# Relativistic Kinetic or Classical?

1. Jun 24, 2014

### MostlyHarmless

1. The problem statement, all variables and given/known data
Suppose you want to make a velocity selector that allows undeflected passage for electrons whose kinetic energy is $5x10^4eV$. The electric field available to you $2x10^5V/m$. What magnetic field will be needed?

2. Relevant equations
$u=\frac{E}{B}$
u is velocity, E is the electric field, B is the magnetic field.

3. The attempt at a solution
So I'm trying to using the Kinetic Energy to solve for u, and then solve for B.

At first I tried using the classical equation for kinetic energy, which gave me ~.44c. But this value for u, would make the Classical formula not very accurate.

So I tried the relativist formula: $E_{kin}=mc^2(\gamma-1)$ and worked my way down to $u = c\sqrt{1-\frac{1}{\frac{E^2}{(mc^2)^2)}+1}}$

In this case, I end up with ~.01c which is not a relativistic speed. So my question is, which should I use? And obviously I'm doing something wrong, any obvious mistakes with that last equation? Is there an easier way to deduce the magnetic field given the kinetic energy and electric field?

2. Jun 24, 2014

### Oxvillian

I think it's just an algebra error. I got a very slightly different formula at the end

3. Jun 24, 2014

### maajdl

First, I had to understand what a velocity selector is.
Here is a clear picture from Wikipedia:

Indeed, the solution boils down to calculating the velocity of a beam of electron at the enrgy of 5.104eV.
The energy is related to the speed by this formula:

Solving, you should find v/c = 0.412687

v²/c² = (1 - 1 / (Ek/mc² + 1)²)

The classical formula is not so far from the relativistic result.
Anyway, the relativistic formula is more general and is always applicable.
The classical formula is an approximation.

Last edited: Jun 24, 2014
4. Jun 29, 2014

### MostlyHarmless

I think in the frustration I broke an algebra rule, and did: ((E/mc^2)+1)=(E/mc2)^2+1^2

>.<