An electron with a kinetic energy of 22.5 ev moves into a region of uniform magnetic field b of magnitude 4.55 x 104 t. the angle between the directions of the magnetic field and the electron's velocity is 65.5 degrees. what is the pitch of the helical path taken by the electron?

9.17 x 10^-10 m

Explanation:

q = 1.6 x 10^-19 C

B = 4.55 x 10^4 T

m = 9.1 x 10^-31 kg

K = 22.5 eV = 22.5 x 1.6 x 10^-19 J = 3.6 x 10^-18 J

θ = 65.5°

Use the formula for the kinetic energy

Let r be the radius of circular path and v be th evelocity

By substituting the values

r = 3.52 x 10^-10 m

Use the formula for the speed

v = 2.82 x 10^6 m/s

The pitch of the helix =

 =

= 9.17 x 10^-10 m

force is mass*acceleration.

we have a very small mass: 11.4*0.001 kg

if we want 0.035n of force, we need an acceleration:

0.035 = 11.4*0.001*a

a = 0.035/(11.4*0.001) m/s^2

i think it would be 5