The force on the electron is given by the Lorentz equation
$\begin{align}\mathbf{F} = q \left(\mathbf{E} + \mathbf{v} \times \mathbf{B} \right) \label{eqn58:1}\end{align}$
The proton's velocity after being accelerated through a potential difference $V$ in the $z$ direction is
$\setcounter{equation}{1}\begin{align}v = \sqrt{\frac{2 q V}{m}} \hat{\mathbf{z}} \label{eqn58:2}\end{align}$
We are told that the electric field is in the $+x$ direction and that the magnetic field is in the $+y$ direction. Therefore,
$\begin{align*}\mathbf{E} &= E \hat{\mathbf{x}} \\\mathbf{B} &= B \hat{\mathbf{y}}\end{align*}$
Because the proton's trajectory is unaffected, we set $\mathbf{F} = 0$ in equation (1).
$\begin{align*}0 = q \left(E \hat{\mathbf{x}} + v \hat{\mathbf{z}} \times B \hat{\mathbf{y}} \right) = q \left(E \hat{\mathbf{...$
So, initially $E = vB$ . But when the potential difference is increased by a factor of $2$ , $v$ is increased by a factor of two (see equation (2)). Therefore, $vB > E$ , so the particle is deflected in the $-\hat{\mathbf{x}}$ direction. Therefore, answer (B) is correct.

Solution to 2001 Problem 58