# How is charge/mass ratio derived in this example?

1. Oct 24, 2012

### SlothNast

How is charge/mass ratio derived in this example?!?!

Hello, I know there are many ways to derive an expression for "q/m", but I am having some troubles with this one. We are given:

(1/2)mv^2 = qV (mass, velocity, charge, voltage)

qvB = (mv^2) / r (charge, velocity, mag field, mass, velocity, radius of arc)

r = (mv) / (qB) (radius, mass, velocity, charge, mag field)

given these equations, we are supposed to come up with:

q/m = (2V) / (B^2 x r^2) (chargemass = 2Voltage / (mag field)^2 times (radius)^2)

Its probably a very simple step, I just don't see it. Any help is greatly appreciated, Thanks!

2. Oct 24, 2012

### Dickfore

Re: How is charge/mass ratio derived in this example?!?!

You have 2 eqns. and 6 quantities (mass, charge, velocity, voltage, radius and mag. field), but charge and mass in both of the eqns. occur as a combination x = q/m. In terms of this ratio, your eqns. read:
(1/2) v2 = x V

x B = v/r

You can use 2 eqns. to eliminate 2 out of 5 unknowns. You do not know x! Also, your final result does not contain velocity v. So, I would suggest you solve the 2nd eqn. for v, substitute that in the 1st eqn. and then solve it for x = q/m. Your answer should depend on r, B and V.

Edit:
Whenever you use 2nd Newton's Law (2nd eqn.) or the work-energy theorem (1St) and the only forces are electromagnetic (proportional to the charge q of the particle), you are guaranteed that charge and mass enter as a combination q/m. This is because each of this equations is of the form l.h.s. = r.h.s. and l.h.s. is proportional to m (the m*a in Newton's Law or (1/2)m*v2 for the kinetic energy in the work-energy theorem), while r.h.s. is proportional to q (for an electromagnetic force, or the work done by the electric field).

Last edited: Oct 24, 2012
3. Oct 24, 2012

### andrien

Re: How is charge/mass ratio derived in this example?!?!

just put the value of velocity from second eqn. into first eqn.(from radius to kinetic energy)

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