Question

Consider some electrons undergoing cyclotron motion. These electrons can be sped up by increasing the magnetic field; the accompanying electric field will impart tangential acceleration. Suppose we want to to keep the radius of the orbit constant during this process. Show that we can do this by designing a magnet that produces a field such that the average field over the area of the orbit is twice the field at the circumference. Assume that the electrons we have start from rest in zero field, you may also assume that the apparatus we have is symmetric about the center of the orbit. You can also safely assume the velocity of the electrons remains well below c and as such can be treated non relativistically.
Hint - Differentiate p = QBR wrt time and utilise F = ma = qE

Answer 1

we have magnetic force f = B q v

and the f = m v² / R

B q v = m v² / R

q B R = m v

differentiating with respect to time is

q R dB/dt = m dv/dt

a = dv/dt

q R dB/dt = m a

F = ma

F = q R dB/dt

we have F = q E

q E = q R dB/dt

E = R dB/dt --------- 1

using equation

E x ( 2 R ) = - d/dt

E = ( -1/2 R) d/dt ------------- 2

using 1 and 2 equations

R dB/dt = ( -1/2 R) d/dt

dB/dt = ( -1/2 R²) d/dt

integrating on both sides

B = - 0.5 ( 1/ R²) + C

at t = 0 and magnetic field B = 0 , C = 0

B (R) = (1/2) x ( / R²)