Relativistic Kinetic or Classical for Velocity Selector Design?

[MostlyHarmless]

Homework Statement

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?

Homework Equations

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

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?

 

[Oxvillian]

and worked my way down to $u = c\sqrt{1-\frac{1}{\frac{E^2}{(mc^2)^2)}+1}}$

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

 

[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 energy 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.

 

[MostlyHarmless]

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

>.<