How Do You Calculate an Electron's Motion in a Magnetic Field?

[exitwound]

Homework Statement

An electron of kinetic energy 1.20keV circles in a plane perpendicular to a uniform magnetic field. The orbit radius is 25.0cm. Find a.) The electron's speed. b.) the magnetic field magnitude c.)the circling frequency, and d) the period of the motion.

Homework Equations

$$F=m\frac{v^2}{r}$$
$$qvB=m\frac{v^2}{r}$$
$$r= \frac{mv}{qB}$$
$$T=\frac {2\pi r}{v}$$

The Attempt at a Solution

1.2 keV = 1.92 x10^-16 Joules

I don't know what to do to find the velocity. I know if it's moving, it's kinetic energy can be linked to the velocity somehow. It's acceleration is v^2/r but that's all I know.

Anyone?

 

[thaer_dude]

k=0.5mv²

you should know that formula

 

[exitwound]

That's what I did FIVE TIMES and kept getting the wrong answer. I do it NOW after you've responded and it's correct. I was absolutely sure that .5mv^2 was the right formula to use. I don't know what I was doing wrong.

$1.92x10^{-16}=.5(9.1x10^{-31})(v^2)$

$v^2=\frac{(1.92x10^{-16})(2)}{(9.1x10^{-31})}$
$v=2.05x10^7$

sorry to bug you.

 

[thaer_dude]

oh well :p