Finding the speed of electrons in a magnetic field So I'm trying to solve this problem in which an electron beam is "drawing a picture" on a TV screen. The electrons are accelerated to a voltage of $3 kV$ by wire coils and are then directed to different points on the screen. The wires produce a magnetic field of up to $0.68 T$.
The following equations are the ones we've been using in class to solve magnetic field problems:
$F=Bqv$, where F is the force of the magnetic field, B is the magnetic field strength, q is the charge and v is the velocity.
$r =\displaystyle{\frac{mV}{qB}}$, where $r$ is the radius of the magnetic field and $V$ is the voltage.
Rearranging these equations to solve for $v$, we get $v=\displaystyle\sqrt{\frac{2q\Delta V}{m}}$.
I thought I was on the right track by dividing (2 $\times$ the charge of an electron $\times$ the voltage) by the mass of an electron, and then square rooting everything. But I got the incorrect answer. Am I using the wrong equation? This question threw me off as it isn't circular motion like many magnetic field problems.
 A: There are two different values represented by $v$ in this problem:


*

*lowercase $v$ is the velocity of the electron

*uppercase $V$ is the voltage that accelerates the electron


In both the force equation $F=qvB$ and the radius equation $r = mv/qB$, $v$ refers to the velocity of the electron. I believe this is where your mistake was, since you said that the $v$ in the radius equation was the voltage.
You were on the right track to find the velocity of the electron using the voltage. The kinetic energy of the electron is
$$KE = \frac{1}{2}mv^2 = qV$$
(notice the difference between $v$ and $V$). Solving for $v$ here gives the same expression you got.
$$v = \sqrt{\frac{2qV}{m}}$$
If you just want the speed of the electron, you're done. If you want the radius of the electron's path, use the expression for $r$ with the velocity just calculated.
A: In the equation provided for $r$, there is a mistake. The equation is got by equating the centripetal force acting on the charge with the magnetic force on the charge since the circular motion is provided by the magnetic field.  
So, $$r=\frac{mv}{qB}$$  
the $v$ is the velocity of charge, not the voltage. To calculate the velocity, you are provided with the electron acceleration. The kinetic energy acquired by an electron when it travels through one volt potential difference is one electron volt. So, equate the kinetic energy to $qV$. From that you will get the required expression for voltage. that equation seems correct in your question
