Question

A positron with kinetic energy 2.40 keV is projected into a uniform magnetic field of magnitude 0.110 T, with its velocity vector making an angle of 80.0° with the field. Find (a) the period, (b) the pitch p, and (c) the radius r of its helical path

answer with units please

Answer 1

We will first find the velocity of the positron
The kinetic energy of the positron is given K = 2.40 keV
K = (1/2) m v²
v = sqrt (2K / m)
v = sqrt (2 x 2.40 x 10³ x 1.6 x 10^-19 / 9.11 x 10^-31)
v = 2.90 x 10⁷ m/s
a) The period of the revolution
T = 2 r / v_h
Where v is the horizontal velocity v_h  = v sin ,   is the angle with magnetic field
T = 2 r / v sin
The radius of the motion
r = m v_h / qB
r = m v sin / q B
Then T becomes
T =   2 m / q B
T = 2 x 3.14 x 9.11 x 10^-31 / (1.6 x 10^-19 x 0.110)
T = 3.25 x 10^-10 s
b) The pitch is the distance traveled along the magnetic field in a time interval of one period
d = v_v x T
Where v_v is the vertical velocity
d = v cos x T
d =  2.90 x 10⁷ m/s x cos 80.0 x 3.25 x 10^-10 s
d = 1.637 x 10^-3 m
c) The radius of the helical path is
r = m v_h / qB = m v sin / q B
r = 9.11 x 10^-31 x 2.90 x 10⁷ x sin 80 / 1.6 x 10^-19 x 0.110
r = 1.478 x 10^-3 m