b) its mass
m =
asked Feb 10, 2012 in Physics

Start by writing the equation for magnetic force:

magnetic force = q*v*B

Now write the equation for centripetal acceleration:

a = v^2 / R

Force = m*a

plugging in a we get:

Force = m* v^2 / R

Now set magnetic force equal to force:

q*v*B  = m* v^2 / R

Now we need to use K.

K = (1/2)m*v^2

multiplying by 2 this becomes:

2K = m*v^2  , which we can now subsitute into the equation above and get:

q*v*B  = 2K / R

Solving for v we get:

v = 2K/ (R*q*B)

Part B, similarly we can calculate mass by using the force equation again:

Force = m*a

and so solving for m we get:

m = Force / a

Where Force = q*v*B furthermore the v can be replaced for what we got in part A.

so Force = q*(2K/ (R*q*B))*B

and a = v^2 / R, furthermore the v can be replaced for what we got in part A.

so a =  (2K/ (R*q*B))^2) / R

plugging into the equation for m we get:

m = *q*(2K/ (R*q*B))*B) / (2K/ (R*q*B))^2) / R

after simplifying we get that:

m = (R*q*B)^2 / (2K)
answered Feb 18, 2012 by ~Expert~ (3,856 points)
selected Feb 18, 2012 by Joey33
a) v=2K/(RqB)

b) (RqB)^2/2K
answered Feb 15, 2012 by ~Rookie~ (80 points)