1. The problem statement, all variables and given/known data An electron accelerated from rest through a voltage of 420 V enters a region of constant magnetic field. If the electron follows a circular path with a radius of 29 cm, what is the magnitude of the magnetic field? 2. Relevant equations r=mv/lqlB K=VQ Conservation of Energy: Total Energy = KE + PE KE = 1/2 mv^2 3. The attempt at a solution So I know I need to use r=mv/qB to find the Magnitude of the Magnetic Field. r= .29m m=9.11x10-31 kg q= 1.6x10-19 v= ??? B= what we're solving for. If I can find the velocity of the electron, then I will be able to solve for the Magnitude of the Magnetic field. I tried looking online for other solutions and someone said this, but I can't really understand it that well: From conservation of energy we can calculate the velocity K = V*q so 1/2*m*v^2 = V*q...so v = sqrt(2*V*q/m) = sqrt(2*410*1.60x10^-19/9.11x10^-31) = 1.20x10^7m/s Now we have F = m*a ....q*v*B = m*v^2/r What is this K=vq equation? I couldn't find it when I was looking for it. It's the Voltage x The Charge? Because we know the particle starts at rest, we know that the potential energy = kinetic energy. But I really don't know how to find the velocity. Help please!