In a vacuum chamber there is a uniform electric field of 2.0 × 10⁴ V/m along +y and a uniform…

Step 1. The drift (a). In a frame moving at $u\hat x$, the electric field becomes $\vec E + \vec u \times \vec B = (E - uB)\hat y$. It vanishes for $u = E/B = 1.00 \times 10^{5}\ \text{m/s}$.

Step 2. The circle (b). In that frame the proton starts with velocity $-u\hat x$ and only $B$ acts, so it moves in a circle of radius $r = \dfrac{m_pu}{eB}$ at the cyclotron rate $\omega = \dfrac{eB}{m_p} = 1.916 \times 10^{7}\ \text{rad/s}$. One arch takes one turn: $T = \dfrac{2\pi}{\omega} = 0.328\ \mu\text{s}$.

Step 3. Height (c). The greatest height is the diameter: $y_{\max} = 2r = \dfrac{2m_pE}{eB^2} = 1.04\ \text{cm}$. Each arch is $2\pi r = 3.28\ \text{cm}$ long along $x$.

Step 4. Speed and energy (d, e). At the top the circular velocity, $u$ along $+x$, adds to the frame's $u$: $v_{\max} = 2E/B = 2.00 \times 10^{5}\ \text{m/s}$. Then $KE = \tfrac12m_pv_{\max}^2 = 209\ \text{eV}$. As a check, the electric field has done work $eEy_{\max}$, the same amount; the magnetic force does no work.

Step 5. The plate (f). The proton misses the plate if $\dfrac{2m_pE}{eB^2} < h$, so $B > \sqrt{\dfrac{2m_pE}{eh}} = 0.228\ \text{T}$.