- #1
Travkid
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Homework Statement
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?
Homework Equations
r=mv/lqlB
K=VQ
Conservation of Energy: Total Energy = KE + PE
KE = 1/2 mv^2
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!