Say I have an electric field and a magnetic field aligned in the y direction of an xyz plane. What strength of $E$ and $M$ field do I need to get a proton to pass through the point $(2 mm, 3 mm, 1 mm)$ if the proton starts out with a velocity $v$ of $10^5 m/s$ in the positive $x$ direction? My solution follows:
First I find the strength of the Magnetic field required to make a particle go in a circle passing through $x=2, z=1$. The radius of the circle must be $2.5 mm$ on the z-axis (simple trig) and the radius must also be equal to $mv/qB$ since $qvB = mv^2/r$. Letting $v = 10^5 m/s$ gives $B = 0.4175 T$.
The time taken for a particle to reach the $x=2,z=1$ is equal to $t = \theta r /v$ where $\theta$ is the angle between the origin, the centre and the point $x=2,z=1$. The Electric field must accelerate the proton in the y-direction in this time $t$. Computing $\theta$ gives $\theta = 0.927$. and since $d = at^2, a = qE/m $ I get $3 \times 10^{-3} \times m / q\theta^2r^2v^2 = E$. Now computing E gives me the strength of the electric field to be $5.8 \times 10^{-16}$.
I am concerned about the final values for $B$ and $E$, they seem really small. Where is my mistake or is the value really just this small? Thanks.
It's a simple mistake $$3\times10^{-3}=at^2=\frac{qE}{m}\frac{\theta^2r^2}{v^2}$$ This implies $$E=3\times10^{-3}\frac{mv^2}{q\theta^2r^2}$$You need to multiply with $v^2$ not divide