# Deriving a formula for magnetic

• May 9th 2008, 05:18 PM
joeblack437
Deriving a formula for magnetic
Hello, it says that we must derive m = qB^2r^2/2V ..(volts)
using formula for kinetic energy and E= kq/r^2

thanks in advance for the help
• May 9th 2008, 05:49 PM
mr fantastic
Quote:

Originally Posted by joeblack437
Hello, it says that we must derive m = qB^2r^2/2V ..(volts)
using formula for kinetic energy and E= kq/r^2

thanks in advance for the help

I can't make sense of your question. Is the charge in a magnetic field? Electric field? Both? What sort of fields? Constant?

Give all the of the question.
• May 9th 2008, 05:53 PM
joeblack437
A mass spectometer accelerates ions through a voltage, V, before the ions enter a uniform magnetic field, B, that makes them turn. Develop the forumla m= qB^2r^2/2V for the mass of the ion, using kinetic and charge energy. the radius of the turn is r and the charge is q
• May 9th 2008, 06:15 PM
mr fantastic
Quote:

Originally Posted by joeblack437
A mass spectometer accelerates ions through a voltage, V, before the ions enter a uniform magnetic field, B, that makes them turn. Develop the forumla m= qB^2r^2/2V for the mass of the ion, using kinetic and charge energy. the radius of the turn is r and the charge is q

K.E. = qV when ion enters magnetic field: $\frac{mv^2}{2} = qV$

$\Rightarrow mv^2 = 2 q V$ .... (1)

In magnetic field: $F = \frac{mv^2}{r} = q v B \Rightarrow mv = r q B$

$\Rightarrow v^2 = \frac{r^2 q^2 B^2}{m^2}$ .... (2)

Substitute (2) into (1): $\frac{r^2 q^2 B^2}{m} = 2 q V$

and I'm sure you can make m the subject and simplify.