Solution to 2001 Problem 58

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.

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