The cyclotron forumla is derived here(adsbygoogle = window.adsbygoogle || []).push({});

See -www.geocities.com/physics_world/cyclotron.htm

That's a lot of work for something so simple. Recall that the relativistic force can be written as

equation. Recall that

F = M_t A_t + A_L M_L

where

M_t = transverse mass

A_t = transverse acceleration

M_L = longitudinal mass

A_L = longitidinal acceleration

It can be shown that transverse mass = (relativistic) mass = m. For the cyclotron motion described in above link A_L = 0 and a = A_t = v^2/r. Therefore

F = dp/dt = M_t A_t = m A_t = ma

where a = transverse acceleration = v^2/r and p = mv = gamma*m_o*v. Therefore it follows that

F = ma = e[E + (v/c)xB]

So, in this particular instance, F = ma is 100% correct if m = relativistic mass and a is transverse acceleration. Plugging vxB = vB (B is now a magnitude rather than a vector)

ma = m(v^2/r) = vB

Canceling a factor of v and moving the r to the otherside gives

mv = vBr

p = m_o*v/sqrt[1-(v/c)^2] = mv

p = vBr

which is the relativistically correct relation for cyclotron motion.

Someone claims that this is not simpler than the normal derivation derived above [Then again the same people didn't know that F = q[E + vxB] either]. Does anyone know of a simpled/easier way to derive the cyclotron formula?

Pete

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Cyclotron formula and Transverse Mass

Can you offer guidance or do you also need help?

**Physics Forums | Science Articles, Homework Help, Discussion**