A vertical wire carries an upward current. An electron beam...

Let the vertical wire be placed along the y-axis with the current $I$ flowing in the $+\hat{j}$ direction (upwards).

Let the electron beam travel horizontally along the x-axis towards the wire. Its velocity vector is $\vec{v} = -v \hat{i}$.

According to the right-hand thumb rule, the magnetic field $\vec{B}$ produced by the current-carrying wire at a point on the positive x-axis is directed outwards, i.e., in the $+\hat{k}$ direction.

The magnetic force acting on the moving electron is given by the Lorentz force formula:

$\vec{F} = q(\vec{v} \times \vec{B})$

Substituting the charge of an electron $q = -e$ and the respective vectors:

$\vec{F} = -e(-v \hat{i} \times B \hat{k})$

$\vec{F} = e v B (\hat{i} \times \hat{k})$

Since $\hat{i} \times \hat{k} = -\hat{j}$, we get:

$\vec{F} = -e v B \hat{j}$

The force is in the $-\hat{j}$ direction, which corresponds to the downward direction. Therefore, the electron beam will be deflected downwards.